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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]  %Headings in Russian
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
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\title[On function spaces. II]{On function spaces. II}
\author{Yu.~L. Ershov}
\address{Yuri Leonidovich Ershov
\newline\hphantom{iii} %${}^{\rm (a)}$
Sobolev Institute of Mathematics,
\newline\hphantom{iii} Acad. Koptyug ave., 4,
\newline\hphantom{iii}
630090, Novosibirsk, Russia}%
\email{ershov@math.nsc.ru}%
\author{M.~V. Schwidefsky}
\address{Marina Vladimirovna Schwidefsky
\newline\hphantom{iii} %${}^{\rm (a)}$
Sobolev Institute of Mathematics,
\newline\hphantom{iii} Acad. Koptyug ave., 4,
\newline\hphantom{iii}
630090, Novosibirsk, Russia}%
\email{semenova@math.nsc.ru}%

\thanks{\sc Ershov, Yu.~L., Schwidefsky, M.~V., On function spaces. II}
\thanks{\copyright \ 2021 Ershov, Yu.~L., Schwidefsky, M.~V.}
\thanks{The research was carried out in the framework of the state contract of Sobolev Institute of Mathematics,
project no. FWNF-2022-0012.}
\thanks{\it Received  February 23, 2022, published  ????.}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract. } For certain properties $\mathfrak{P}$ of topological $T_0$-spaces, we prove that
a $T_0$-space $\mathbb{Y}$ has property $\mathfrak{P}$ if and only if the function space
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ endowed with a particular topology $\mathcal{T}$ possesses
$\mathfrak{P}$ for some $T_0$-space $\mathbb{X}$.
\medskip

\noindent{\bf Keywords:} $A$-space, core-compact space, $d$-space, essentially complete space, function space, injective space, sober space, $T_0$-space.
 \end{quote}
}

\section{Introduction}
%%%% to write

\noindent
This paper continues \cite{CXY}.
In \cite{CXY}, the interplay of different topological properties for a $T_0$-space $\mathbb{Y}$ and its function space $\mathbb{C}(\mathbb{X},\mathbb{Y})$
(endowed with the pointwise convergence topology) was investigated.
Here, we extend some results from \cite{CXY} as well as from \cite{Top} considering different topologies
on the set $C(\mathbb{X},\mathbb{Y})$ of continuous functions from
a $T_0$-space $\mathbb{X}$ to a $T_0$-space $\mathbb{Y}$.

Our main results are Theorems \ref{T:FunD}, \ref{T:c10}, \ref{T:CXY-1}, \ref{T:FA}, \ref{T:FA0}, \ref{T:CXY-4}, \ref{T:CXY-3},
Propositions \ref{P:FunA} and \ref{P:del}, and
Corollaries \ref{C:ts-0}, \ref{C:1}, and \ref{C:2}.

For all the notions and notation which are not defined here, we refer to the monographs \cite{Top,Comp,G} as well as to \cite{CXY}.

\section{Topologies on $C(\mathbb{X},\mathbb{Y})$}

\noindent
Given topological $T_0$-spaces $\mathbb{X}$ and $\mathbb{Y}$, let $C(\mathbb{X},\mathbb{Y})$ denote the set of all continuous functions from
$\mathbb{X}$ to $\mathbb{Y}$. An arbitrary function $f\in C(\mathbb{X},\mathbb{Y})$ can be considered as an element
of the Cartesian power $Y^{X}$. \emph{The pointwise convergence topology}
on $C(\mathbb{X},\mathbb{Y})$ is induced by the Tychonoff topology on $Y^{X}$. Therefore, the collection of all sets of the form
\[
V_{x,U}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid f(x)\in U\bigr\},
\ \text{where}\ x\in X\ \text{and}\ \varnothing\ne U\in\mathcal{T}(\mathbb{Y}),
\]
forms a subbasis of the pointwise convergence topology. In what follows, we denote this topology by $\mathcal{P}$.

If $\mathcal{T}$ is a topology on $C(\mathbb{X},\mathbb{Y})$ then we denote the space $\bigl\langle C(\mathbb{X},\mathbb{Y}),\mathcal{T}\bigr\rangle$
by $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ and by $\mathbb{C}(\mathbb{X},\mathbb{Y})$ when $\mathcal{T}=\mathcal{P}$.
The specialization order on $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is denoted by $\leq_\mathcal{T}$ or just by $\leq$ in case when $\mathcal{T}=\mathcal{P}$.
For $f$, $g\in C(\mathbb{X},\mathbb{Y})$, the relation $f\leq_{\mathcal{P}}g$ holds if and only if $f(x)\leq_{\mathbb{Y}}g(x)$ for all $x\in X$.
We write $f\leq g$ instead of $f\leq_{\mathcal{P}}g$. It is clear that $f\leq_\mathcal{T}g$ implies that $f\leq g$ whenever $\mathcal{P}\subseteq\mathcal{T}$.

We also consider the mapping
\[
\xi\colon\mathbb{Y}\to C(\mathbb{X},\mathbb{Y});\quad\xi\colon y\mapsto\xi_y,\quad
\text{where}\ \xi_y(x)=y\ \text{for all}\ x\in X.
\]

\begin{lem}\label{L:00}
Let $\mathbb{X},\mathbb{Y}$ be $T_0$-spaces. If $\mathcal{T}$ is a topology on $C(\mathbb{X},\mathbb{Y})$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\leq_\mathcal{P})$ then $\leq_\mathcal{T}$
coincides with $\leq_\mathcal{P}$. In this case, the following holds.
\begin{enumerate}
\item
$\mathbb{Y}$ contains a least element with respect to the specialization order
$\leq_\mathbb{Y}$ if and only if $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ contains a least element with respect to $\leq_\mathcal{T}$ for some
$($equivalently, for each$)$ $T_0$-space $\mathbb{X}$.
\item
$\mathbb{Y}$ contains a greatest element with respect to $\leq_\mathbb{Y}$ if and only if
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ contains a greatest element with respect to $\leq_\mathcal{T}$ for some
$($equivalently, for each$)$ $T_0$-space $\mathbb{X}$.
\end{enumerate}
\end{lem}

\begin{proof}
The pointwise convergence topology $\mathcal{P}$ is finer than
the Alexandroff topology $\mathcal{T}_A(\leq_\mathcal{P})$, see for example Lemma 1.2.8 in \cite{Top}.
As the specialization order of $\mathcal{T}_A(\leq_\mathcal{P})$ coincides with $\leq_\mathcal{P}$, we obtain that
$\leq_\mathcal{T}$ coincides with $\leq_\mathcal{P}$; we denote this order simply by $\leq$.

(1) Let $\langle Y;\leq_\mathbb{Y}\rangle$ have a least element $\bot$. It is clear that $\xi_\bot\in C(\mathbb{X},\mathbb{Y})$ is a least element
in $\bigl\langle C(\mathbb{X},\mathbb{Y});\leq\bigr\rangle$.
Conversely, suppose that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\leq)$ and
$o\in C(\mathbb{X},\mathbb{Y})$ is a least element in $\bigl\langle C(\mathbb{X},\mathbb{Y});\leq\bigr\rangle$.
Choose arbitrary elements $x,x^\prime\in X$ and suppose that $o(x)\in U\in\mathcal{T}(\mathbb{Y})$.
Then $o\in V_{x,U}\in\mathcal{P}\subseteq\mathcal{T}$.
If $y\in Y$ then $o\leq\xi_y$, whence $\xi_y\in V_{x,U}$ and $y\in U$.
As we can choose $y$ arbitrary, we conclude that $U=Y$ and $o(x^\prime)\in U$.
Thus, we proved that $o(x)\leq_\mathbb{Y} o(x^\prime)$.
Symmetrically, $o(x^\prime)\leq_\mathbb{Y} o(x)$, whence $o=\xi_\bot$ for some $\bot\in Y$.
Since $\xi_\bot=o\leq\xi_y$, we conclude that $\bot\leq y$ for all $y\in Y$.
Therefore, $\langle Y;\leq_\mathbb{Y}\rangle$ has a least element.

The proof of (2) is similar but simpler.
\end{proof}

\noindent
Given topological spaces $\mathbb{X}$, $\mathbb{Y}$, $\mathbb{Z}$, let
$M(\mathbb{X},\mathbb{Y})$ denote the set of all functions from $X$ to $Y$.
We put
\begin{align*}
\lambda\colon &M(\mathbb{Z}\times\mathbb{X},\mathbb{Y})\to M\bigl(\mathbb{Z},M(\mathbb{X},\mathbb{Y})\bigr),\quad
\lambda(f)\colon z\mapsto f(z,x);\\
\lambda^\ast\colon &M\bigl(\mathbb{Z},M(\mathbb{X},\mathbb{Y})\bigr)\to M(\mathbb{Z}\times\mathbb{X},\mathbb{Y}),\quad
\lambda^\ast(g)\colon(z,x)\mapsto[g(z)](x).
\end{align*}
It is not hard to verify that $\lambda$ and $\lambda^\ast$ are mutually inverse mappings.
Hence they establish a one-to-one correspondence between sets
$M(\mathbb{Z}\times\mathbb{X},\mathbb{Y})$ and $M\bigl(\mathbb{Z},M(\mathbb{X},\mathbb{Y})\bigr)$.

For the following definition, we refer to \cite{AD51}. A topology $\mathcal{T}$ on $C(\mathbb{X},\mathbb{Y})$ is \emph{proper}, if
$\lambda\bigl(C(\mathbb{Z}\times\mathbb{X},\mathbb{Y})\bigr)\subseteq C\bigl(\mathbb{Z},\mathbb{C}_{\mathcal{T}}(\mathbb{X},\mathbb{Y})\bigr)$
for an arbitrary space $\mathbb{Z}$.
A topology $\mathcal{T}$ on $C(\mathbb{X},\mathbb{Y})$ is \emph{admissible}, if
$\lambda^\ast\bigl(C\bigl(\mathbb{Z},\mathbb{C}_{\mathcal{T}}(\mathbb{X},\mathbb{Y})\bigr)\bigr)\subseteq C(\mathbb{Z}\times\mathbb{X},\mathbb{Y})$
for an arbitrary space $\mathbb{Z}$. A topology $\mathcal{T}$  on $C(\mathbb{X},\mathbb{Y})$ is \emph{exponential},
if $\mathcal{T}$ is both proper and admissible.

The \emph{compact-open topology} $\mathcal{K}$ on $C(\mathbb{X},\mathbb{Y})$ is defined by the subbasis of open sets
\[
V_{K,U}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid f(K)\subseteq U\bigr\},
\ \text{where}\ K\subseteq X\ \text{is compact and}\ \varnothing\ne U\in\mathcal{T}(\mathbb{Y}),
\]
The \emph{Isbell topology} $\mathcal{I}$ on $C(\mathbb{X},\mathbb{Y})$ is defined by the subbasis of open sets
\[
V_{\mathcal{H},U}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid f^{-1}(U)\in\mathcal{H}\bigr\},
\]
where $\mathcal{H}$ is a Scott-open set in $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$ and $\varnothing\ne U\in\mathcal{T}(\mathbb{Y})$.
The \emph{core-open topology} $\mathcal{C}$ on $C(\mathbb{X},\mathbb{Y})$ is defined by the subbasis of open sets
\[
V_{U,W}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid U\ll f^{-1}(W)\bigr\},
\ \text{where}\ \varnothing\ne U\in\mathcal{T}(\mathbb{X})\ \text{and}\ \varnothing\ne W\in\mathcal{T}(\mathbb{Y}).
\]
Let $\mathcal{T}$ be a $T_0$-topology on $\mathcal{T}(\mathbb{X})$ such that the specialization order $\leq_\mathcal{T}$ coincides with $\subseteq$.
Consider the topology on $C(\mathbb{X},\mathbb{Y})$ defined by the subbasis of open sets
\[
V_{\mathcal{O},U}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid f^{-1}(U)\in\mathcal{O}\bigr\},
\]
where $\mathcal{O}\in\mathcal{T}$ and $U\in\mathcal{T}(\mathbb{Y})$.
We say that this topology is \emph{induced by} $\mathcal{T}$ and denote it by $\mathcal{T}^\sharp$.
It is clear that $\mathcal{T}_0\subseteq\mathcal{T}_1$ implies that $\mathcal{T}_0^\sharp\subseteq\mathcal{T}_1^\sharp$.
Moreover, it follows from the definition above that $\mathcal{I}$ is induced by the Scott topology on $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$.

\begin{lem}\label{L:A}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces. Then $\mathcal{T}_A(\subseteq)^\sharp\subseteq\mathcal{T}_A(\leq)$.
\end{lem}

\begin{proof}
It suffices to show that $V_{\mathcal{O},W}\in\mathcal{T}_A(\leq)$ for all upper cones $\mathcal{O}$ in $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$
and all open sets $W\in\mathcal{T}(\mathbb{Y})$.
Indeed, let $f\in V_{\mathcal{O},W}$ and let $f\leq g$ in $C(\mathbb{X},\mathbb{Y})$.
We prove first that $f^{-1}(W)\subseteq g^{-1}(W)$. Indeed, let $x\in f^{-1}(W)$; then $f(x)\in W$.
Since $W$ is an upper cone with respect to $\leq_\mathbb{Y}$,
we obtain that $g(x)\in W$ and thus, $x\in g^{-1}(W)$ which is our desired conclusion.

Since $f^{-1}(W)\in\mathcal{O}$ and $f^{-1}(W)\subseteq g^{-1}(W)$, we conclude that $g^{-1}(W)\in\mathcal{O}$ as $\mathcal{O}$ is an upper cone with respect to $\subseteq$.
This yields that $g\in V_{\mathcal{O},W}$ and $V_{\mathcal{O},W}\in\mathcal{T}_A(\leq)$.
\end{proof}

\noindent
In what follows, we consider also the \emph{Scott topology} $\mathcal{S}=\mathcal{T}_S(\leq_\mathcal{P})$ on $C(\mathbb{X},\mathbb{Y})$.

\begin{lem}\label{L:xi}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces.
%and let $\mathcal{U}$ be a topology on $\mathcal{T}(\mathbb{X})$ such that $\mathcal{U}\subseteq\mathcal{T}_A(\subseteq)$.
If $\mathcal{T}$ is a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$ then
$\xi$ is a homeomorphic embedding of $\mathbb{Y}$ into $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$.
\end{lem}

\begin{proof}
It is straightforward to verify that $\xi(Y)\cap V_{x,U}=\xi(U)$ for all $x\in X$ and $U\in\mathcal{T}(\mathbb{Y})$.
Therefore, $\xi$ is open.
Moreover, for all $W\in\mathcal{T}(\mathbb{Y})$ and all $y\in Y$, we have
\[
\xi_y^{-1}(W)=
\begin{cases}
X,\quad&\text{if}\ y\in W;\\
\varnothing,\quad&\text{if}\ y\notin W.
\end{cases}
\]
Furthermore, let $\mathcal{O}\in\mathcal{T}_A(\subseteq)$ and let $W\in\mathcal{T}(\mathbb{Y})$.
Then we have
\begin{align*}
\xi^{-1}(V_{\mathcal{O},W})&=\bigl\{y\in Y\mid\xi_y\in V_{\mathcal{O},W}\bigr\}=
\bigl\{y\in Y\mid\xi_y^{-1}(W)\in\mathcal{O}\bigr\}=\\
&=
\begin{cases}
Y,\quad&\text{if}\ \mathcal{O}=\mathcal{T}(\mathbb{X});\\
W,\quad&\text{if}\ \mathcal{O}\notin\bigl\{\mathcal{T}(\mathbb{X}),\varnothing\bigr\};\\
\varnothing,\quad&\text{if}\ \mathcal{O}=\varnothing.
\end{cases}
\end{align*}
In any case, $\xi^{-1}(V_{\mathcal{O},W})\in\mathcal{T}(\mathbb{Y})$.
Therefore, $\xi^{-1}(V)\in\mathcal{T}(\mathbb{Y})$ for all $V\in\mathcal{T}_A(\subseteq)^\sharp$,
whence the desired statement follows.
\end{proof}

\noindent
Let $\mathbb{X}$ be a $T_0$-space.
If $\langle I;\leqslant\rangle$ is an up-directed poset and $\mathfrak{x}=\{x_i\mid i\in I\}\subseteq X$ then we say that $\mathfrak{x}$ is a \emph{net} in $\mathbb{X}$.
If $A\subseteq X$, then $\mathfrak{x}$ is \emph{eventually in} $A$ if there is $i\in I$ such that $x_j\in A$ for all $j\geqslant i$.
Then $\mathfrak{x}$ converges to some $x\in X$ if $x\in U\in\mathcal{T}(\mathbb{X})$ implies that $\mathfrak{x}$ is eventually in $U$ for all $U\in\mathcal{T}(\mathbb{X})$.
Further, if $\mathbb{Y}$ is also a $T_0$-space and $\mathfrak{f}=\{f_j\mid j\in J\}$ is a net in $C(\mathbb{X},\mathbb{Y})$
for some up-directed poset $\langle J;\sqsubseteq\rangle$, then $\mathfrak{f}$ \emph{converges continuously to} $f\in C(\mathbb{X},\mathbb{Y})$ if
for arbitrary $x\in X$, the net $\bigl\{f_j(x_i)\mid(i,j)\in I\times J\bigr\}$ converges to $f(x)$ whenever $\mathfrak{x}=\{x_i\mid i\in I\}$ converges to $x$.

The \emph{natural topology} $\mathcal{T}_\ast$ on $C(\mathbb{X},\mathbb{Y})$ consists of those sets
$U\subseteq C(\mathbb{X},\mathbb{Y})$ which satisfy the following condition for all nets $\mathfrak{f}$ in $C(\mathbb{X},\mathbb{Y})$ and all
$f\in C(\mathbb{X},\mathbb{Y})$:
\[
\text{if}\ \mathfrak{f}\ \text{converges continuously to}\ f\in U\ \text{then}\ \mathfrak{f}\ \text{is eventually in}\ U.
\]
For the following facts, we refer to Chapters 6 and 15 in \cite{Top} as well as to Sections 5.3--5.4 in \cite{G} and to Section II-4 in \cite{Comp}.

\begin{prop}\label{P:0}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces.
\begin{enumerate}
\item
If $\mathcal{T}_0$ is a proper and $\mathcal{T}_1$ is an admissible topology on $C(\mathbb{X},\mathbb{Y})$
then $\mathcal{T}_0\subseteq\mathcal{T}_1$.
\item
If $\mathcal{T}_0\subseteq\mathcal{T}_1$ and $\mathcal{T}_1$ is a proper topology on $C(\mathbb{X},\mathbb{Y})$
then $\mathcal{T}_0$ is also a proper topology on $C(\mathbb{X},\mathbb{Y})$.
\item
If $\mathcal{T}_0\subseteq\mathcal{T}_1$ and $\mathcal{T}_0$ is an admissible topology on $C(\mathbb{X},\mathbb{Y})$
then $\mathcal{T}_1$ is also an admissible topology on $C(\mathbb{X},\mathbb{Y})$.
\item
$\mathcal{P}\subseteq\mathcal{K}\subseteq\mathcal{I}\subseteq\mathcal{T}_\ast$ and $\mathcal{K}\subseteq\mathcal{C}$.
\item
The natural topology $\mathcal{T}_\ast$ is the finest proper topology on $C(\mathbb{X},\mathbb{Y})$.
\end{enumerate}
\end{prop}

\noindent
For the next statement, we refer to \cite{DK} and to \cite[Theorem 6.2.1]{Top}, see also \cite[Theorem 5.4.4]{G}.

\begin{thm}\label{T:cc}
The following conditions are equivalent for a $T_0$-space $\mathbb{X}$.
\begin{enumerate}
\item
$\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$ is a continuous poset.
\item
$\langle\mathcal{T}(\mathbb{X}),\mathcal{T}_S(\subseteq)\rangle$ is an $\alpha$-space.
\item
For an arbitrary $T_0$-space $\mathbb{Y}$, there is an exponential topology on $C(\mathbb{X},\mathbb{Y})$.
\item
There is an exponential topology on $C(\mathbb{X},\mathbb{S})$.
\end{enumerate}
\end{thm}

\noindent
A topological space $\mathbb{X}$ which satisfies the equivalent conditions of Theorem \ref{T:cc} is called a \emph{core-compact space}.
A topological space $\mathbb{X}$ is \emph{locally compact}, if for each $x\in X$ and for each $U\in\mathcal{T}(\mathbb{X})$ with $x\in U$, there is a compact set $K\subseteq X$ and an open set $V\in\mathcal{T}(\mathbb{X})$ such that $x\in V\subseteq K\subseteq U$.

For the next statement, we refer to \cite{Ba77}, \cite[Lemma II-4.2]{Comp}, to \cite[Theorem 6.3.3]{Top}, and to \cite[Section 5.4]{G}.

\begin{prop}\label{P:1}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces.
\begin{enumerate}
\item
If $\mathbb{X}$ is core-compact then $\mathcal{C}=\mathcal{I}=\mathcal{T}_\ast$ and this topology is exponential on $C(\mathbb{X},\mathbb{Y})$.
\item
If $\mathbb{X}$ is locally compact then $\mathcal{K}=\mathcal{C}=\mathcal{I}=\mathcal{T}_\ast$ and this topology is exponential on $C(\mathbb{X},\mathbb{Y})$.
\item
If $\mathbb{X}$ is an $\alpha^\ast$-space then $\mathcal{P}=\mathcal{K}=\mathcal{C}=\mathcal{I}=\mathcal{T}_\ast$ and this topology is exponential on $C(\mathbb{X},\mathbb{Y})$.
\end{enumerate}
\end{prop}

\begin{cor}
For a core-compact space $\mathbb{X}$ and a $T_0$-space $\mathbb{Y}$ the exponential topology on $C(\mathbb{X},\mathbb{Y})$ is defined by the subbasis
of open sets
\[
V_{U,W}=\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid U\prec f^{-1}(W)\bigr\},
\]
where $\varnothing\ne U\in\mathcal{T}(\mathbb{X})$ and $\varnothing\ne W\in\mathcal{T}(\mathbb{Y})$.
\end{cor}



\section{$d$-spaces $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$}

\noindent
The proof of the following lemma repeats the proof of \cite[Lemma 2]{CXY}.

\begin{lem}\label{L:FunD1}
Let $\mathbb{X}$ and $\mathbb{Y}$ be $T_0$-spaces, let
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\leq_\mathcal{P})$, and let
\[
F=\{f_i\in C(\mathbb{X},\mathbb{Y})\mid i\in I\}
\]
be an up-directed set with respect to $\leq_\mathcal{P}$.
If a function $f\colon X\to Y$ is such that $f(x)=\sup_\mathbb{Y}\{f_i(x)\mid i\in I\}$ and $f(x)\in\cl_\mathbb{Y}\{f_i(x)\mid i\in I\}$
for all $x\in X$, then $f=\sup F$ in $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$.
\end{lem}

\begin{thm}\label{T:FunD}
For $T_0$-spaces $\mathbb{X}$ and $\mathbb{Y}$,
let $\mathcal{T}$ satisfy one of the following conditions:
\begin{itemize}
\item[(a)]
$\mathcal{T}$ is a proper topology on $C(\mathbb{X},\mathbb{Y})$ such that $\mathcal{P}\subseteq\mathcal{T}$;
\item[(b)]
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{S}$.
\end{itemize}
Then $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $d$-space if and only if
$\mathbb{Y}$ is a $d$-space.
\end{thm}

\begin{proof}
According to \cite[Lemma 6.1.9]{Top}, $\mathcal{T}_A(\leq_\mathcal{P})$ is an admissible topology.
By Proposition \ref{P:0}(1), we obtain that $\mathcal{T}\subseteq\mathcal{T}_A(\leq_\mathcal{P})$.
Therefore, $\leq_\mathcal{T}$ and $\leq$ agree by Lemma \ref{L:00}.

Suppose first that $\mathbb{Y}$ is a $d$-space and consider the case when $\mathcal{T}$ is a proper topology finer than $\mathcal{P}$ (condition (a)).
According to Lemma \ref{L:FunD1}, each nonempty set
$F=\{f_i\in C(\mathbb{X},\mathbb{Y})\mid i\in I\}$ which is up-directed with respect to $\leq$ has a least upper bound $f$.
It remains to prove that $f\in\cl_\mathcal{T}F$.

We establish that $F$ converges continuously to $f$.
Let a net $\mathfrak{x}=\{x_j\mid j\in J\}\subseteq X$ with $J$ being up-directed converge to some $x\in X$ and let $f(x)\in U\in\mathcal{T}(\mathbb{Y})$.
Then $x\in f^{-1}(U)\in\mathcal{T}(\mathbb{X})$ and thus $\mathfrak{x}$ is eventually in $f^{-1}(U)$.
This means that $\{f(x_j)\mid j\in J\}$ is eventually in $U$.
Since $f(x_j)\in\cl_\mathbb{Y}\{f_i(x_j)\mid i\in I\}$ for all $j\in J$ by Lemma \ref{L:FunD1} and $F$ is up-directed with respect to $\leq$,
we conclude that $\{f_i(x_j)\mid(i,j)\in I\times J\}$ is eventually in $U$ which proves our desired statement.

Let $f\in W\in\mathcal{T}$. By Proposition \ref{P:0}(5), we conclude that $W\in\mathcal{T}_\ast$. Since
$F$ converges continuously to $f$, we obtain that $F$ is eventually in $W$.
This yields that $f$ is a limit point for $F$ and
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $d$-space.

Suppose next that $\mathbb{Y}$ is a $d$-space and consider the case when $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{S}$ (condition (b)).
In the notation above, we have to prove again that $f\in\cl_\mathcal{T}F$.
Indeed, if $f\in W\in\mathcal{T}\subseteq\mathcal{S}$ then by the definition of the Scott topology, we obtain that
$f_i\in W$ for some $i\in I$ which is our desired conclusion.

Suppose now that $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $d$-space.
In what follows, we use only the assumption that $\mathcal{P}\subseteq\mathcal{T}$.
Let a set $D\subseteq Y$ be up-directed with respect to the specialization order.
Then the set $\xi(D)=\{\xi_d\mid d\in D\}$ is also up-directed with respect to $\leq$.
According to our assumption, there exists $f=\sup\xi(D)\in C(\mathbb{X},\mathbb{Y})$ and $f\in\cl_\mathcal{T}\xi(D)$. We show that
$f\in\xi(Y)$.  Indeed, let $f(x_0)\in U\in\mathcal{T}(\mathbb{Y})$. We also choose an arbitrary element $x_1\in X$.
Then $f\in V_{x_0,U}\in\mathcal{P}\subseteq\mathcal{T}$.
Thus, there exists $d\in D$ such that $\xi_d\in V_{x_0,U}$, whence $f(x_1)\geq\xi_d(x_1)=d=\xi_d(x_0)\in U$.
This yields that $f(x_1)\in U$ whence $f(x_0)\leq f(x_1)$.
A similar argument shows that $f(x_1)\leq f(x_0)$.
Therefore $f=\xi_y$ for some $y\in Y$.
It is clear that $y=\sup_\mathbb{Y}D$. If $y\in U\in\mathcal{T}(\mathbb{Y})$, then
$\xi_y\in V_{x,U}\in\mathcal{P}\subseteq\mathcal{T}$, where $x\in X$. As $\xi_y\in\cl_\mathcal{T}\xi(D)$,
we conclude that $\xi_d\in V_{x,U}$ for some $d\in D$. Thus, $d=\xi_d(x)\in U$ and $y\in\cl_\mathbb{Y}D$.
\end{proof}

\noindent
The fact that $\mathbb{C}(\mathbb{X},\mathbb{Y})$ is a $d$-space whenever $\mathbb{Y}$ is a $d$-space
was established in \cite{E99}.
The fact that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is a $d$-space whenever $\mathbb{Y}$ is a $d$-space
is established in \cite[Lemma II-4-3(i)]{Comp}.
Theorem 3.3 in \cite{BQW} shows that $\mathbb{Y}$ is a $d$-space provided that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is a $d$-space
for some $T_0$-space $\mathbb{X}$.
For the case $\mathcal{T}=\mathcal{P}$, Theorem \ref{T:FunD} was proved in \cite{CXY}, see also \cite[Corollary 8.4.2]{Top}.


\section{Sobriety in function spaces}

\noindent
The proof of the following statement almost repeats the proof of \cite[Theorem 6]{CXY}. Nonetheless, we give it here for the sake of completeness.

\begin{prop}\label{P:sob}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a sober space then $\mathbb{Y}$ is also sober.
\end{prop}

\begin{proof}
Let $\mathbb{Z}$ be a $T_0$-space and let a subspace $\mathbb{Z}_0\leq\mathbb{Z}$ be such that $Z=\sob_\mathbb{Z}Z_0$.
Consider an arbitrary continuous function $f_0\colon\mathbb{Z}_0\to\mathbb{Y}$.
By Lemma \ref{L:xi}, $\xi\colon\mathbb{Y}\to\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is continuous.
Thus, the function $\xi f_0\colon\mathbb{Z}_0\to\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is continuous.
As $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a sober space, there is a continuous function
$f\colon\mathbb{Z}\to\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ such that $f|_{\mathbb{Z}_0}=\xi f_0$.
We prove that $f(Z)\subseteq\xi(Y)$.
Indeed, let $z\in Z$; then $z=\sup_\mathbb{Z}(\dnw z\cap Z_0)$ and $z\in\cl_\mathbb{Z}(\dnw z\cap Z_0)$.
It is straightforward to verify that $f(z)\in\cl_\mathbb{Y}f(\dnw z\cap Z_0)$ and $f(z)=\sup f(\dnw z\cap Z_0)$,
see Lemma 1.5.3 in \cite{Top}.
Let $x_0$, $x_1\in X$ and let $[f(z)](x_0)\in U\in\mathcal{T}(\mathbb{Y})$.
Then $f(z)\in V_{x_0,U}\in\mathcal{P}\subseteq\mathcal{T}$, whence $z\in f^{-1}(V_{x_0,U})\in\mathcal{T}(\mathbb{Z})$.
Therefore there is $z_0\in\dnw z\cap Z_0$ such that $f(z_0)\in V_{x_0,U}$.
Since the function $f(z_0)$ is constant, $f(z_0)\in V_{x_1,U}\in\mathcal{P}\subseteq\mathcal{T}$. But then $f(z)\in V_{x_1,U}$,
as $f(z_0)\leq f(z)$. Therefore $[f(z)](x_1)\in U$.
Similarly, $[f(z)](x_1)\in U$ implies that $[f(z)](x_0)\in U$; that is,
$[f(z)](x_0)=[f(z)](x_1)$, which is our desired conclusion.
Inclusion $f(Z)\subseteq\xi(Y)$ implies that $(\xi^{-1}f)|_{\mathbb{Z}_0}=f_0$.
In view of \cite[Theorem 5.3.2]{Top}, $\mathbb{Y}$ is a sober space.
\end{proof}

\noindent
For the following statement, we refer to \cite[Proposition 16]{CXY} or to \cite[Proposition 6.1.15]{Top}.

\begin{prop}\label{P:c9}
Let $\mathcal{T}$ be an exponential topology on $C(\mathbb{X},\mathbb{Y})$.
\begin{enumerate}
\item
If $\mathbb{Y}$ is $[$densely$]$ injective, then
$\mathbb{C}_{\mathcal{T}}(\mathbb{X},\mathbb{Y})$ is $[$densely$]$ injective.
\item
If $\mathbb{Y}$ is sober, then
$\mathbb{C}_{\mathcal{T}}(\mathbb{X},\mathbb{Y})$ is sober.
\end{enumerate}
\end{prop}

\noindent
The following statement generalizes Theorem 6 in \cite{CXY}.

\begin{thm}\label{T:c10}
For a $T_0$-space $\mathbb{Y}$, the following conditions are equivalent.
\begin{enumerate}
\item
$\mathbb{Y}$ is sober.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is sober for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is sober for some core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is sober for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is sober for each $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is sober for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is sober for some $[$core-compact$]$ $T_0$-space $\mathbb{X}$ and some topology $\mathcal{T}$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
\end{enumerate}
\end{thm}

\begin{proof}
(1) implies (2) by Proposition \ref{P:c9}(2).
(2) implies (3) and (3) implies (4) in a trivial way. (4) implies (1) by Proposition \ref{P:sob}.
Statements (1), (5), and (6) are equivalent by \cite[Theorem 6]{CXY}.
Furthermore, (3) obviously implies (7) and (7) implies (1) by Proposition \ref{P:sob}.
\end{proof}


\section{Essential completeness in function spaces}

\noindent
As in \cite{E99-1}, we consider the following properties of a $T_0$-space $\mathbb{X}$.
\begin{enumerate}
\item[$(\mathrm{H}_0)$]
$\mathbb{X}$ has a least element $0$ with respect to the specialization order $\leq$.
\item[$(\mathrm{H}_1)$]
$\mathbb{X}$ is a join semilattice with respect to $\leq$; $\vee$ denotes the join operation in $X$.
\item[$(\mathrm{H}_2)$]
The join operation $\vee\colon\mathbb{X}^2\to\mathbb{X}$ is continuous; that is, $\bigl\langle X,\vee,\mathcal{T}(\mathbb{X})\bigr\rangle$
is a topological join-semilattice.
\end{enumerate}

\noindent
In \cite{E99-1}, the following statement was proved, see also \cite[Corollary 10.4.2]{Top}.

\begin{thm}\label{T:EssCom}\cite[Corollary 10.4.2]{Top}
A $T_0$-space $\mathbb{X}$ is essentially complete if and only if $\mathbb{X}$ is a $d$-space with properties $(\mathrm{H}_0)$--$(\mathrm{H}_2)$.
\end{thm}

\begin{prop}\label{P:H}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$. If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
has the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$ then $\mathbb{Y}$ also possesses
the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$.
\end{prop}

\begin{proof}
According to our assumption, for arbitrary $y_0$, $y_1\in Y$, there is $f=\xi_{y_0}\vee\xi_{y_1}\in C(\mathbb{X},\mathbb{Y})$.
We prove that $f$ is a constant function.
Indeed, let $f(x_0)\in U\in\mathcal{T}(\mathbb{Y})$;
then $f\in V_{x_0,U}\in\mathcal{P}\subseteq\mathcal{T}$.
We may assume without loss of generality that $U\ne Y$.
We choose an arbitrary element $x_1\in X$.
Since $\vee$ is a continuous function, there are $V_0,V_1\in\mathcal{T}$ such that
$\xi_{y_0}\in V_0$, $\xi_{y_1}\in V_1$, and $V_0\cap V_1\subseteq V_{x_0,U}$.
As $U\ne Y$, we conclude that $V_{x_0,U}\ne C(\mathbb{X},\mathbb{Y})$.
There are three possible cases.

\Case{1}: $V_{1-i}=C(\mathbb{X},\mathbb{Y})$ for some $i<2$.
In this case, $\xi_{y_i}\in V_i=V_0\cap V_1\subseteq V_{x_0,U}$ and thus $y_i\in U$.
Therefore, $\xi_{y_i}\in V_{x_1,U}$. As $\xi_{y_i}\leq f$, we obtain that $f\in V_{x_1,U}$ and $f(x_1)\in U$.

\Case{2}: $V_0,V_1\ne C(\mathbb{X},\mathbb{Y})$.
Since $\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$ and $V_0,V_1\in\mathcal{T}$, we conclude that there are $k_0>0$ and $k_1>0$, there are
upper cones $\mathcal{H}_{i0}$, \ldots, $\mathcal{H}_{ik_i}$ in $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$,
and there are open sets $W_{i0},\ldots,W_{ik_i}\in\mathcal{T}(\mathbb{Y})$, $i<2$, such that
\[
\xi_{y_i}\in V_{\mathcal{H}_{i0},W_{i0}}\cap\ldots\cap V_{\mathcal{H}_{ik_i},W_{ik_i}}\subseteq V_i\quad\text{for all}\ i<2.
\]
Therefore, $\xi^{-1}_{y_i}(W_{ij})\in\mathcal{H}_{ij}$ for all $i<2$ and all $j\leqslant k_i$.
We put $U_0=W_{00}\cap\ldots\cap W_{0k_0}$ and $U_1=W_{10}\cap\ldots\cap W_{1k_1}$; then $U_0,U_1\in\mathcal{T}(\mathbb{Y})$.
As $\varnothing\notin\{V_0,V_1\}$, we conclude that $\mathcal{H}_{ij}\ne\varnothing$ for all $i<2$ and all $j\leqslant k_i$.
As $V_0,V_1\ne C(\mathbb{X},\mathbb{Y})$, we may assume that $\mathcal{H}_{ij}\ne\mathcal{T}(\mathbb{X})$ for all $i<2$ and all $j\leqslant k_i$.
This yields that $\xi^{-1}_{y_i}(W_{ij})\ne\varnothing$ whence $\xi^{-1}_{y_i}(W_{ij})=X$ for all $i<2$ and all $j\leqslant k_i$.
Hence, $y_i\in W_{ij}$ for all $i<2$ and all $j\leqslant k_i$ and therefore, $y_0\in U_0$ and $y_1\in U_1$.

Since $\xi_{y_0}$, $\xi_{y_1}$ are constant functions, we have
$\xi_{y_0}\in V_{x_1,U_0}$ and $\xi_{y_1}\in V_{x_1,U_1}$.
Therefore $f\in V_{x_1,U_0}\cap V_{x_1,U_1}=V_{x_1,U_0\cap U_1}$, whence $f(x_1)\in U_0\cap U_1\in\mathcal{T}(\mathbb{Y})$.
Moreover, if $y^\prime\in U_0\cap U_1$ then
\[
\xi_{y^\prime}\in V_{\mathcal{H}_{00},W_{00}}\cap\ldots\cap V_{\mathcal{H}_{0k_0},W_{0k_0}}\cap V_{\mathcal{H}_{10},W_{10}}
\cap\ldots\cap V_{\mathcal{H}_{1k_1},W_{1k_1}}\subseteq V_0\cap V_1\subseteq V_{x_0,U},
\]
whence $y^\prime\in U$.
We have therefore proved that $f(x_1)\in U_0\cap U_1\subseteq U$.
This means that $f(x_0)\leq f(x_1)$.
A symmetric argument shows that $f(x_1)\leq f(x_0)$.
We obtain that $f=\xi_y$ for some $y\in Y$, whence $y=y_0\vee y_1$ in $\mathbb{Y}$.
Thus, $\mathbb{Y}$ has the property $(\mathrm{H}_1)$.

Finally, we prove that $\mathbb{Y}$ has the property $(\mathrm{H}_2)$.
Suppose that $y=y_0\vee y_1\in U\in\mathcal{T}(\mathbb{Y})$.
Then $\xi_{y_0}\vee\xi_{y_1}=\xi_y\in V_{x,U}$ for each $x\in X$.
Since $\vee$ is a continuous function on $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$, there are open sets $V_0,V_1\in\mathcal{T}$ such that
$\xi_{y_0}\in V_0$, $\xi_{y_1}\in V_1$, and $V_0\cap V_1\subseteq V_{x_0,U}$.
As $\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$, we conclude that there are $k_0>0$ and $k_1>0$, there are
upper cones $\mathcal{H}_{i0}$, \ldots, $\mathcal{H}_{ik_i}$ in $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$,
and there are open sets $W_{i0},\ldots,W_{ik_i}\in\mathcal{T}(\mathbb{Y})$, $i<2$, such that
\[
\xi_{y_i}\in V_{\mathcal{H}_{i0},W_{i0}}\cap\ldots\cap V_{\mathcal{H}_{ik_i},W_{ik_i}}\subseteq V_i\quad\text{for all}\ i<2.
\]
Therefore, $\xi^{-1}_{y_i}(W_{ij})\in\mathcal{H}_{ij}$ for all $i<2$ and all $j\leqslant k_i$.
In particular, $\mathcal{H}_{ij}\ne\varnothing$ for all $i<2$ and all $j\leqslant k_i$.
We put
\[
U_0=\bigcap\bigl\{W_{0j}\mid j\leqslant k_0,\ \mathcal{H}_{0j}\ne\mathcal{T}(\mathbb{X})\bigr\}\ \text{and}\
U_1=\bigcap\bigl\{W_{1j}\mid j\leqslant k_1,\ \mathcal{H}_{1j}\ne\mathcal{T}(\mathbb{X})\bigr\}.
\]
Then we have that $U_0,U_1\in\mathcal{T}(\mathbb{Y})$.
Moreover, $\xi^{-1}_{y_i}(W_{ij})\ne\varnothing$ whence $\xi^{-1}_{y_i}(W_{ij})=X$ for all $i<2$ and all $j\leqslant k_i$
such that $\mathcal{H}_{ij}\ne\mathcal{T}(\mathbb{X})$.
As above, we obtain that $y_0\in U_0$ and $y_1\in U_1$ and $U_0\cap U_1\subseteq U$.
Therefore, $\mathbb{Y}$ has the property $(\mathrm{H}_2)$.
\end{proof}

\noindent
The following statement has quite a straightforward proof and is to find as Proposition 3.2 in \cite{B}.

\begin{prop}\label{P:B}
Let $\mathbb{X}$ be a core-compact space and let $\mathcal{B}\subseteq\mathcal{T}(\mathbb{X})$ be an additive basis of $\mathcal{T}(\mathbb{X})$.
If open sets $U,U^\prime,V_0,\ldots,V_n\in\mathcal{B}$, $n<\omega$, are such that $U\prec U^\prime\prec V_0\cup\ldots\cup V_n$
then there are $W_0,\ldots,W_n\in\mathcal{B}$ such that
\[
W_i\prec V_i\ \text{for all}\ i\leqslant n\ \text{and}\ U\subseteq W_0\cup\ldots\cup W_n\subseteq U^\prime.
\]
\end{prop}

\begin{proof}
For each $x\in U^\prime$, there is $i\leqslant n$ such that $x\in U^\prime\cap V_i\in\mathcal{T}(\mathbb{X})$.
As $\mathbb{X}$ is core-compact and as $\mathcal{B}$ is a basis, there is $W_x\in\mathcal{B}$
such that $x\in W_x\prec U^\prime\cap V_i$.
Hence, $U\prec U^\prime\subseteq\bigcup_{x\in U^\prime}W_x$.
This yields that $U\subseteq\bigcup_{j\leqslant k}W_{x_j}\subseteq U^\prime$ for some $x_0,\ldots,x_k\in U^\prime$.
For each $i\leqslant n$, we put $W_i=\bigcup\{W_{x_j}\mid j\leqslant k,\ W_{x_j}\prec U^\prime\cap V_i\}$.
Then it is clear that $W_i\prec V_i$ for all $i\leqslant n$ and that $U\subseteq W_0\cup\ldots\cup W_n\subseteq U^\prime$.
\end{proof}

\begin{prop}\label{P:ec}
Let $\mathbb{X}$ be a core-compact space and let $\mathbb{Y}$ possess the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$.
Then $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ also possesses the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$.
\end{prop}

\begin{proof}
Let $f_0,f_1\in C(\mathbb{X},\mathbb{Y})$.
For each $x\in X$, we put
\[
g(x)=f_0(x)\vee f_1(x).
\]
We prove that $g\in C(\mathbb{X},\mathbb{Y})$. Indeed, let $V\in\mathcal{T}(\mathbb{Y})$ and let $x\in g^{-1}(V)$.
As $f_0(x)\vee f_1(x)\in V$ and $\vee$ is continuous in $\mathbb{Y}$, we conclude that there are open sets $V_0,V_1\in\mathcal{T}(\mathbb{Y})$
such that $f_0(x)\in V_0$, $f_1(x)\in V_1$, and $V_0\cap V_1\subseteq V$.
But then
\[
x\in U=f_0^{-1}(V_0)\cap f_1^{-1}(V_1)\in\mathcal{T}(\mathbb{X}).
\]
If $z\in U$ then $g(z)=f_0(z)\vee f_1(z)\geq f_0(z)\in V_0$ and $g(z)\geq f_1(z)\in V_1$,
whence $g(z)\in V_0\cap V_1\subseteq V$ which implies that $x\in U\subseteq g^{-1}(V)$.
The fact that $U\in\mathcal{T}(\mathbb{X})$ proves that $g^{-1}(V)\in\mathcal{T}(\mathbb{Y})$.
Therefore, $g$ is continuous and $g=f_0\vee f_1$ in $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$.
Hence, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ has the property $(\mathrm{H}_1)$.

To prove that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ has the property $(\mathrm{H}_2)$, we consider two functions
$f_0,f_1\in C(\mathbb{X},\mathbb{Y})$ and open sets $U\in\mathcal{T}(\mathbb{X})$, $W\in\mathcal{T}(\mathbb{Y})$ such that
\[
g=f_0\vee f_1\in V_{U,W}.
\]
This means that $U\prec g^{-1}(W)$.
Since $\mathbb{Y}$ possesses $(\mathrm{H}_2)$, for each element $x\in g^{-1}(W)$, there are open sets $V_{0x},V_{1x}\in\mathcal{T}(\mathbb{Y})$ such that
$f_0(x)\in V_{0x}$, $f_1(x)\in V_{1x}$, and $V_{0x}\cap V_{1x}\subseteq W$.
We claim that
\[
g^{-1}(W)=\bigcup\bigl\{f_0^{-1}(V_{0x})\cap f_1^{-1}(V_{1x})\mid x\in g^{-1}(W)\bigr\}.
\]
Indeed it is clear that $g^{-1}(W)\subseteq\bigcup\bigl\{f_0^{-1}(V_{0x})\cap f_1^{-1}(V_{1x})\mid x\in g^{-1}(W)\bigr\}$.
To prove the reverse inclusion, we consider an arbitrary element $x\in g^{-1}(W)$ and establish that
$U(x)=f_0^{-1}(V_{0x})\cap f_1^{-1}(V_{1x})\subseteq g^{-1}(W)$.
Indeed, let $z\in U(x)$.
Then $g(z)\geq f_0(z)\in V_{0x}$ and $g(z)\geq f_1(z)\in V_{1x}$,
whence $g(z)\in V_0\cap V_1\subseteq W$ which implies that $z\in g^{-1}(W)$.

Furthermore, since $U\prec g^{-1}(W)=\bigcup\bigl\{U(x)\mid x\in g^{-1}(W)\bigr\}$, we conclude that there are $x_0,\ldots,x_k\in g^{-1}(W)$ such that
$U\prec U(x_0)\cup\ldots\cup U(x_k)$.
As $\mathbb{X}$ is core-compact, there is $V\in\mathcal{T}(\mathbb{X})$ such that
\[
U\prec V\prec U(x_0)\cup\ldots\cup U(x_k).
\]
Taking $\mathcal{B}=\mathcal{T}(\mathbb{X})$ and applying Proposition \ref{P:B}, we obtain that there are open sets
$S_0,\ldots,S_k$, $T_0,\ldots,T_k\in\mathcal{T}(\mathbb{X})$ such that
\[
S_i\prec T_i\prec U(x_i)\ \text{for all}\ i\leqslant k\ \text{and}\ U\subseteq S_0\cup\ldots\cup S_n\subseteq V.
\]
For each $i\leqslant k$, we have $T_i\prec U(x_i)=f_0^{-1}(V_{0x_i})\cap f_1^{-1}(V_{1x_i})$
whence
\begin{align*}
&f_0\in W_0=V_{T_0,V_{0x_0}}\cap\ldots\cap V_{T_k,V_{0x_k}}\in\mathcal{I};\\
&f_1\in W_1=V_{T_0,V_{1x_0}}\cap\ldots\cap V_{T_k,V_{1x_k}}\in\mathcal{I}.
\end{align*}
To establish $(\mathrm{H}_2)$, it suffices to prove that $W_0\cap W_1\subseteq V_{U,W}$.
Indeed, if $h\in W_0\cap W_1$ then $h\in V_{T_i,V_{0x_i}}\cap V_{T_i,V_{1x_i}}$ for each $i\leqslant k$.
This implies that
\[
S_i\prec T_i\subseteq h^{-1}(V_{0x_i})\cap h^{-1}(V_{1x_i})=h^{-1}(V_{0x_i}\cap V_{1x_i})\subseteq h^{-1}(W)
\]
for all $i\leqslant k$.
Hence, $U\subseteq S_0\cup\ldots\cup S_k\prec h^{-1}(W)$ yielding $h\in V_{U,W}$,
which is our desired conclusion.
\end{proof}

\begin{cor}\label{C:ts-0}
For a $T_0$-space $\mathbb{Y}$, the following statements are equivalent.
\begin{enumerate}
\item
$\mathbb{Y}$ is a topological semilattice with respect to $\leq_\mathbb{Y}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for some core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for each $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is a topological semilattice with respect to $\leq$ for some $[$core-com\-pact$]$ $T_0$-space $\mathbb{X}$ and some topology $\mathcal{T}$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
\end{enumerate}
\end{cor}

\begin{proof}
(1) implies (2) by Proposition \ref{P:ec}.
(2) implies (3) and (3) implies (4) in a trivial way. (4) implies (1) by Proposition \ref{P:H}.
Statements (1), (5), and (6) are equivalent by \cite[Proposition 9]{CXY}.
Furthermore, (3) obviously implies (7) and (7) implies (1) by Proposition \ref{P:H}.
\end{proof}

\noindent
The next theorem is a generalization of Theorem 10 from \cite{CXY}.

\begin{thm}\label{T:CXY-1}
For a $T_0$-space $\mathbb{Y}$, the following statements are equivalent.
\begin{enumerate}
\item
$\mathbb{Y}$ is essentially complete.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is essentially complete for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is essentially complete for some core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is essentially complete for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is essentially complete for each $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is essentially complete for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is essentially complete for some $[$core-com\-pact$]$ $T_0$-space $\mathbb{X}$ and some topology $\mathcal{T}$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
\end{enumerate}
\end{thm}

\begin{proof}
If $\mathbb{X}$ is a core-compact space and $\mathbb{Y}$ is an essentially complete space,
then $\mathbb{Y}$ is a $d$-space and possesses the properties $(\mathrm{H}_0)$--$(\mathrm{H}_2)$
by Theorem \ref{T:EssCom}. By Theorem \ref{T:FunD}, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is a $d$-space.
By Proposition \ref{P:ec}, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ has the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$.
By Lemma \ref{L:00} and Lemma \ref{L:A}, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ has the property $(\mathrm{H}_0)$.
Applying Theorem \ref{T:EssCom} again, we obtain that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is essentially complete.
Therefore, (1) implies (2).

It is straightforward that (2) implies (3), (3) implies (4), and (3) implies (7).

Suppose now that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is an essentially complete space for some $T_0$-spaces $\mathbb{X}$ and $\mathbb{Y}$.
According to Theorem \ref{T:EssCom}, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is a $d$-space and has the properties $(\mathrm{H}_0)$--$(\mathrm{H}_2)$.
By the proof of Theorem \ref{T:FunD}, $\mathbb{Y}$ is also a $d$-space.
By Proposition \ref{P:H}, $\mathbb{Y}$ has the properties $(\mathrm{H}_1)$--$(\mathrm{H}_2)$.
By Lemma \ref{L:00} and Lemma \ref{L:A}, $\mathbb{Y}$ has the property $(\mathrm{H}_0)$.
By Theorem \ref{T:EssCom}, $\mathbb{Y}$ is essentially complete and (4) implies (1).
Then (7) implies (1) in a similar way.

Statements (1) and (5)--(6) are equivalent by \cite[Theorem 10]{CXY}.
\end{proof}


\section{$A$-spaces}

\begin{definition}\label{D:AS}
A $T_0$-space $\mathbb{X}$ is an $A$-\emph{space} [an $f$-\emph{space}] if it has a basic subspace $\mathbb{X}_0$ [consisting of compact elements, respectively]
which is a \emph{parus} with respect to the specialization order $\leq_\mathbb{X}$; that is,
if $x,y\in X_0$ and $x,y\leq a$ for some $a\in X$, then there exists $x\vee y\in X_0$.
In this case, $\mathbb{X}_0$ is called an $A$-\emph{basic subspace of} $\mathbb{X}$ [an $f$-\emph{basic subspace of} $\mathbb{X}$, respectively].
%A $T_0$-space $\mathbb{X}$ is an $S$-\emph{space} [an $s$-\emph{space}] if it has a basic subspace $\mathbb{X}_0$ [consisting of compact elements, respectively]
%which is an $S$-\emph{set}; that is,
%$X_0\cap\dnw a$ is a join-semilattice with a least element with respect to the specialization order $\leq_\mathbb{X}$ for each $a\in X$.
%In this case, $\mathbb{X}_0$ is called an $S$-\emph{basic subspace of} $\mathbb{X}$ [an $s$-\emph{basic subspace of} $\mathbb{X}$, respectively].
\end{definition}

\noindent
The following theorem was established in \cite{Top}.

\begin{thm}\label{T:A-top}\cite[Theorem 7.3.4]{Top}
Let $\mathbb{X}$ be an $\alpha^\ast$-space and let $\mathbb{Y}$ be an $A$-space with a least element.
Then $\mathbb{C}(\mathbb{X},\mathbb{Y})$ is also an $A$-space with a least element.
\end{thm}

\noindent
The next statement shows that the requirement in Theorem \ref{T:A-top} on $\mathbb{Y}$ to have a least element can be weakened under certain circumstances.
The proof of Theorem \ref{T:FA} follows the same lines as the one of Theorem 7.3.4 in \cite{Top}.
In the proof, we also use some ideas from \cite{LL}.

\begin{thm}\label{T:FA}
Let $\mathbb{X}$ be a core-compact space
and let $\mathbb{Y}$ be a $T_0$-space such that $\dnw y$ has a least element for all $y\in Y$.
Then $\dnw f$ has a least element with respect to $\leq_\mathcal{P}$ for each $f\in C(\mathbb{X},\mathbb{Y})$.
Moreover, if $\mathbb{Y}$ is an $A_d$-space then $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$
is an $A_d$-space.
%the following statements hold.
%%\begin{enumerate}
%\item
%If $\mathbb{Y}$ is an $A_d$-space then $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$
%is an $A_d$-space.
%\item
%If $\mathbb{Y}$ is an $S_d$-space then $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$
%is an $S_d$-space.
%\end{enumerate}
\end{thm}

\begin{proof}
As usual, we write $\leq$ instead of $\leq_\mathcal{P}$ for the sake of simplicity.
%It follows from Theorem \ref{T:FunD} that, under our assumptions about $\mathbb{Y}$,
%$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $d$-space for each
%topology $\mathcal{T}$ on $C(\mathbb{X},\mathbb{Y})$
%such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{I}$.
%Therefore, $\mathcal{I}\subseteq\mathcal{S}$ by Lemma 1.8.6(i) in \cite{Top}.
For each $y\in Y$, let $\bot_y$ denote the least element of the set $\dnw y$ in $\mathbb{Y}$ and let
$\mathbb{Y}_0$ be an $A$-basic subspace in $\mathbb{Y}$.

According to Theorem \ref{T:FunD}, $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$ is a $d$-space.

\setcounter{claim}{0}
\begin{claim}
Let $\mathbb{Z}$ be an arbitrary $\alpha$-space and let $z\in Z$ be such that $\dnw_\mathbb{Z}z$ has a least element $\bot_z$.
Then $\bot_z\prec_\mathbb{Z}\bot_z$.
\end{claim}

\begin{scproof}
Indeed, let $a\in\upw\bot_z$. Since $a\in Z$ and
$\mathbb{Z}$ is a basic subspace in $\mathbb{Z}$, we conclude that there is $b\in Z$ such that
$b\prec a$. But then $b\in\dnw a$ and thus $\bot_z\leq_\mathbb{Z}b\prec a$.
This implies that $a\in\inte\upw b\subseteq\upw\bot_z$ which yields that $\upw\bot_z\in\mathcal{T}(\mathbb{Z})$.
\end{scproof}

\noindent
It follows from Claim 1 that $\bot_y\in Y_0$ for all $y\in Y$.

\begin{claim}
$\dnw f$ has a least element with respect to $\leq$ for each $f\in C(\mathbb{X},\mathbb{Y})$.
\end{claim}

\begin{scproof}
Consider the mapping
\[
\bot_f\colon X\to Y,\quad
\bot_f\colon x\mapsto\bot_{f(x)}.
\]
It is straightforward that $\bot_f\leq g$ for each $g\leq f$.
So, it remains to prove that $\bot_f$ is continuous. Indeed, let
$V\in\mathcal{T}(\mathbb{Y})$; then applying Claim 1 and using the continuity of $f$, we have
\[
\bot_f^{-1}(V)=\{x\in X\mid\bot_{f(x)}\in V\}=\bigcup\bigl\{f^{-1}(\upw\bot_y)\mid y\in Y,\ \bot_y\in V\bigr\}\in\mathcal{T}(\mathbb{X}),
\]
which completes the proof.
\end{scproof}

%\begin{claim}
%If $\mathbb{X}$ is compact then $\upw\xi_y\in\mathcal{K}\subseteq\mathcal{I}$ for each compact element $y\in Y$.
%\end{claim}

%\begin{scproof}
%We have
%\begin{align*}
%\upw\xi_y=&\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid y\leq f(x)\ \text{for all}\ x\in X\bigr\}=\\
%=&\bigl\{f\in C(\mathbb{X},\mathbb{Y})\mid f^{-1}(\upw y)=X\bigr\}=V_{X,\upw y}\in\mathcal{K}\subseteq\mathcal{I},
%\end{align*}
%which proves the desired statement.
%\end{scproof}

\noindent
Similarly to the proof of \cite[Theorem 7.3.4]{Top}, we consider the set
\begin{align*}
\mathcal{W}=&\Bigl\{\langle f,\varnothing,\varnothing\rangle\mid f\in C(\mathbb{X},\mathbb{Y})\Bigr\}\cup\\
\cup&\Bigl\{\bigl\langle f,\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\mid \\
&f\in C(\mathbb{X},\mathbb{Y}),\ n<\omega,\ U_0,\ldots, U_n\in\mathcal{T}(\mathbb{X}){\setminus}\{\varnothing\},
\ y_0,\ldots,y_n\in Y_0\Bigr\},
\end{align*}
where each $\bigl\langle f,\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$ possesses the following properties:
\begin{enumerate}
\item[($\ast$)]
the set $\{U_0,\ldots, U_n\}$ is closed under nonempty intersections;
\item[(${\ast}{\ast}$)]
if $U_i\subseteq U_j$ for some $i,j\leqslant n$ then $y_j\leq_\mathbb{Y}y_i$.
\item[(${\ast}{\ast}{\ast}$)]
$\bot_{f(x)}\leq y_i$ for all $x\in\bigcup_{j\leqslant n}U_j$, where $i\leqslant n$ is such that $U_i=\bigcap\{U_j\mid j\leqslant n,\ x\in U_j\}$.
\end{enumerate}
For each $W=\bigl\langle f,\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$, we define a mapping $f_W\colon X\to Y$
by the following rule:
\[
f_W(x)=
\begin{cases}
y_i,\quad&\text{if}\ U_i=\bigcap\{U_j\mid j\leqslant n,\ x\in U_j\};\\
\bot_{f(x)},\quad&\text{if}\ x\notin\bigcup_{j\leqslant n}U_j.
\end{cases}
\]
It follows from the definition that $f_W(X)\subseteq Y_0$.

\begin{claim}
$f_W\in C(\mathbb{X},\mathbb{Y})$ for each $W\in\mathcal{W}$.
\end{claim}

\begin{scproof}
Let
$W=\bigl\langle f,\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle$.
Since $\mathbb{Y}_0$ is a basic subspace of $\mathbb{Y}$, it suffices to show that $f_W^{-1}(\inte\upw a)\in\mathcal{T}(\mathbb{X})$ for all $a\in Y_0$.
Indeed, for a fixed element $a\in Y_0$, the following three cases are possible.

\Case{1}: $\bot_{f(x)}\in\inte\upw a$ for some $x\in X$.
We have in this case that $a\prec\bot_{f(x)}$ whence $a=\bot_{f(x)}$ as $\bot_{f(x)}$ is a least element in $\dnw f(x)$.
%Let $I=\{i\leqslant n\mid a\leq y_i\}$ and let $U^\prime=\bigcup_{i\in I}U_i$;
%of course, $U=\varnothing$ if $I=\varnothing$.
We claim that
$f_W^{-1}(\inte\upw a)=f^{-1}(\inte\upw a)$.

Indeed, let $x^\prime\in f_W^{-1}(\upw\bot_{f(x)})=f_W^{-1}(\inte\upw a)$. This means that $\bot_{f(x)}\leq_\mathbb{Y}f_W(x^\prime)$. There are two subcases.

\Case{1.1}: $x^\prime\in\bigcup_{i\leqslant n}U_i$. Let $j\leqslant n$ be such that $U_j=\bigcap\{U_i\mid i\leqslant n,\ x^\prime\in U_i\}$.
Then $\bot_{f(x)}\leq f_W(x^\prime)=y_j$.
As $\bot_{f(x^\prime)}\leq y_j$ by (${\ast}{\ast}{\ast}$), we conclude that
$a=\bot_{f(x)}=\bot_{f(x^\prime)}\leq f(x^\prime)$. This implies that $x^\prime\in f^{-1}(\inte\upw a)$.

\Case{1.2}: $x^\prime\notin\bigcup_{i\leqslant n}U_i$. In this case,
$\bot_{f(x)}\leq f_W(x^\prime)=\bot_{f(x^\prime)}$ whence $a=\bot_{f(x)}\prec\bot_{f(x)}=\bot_{f(x^\prime)}\leq f(x^\prime)$.
Therefore, $x^\prime\in f^{-1}(\inte\upw a)$ again.

Thus, $f_W^{-1}(\inte\upw a)\subseteq f^{-1}(\inte\upw a)$. To prove the reverse inclusion, suppose that $x^\prime\in f^{-1}(\inte\upw a)$.
This yields that $a=\bot_{f(x)}\leq f(x^\prime)$ whence $\bot_{f(x)}=\bot_{f(x^\prime)}$.
If $x^\prime\in\bigcup_{i\leqslant n}U_i$ then there is $j\leqslant n$ such that $U_j=\bigcap\{U_i\mid i\leqslant n,\ x^\prime\in U_i\}$.
Thus, $a=\bot_{f(x)}=\bot_{f(x^\prime)}\leq y_j=f_W(x^\prime)$ by (${\ast}{\ast}{\ast}$).
If $x\notin\bigcup_{i\leqslant n}U_i$ then $a=\bot_{f(x)}=\bot_{f(x^\prime)}=f_W(x^\prime)$ and $x^\prime\in f_W^{-1}(\inte\upw a)$ in any case.

Therefore, $f^{-1}(\inte\upw a)\subseteq f_W^{-1}(\inte\upw a)$ and $f_W^{-1}(\inte\upw a)=f^{-1}(\inte\upw a)\in\mathcal{T}(\mathbb{X})$ in Case 1.

\Case{2}: $\{x\in X\mid a\prec\bot_{f(x)}\}=\varnothing$ and $I=\{i\leqslant n\mid a\prec y_i\}\ne\varnothing$.
We put $U=\bigcup_{i\in I}U_i$; then $U\in\mathcal{T}(\mathbb{X})$. We show that $U=f_W^{-1}(\inte\upw a)$.
Indeed, if $x\in U$ then $x\in U_j$ for some $j\in I$.
But then $f_W(x)=y_i$ for some $i\leqslant n$ such that $x\in U_i\subseteq U_j$.
Applying (${\ast}{\ast}$), we obtain that $a\prec y_j\leq y_i=f_W(x)$.
This means that $x\in f_W^{-1}(\inte\upw a)$ and $U\subseteq f_W^{-1}(\inte\upw a)$.
Conversely, let $x\in f_W^{-1}(\inte\upw a)$.
As $\{x\in X\mid a\prec\bot_{f(x)}\}=\varnothing$, we conclude that $y_i=f_W(x)\in\inte\upw a$ whence $i\in I$. It follows that
$x\in U_i\subseteq U$. Therefore, $f_W^{-1}(\inte\upw a)\subseteq U$ which is our desired conclusion.

\Case{3}: $\{x\in X\mid a\prec\bot_{f(x)}\}=\{i\leqslant n\mid a\prec y_i\}=\varnothing$.
In this case, $f_W^{-1}(\inte\upw a)=\varnothing\in\mathcal{T}(\mathbb{X})$.
\end{scproof}

\noindent
The following statement is a corollary of the definition of $f_W$.

\begin{claim}
For $f\in C(\mathbb{X},\mathbb{Y})$ and $W=\bigl\langle f,\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$,
we have $f_W\leq f$ if and only if $y_i\leq f(x)$ for all
$x\in\bigcup_{j\leqslant n}U_j$, where $i\leqslant n$ is such that $U_i=\bigcap\{U_j\mid j\leqslant n,\ x\in U_j\}$.
\end{claim}

\begin{claim}
Let $f\in C(\mathbb{X},\mathbb{Y})$ and let $W=\bigl\langle f,\{U\},\{y\}\bigr\rangle$ be such that
$y\in Y_0$, $U\in\mathcal{T}(\mathbb{X})$, and $U\prec f^{-1}(\inte\upw y)$.
Then $f_W\ll f$.
\end{claim}

\begin{scproof}
First of all, we notice that $W\in\mathcal{W}$ provided that $y\prec f(x)$ for all
$x\in U$. Indeed, for each $x\in U$, we have $\bot_{f(x)}=\bot_y\leq y\prec f(x)$ which implies that $W$ possesses the property (${\ast}{\ast}{\ast}$).
The conditions (${\ast}$)--(${\ast}{\ast}$) are satisfied in a trivial way.
Moreover, it is straightforward to see that $f_W\leq f$.

Suppose that $f\leq g=\bigvee_{i\in I}g_i$ for some family $\{g_i\in C(\mathbb{X},\mathbb{Y})\mid i\in I\}$ which is up-directed with respect to $\leq_\mathcal{P}$.
According to Theorem \ref{T:FunD}, $h(x)=\bigvee_{i\in I}g_i(x)$ exists for all $x\in X$.
It is clear that $h(x)\leq g(x)$ for all $x\in X$. Conversely, if $g(x)\in U\in\mathcal{T}(\mathbb{Y})$ then $g\in V_{x,U}\in\mathcal{P}\subseteq\mathcal{S}$.
Since $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$ is a $d$-space, we conclude that $g_i\in V_{x,U}$ for some $i\in I$, whence $h(g)\geq g_i(x)\in U$.
Therefore, $h(x)\in U$ and $g(x)\leq h(x)$; that is, $g(x)=\bigvee_{i\in I}g_i(x)$ for all $x\in X$.
Hence, we have by our assumption that
\[
U\prec f^{-1}(\inte\upw y)\subseteq g^{-1}(\inte\upw y)=\bigcup_{i\in I}g_i^{-1}(\inte\upw y).
\]
This yields that $U\subseteq g_i^{-1}(\inte\upw y)$ and $g_i(U)\subseteq\upw y$ for some $i\in I$. We claim that $f_W\leq g_i$.
Indeed, if $x\in U$ then $f_W(x)=y\leq g_i(x)$.
Suppose now that $x\notin U$. Since $f(x)\leq g(x)$ and $g_i(x)\leq g(x)$, we conclude that
$\bot_{f(x)}=\bot_{g(x)}=\bot_{g_i(x)}$.
Therefore, $f_W(x)=\bot_{f(x)}=\bot_{g_i(x)}\leq g_i(x)$ and $f_W\leq g_i$ which proves that $f_W\ll f$.
\end{scproof}

\begin{claim}
We have
\[
f=\bigvee\Bigl\{f_{\bigl\langle f,\{U\},\{y\}\bigr\rangle}\mid y\prec f(x)\ \text{for some}\ x\in X,\ U\prec f^{-1}(\inte\upw y)\Bigr\}
\]
for each function $f\in C(\mathbb{X},\mathbb{Y})$.
In particular, $\bigl\langle C(\mathbb{X},\mathbb{Y});\leq_\mathcal{P}\bigr\rangle$ is a continuous poset.
\end{claim}

\begin{scproof}
According to Claim 5, $f$ is an upper bound for the family
\[
\mathcal{F}_f=\Bigl\{f_{\bigl\langle f,\{U\},\{y\}\bigr\rangle}\mid y\prec f(x)\ \text{for some}\ x\in X,\ U\prec f^{-1}(\inte\upw y)\Bigr\}.
\]
Let $g$ be another upper bound for $\mathcal{F}_f$.
It suffices to establish that $f\leq g$.
Indeed, let $x\in X$. As $\mathbb{Y}_0$ is a basic subspace of $\mathbb{Y}$, we conclude that $f(x)=\bigvee\{y\in Y_0\mid y\prec f(x)\}$.
Fix an element $y\in Y_0$ such that $y\prec f(x)$.
Then we have $x\in f^{-1}(\inte\upw y)\in\mathcal{T}(\mathbb{X})$.
As $\mathbb{X}$ is core-compact, the poset $\bigl\langle\mathcal{T}(\mathbb{X});\subseteq\bigr\rangle$ is continuous.
Therefore, there is a nonempty set $U_y\in\mathcal{T}(\mathbb{X})$ such that $x\in U_y\prec f^{-1}(\inte\upw y)$.
We put $W(y)=\bigl\langle f,\{U_y\},\{y\}\bigr\rangle$.
Then we have $f_{W(y)}\in\mathcal{F}_f$ and thus, $y=f_{W(y)}(x)\leq g(x)$.
This implies that $f(x)=\bigvee\{y\in Y_0\mid y\prec f(x)\}\leq g(x)$.
Hence, $f\leq g$ which is our desired conclusion.

The last statement follows from Claim 5.
\end{scproof}

\noindent
From Claim 6 and Lemma 1.10.5 in \cite{Top}, we obtain the following statement.

\begin{claim}
Let $f\in C(\mathbb{X},\mathbb{Y})$ and let $W=\bigl\langle f,\{U\},\{y\}\bigr\rangle$ be such that
$y\in Y_0$, $U\in\mathcal{T}(\mathbb{X})$, and $U\prec f^{-1}(\inte\upw y)$.
Then $f_W\prec_\mathcal{S}f$.
\end{claim}

\noindent
We put $C_0=\{f_W\mid W\in\mathcal{W}\}$.

\begin{claim}
If $Y_0$ is a parus in $\langle Y;\leq_\mathbb{Y}\rangle$ then
$\mathbb{C}_0$ is a parus in $\bigl\langle C(\mathbb{X},\mathbb{Y});\leq_\mathcal{P}\bigr\rangle$.
\end{claim}

\begin{scproof}
We consider arbitrary sequences
\begin{align*}
&W_0=\bigl\langle g_0,\{U_0,\ldots, U_m\},\{a_0,\ldots,a_m\}\bigr\rangle\in\mathcal{W},\\
&W_1=\bigl\langle g_1,\{V_0,\ldots, V_n\},\{b_0,\ldots,b_n\}\bigr\rangle\in\mathcal{W}
\end{align*}
and assume that $f_0=f_{W_0}$, $f_1=f_{W_1}\leq h$ for some $h\in C(\mathbb{X},\mathbb{Y})$.
As $\mathbb{Y}_0$ is a parus in $\mathbb{Y}$, for all $x\in X$, there is $f_0(x)\vee f_1(x)\in Y_0$.
We put
\begin{align*}
U=&\{U_0,\ldots, U_m\}\cup\{V_0,\ldots, V_n\}\cup
\{U_i\cap V_j\mid i\leqslant m,\ j\leqslant n,\ U_i\cap V_j\ne\varnothing\}=\\
=&\{Z_0,\ldots, Z_k\};\\
F=&\{y_0,\ldots,y_k\};\\
W=&\langle h,U,F\rangle,
\end{align*}
where $y_i$, $i\leqslant k$, is defined in the following way.
For each $i\leqslant k$, we have $\bigcap\{U_j\mid j\leqslant m,\ Z_i\subseteq U_j\}\ne\varnothing$.
Hence, $\bigcap\{U_j\mid j\leqslant m,\ Z_i\subseteq U_j\}=U_{m_i}$ for some $m_i\leqslant m$.
Symmetrically, $\bigcap\{V_j\mid j\leqslant n,\ Z_i\subseteq V_j\}\ne\varnothing$,
whence $\bigcap\{V_j\mid j\leqslant n,\ Z_i\subseteq V_j\}=V_{n_i}$ for some $n_i\leqslant n$.
Thus for each element $x\in Z_i$, we have $a_{m_i}\leq f_0(x)\leq h(x)$
and $b_{n_i}\leq f_1(x)\leq h(x)$.
As $\mathbb{Y}_0$ is a parus in $\mathbb{Y}$, there is an element $y_i=a_{m_i}\vee b_{n_i}\in Y_0$.

By our definition, $W$ possesses the property ($\ast$).
Suppose that $Z_{i_0}\subseteq Z_{i_1}$ for some $i_0,i_1\leqslant k$.
For each $j<2$, let $m_j\leqslant m$ and $n_j\leqslant n$ be such that
\[
U_{m_j}=\bigcap\{U_t\mid t\leqslant m,\ Z_{i_j}\subseteq U_t\}\quad\text{and}\quad V_{n_j}=\bigcap\{V_t\mid t\leqslant n,\ Z_{i_j}\subseteq V_t\}.
\]
Since $Z_{i_0}\subseteq Z_{i_1}$, we obtain that
$U_{m_0}\subseteq U_{m_1}$ and $V_{m_0}\subseteq V_{m_1}$.
We have therefore by our definition that
$y_{i_0}=a_{m_0}\vee b_{n_0}\geq a_{m_1}\vee b_{n_1}=y_{i_1}$ as $W_0$ and $W_1$ have the property (${\ast}{\ast}$).
This yields that $W$ possesses the property (${\ast}{\ast}$).
To establish that $W$ also possesses the property (${\ast}{\ast}{\ast}$), we consider an arbitrary element $x\in\bigcup_{i\leqslant k}Z_i$.
Let $j\leqslant k$ be such that $Z_j=\bigcap\{Z_t\mid t\leqslant k,\ x\in Z_t\}$ and let
\[
U_{m_j}=\bigcap\{U_t\mid t\leqslant m,\ Z_j\subseteq U_t\};\quad V_{n_j}=\bigcap\{V_t\mid t\leqslant n,\ Z_j\subseteq V_t\}.
\]
If $x\in U_s$ for some $s\leqslant m$, then $Z_j\subseteq U_s$ as $U_s\in U$. This implies that $x\in Z_j\subseteq U_{m_j}\subseteq U_s$.
Similarly, $x\in V_s$ for some $s\leqslant n$ implies that $x\in Z_j\subseteq V_{n_j}\subseteq V_s$.
Therefore, we obtain that
\[
U_{m_j}=\bigcap\{U_t\mid t\leqslant m,\ x\in U_t\};\quad V_{n_j}=\bigcap\{V_t\mid t\leqslant n,\ x\in V_t\}.
\]
As $W_0$ and $W_1$ possess the property (${\ast}{\ast}{\ast}$), we obtain
\begin{align*}
&y_j=a_{m_j}\vee b_{n_j}\geq a_{m_j}=f_0(x)\geq\bot_{g_0(x)};\\
&y_j=a_{m_j}\vee b_{n_j}\geq b_{n_j}=f_1(x)\geq\bot_{g_1(x)};\\
&h(x)\geq f_0(x)=a_{m_j};\\
&h(x)\geq f_1(x)=b_{n_j}.
\end{align*}
Hence $\bot_{g_0(x)}=\bot_{h(x)}=\bot_{g_1(x)}$ and
\[
h(x)\geq y_j=f_W(x)\geq\bot_{h(x)}=\bot_{g_0(x)}=\bot_{g_1(x)}.
\]
This proves that $W$ also has the property (${\ast}{\ast}{\ast}$).

Finally, we prove that $f_W=f_0\vee f_1$ in $\mathbb{C}(\mathbb{X},\mathbb{Y})$.
We prove first that $f_0,f_1\leq f_W$.
Indeed, let $x\in X$. Two cases are possible. In what follows, we use the notation introduced above.

\Case{1}: $x\in\bigcup_{i\leqslant k}Z_k$. In this case, $f_0(x)=a_{m_j}$, $f_1(x)=b_{n_j}$, $f_W(x)=y_j=a_{m_j}\vee b_{n_j}\geq a_{m_j}=f_0(x)$.
Similarly, $f_W(x)\geq b_{n_j}=f_1(x)$.

\Case{2}: $x\notin\bigcup_{i\leqslant k}Z_i$.
This implies that $x\notin\bigcup_{i\leqslant m}U_i\cup\bigcup_{i\leqslant n}V_i$.
Therefore,
\[
\bot_{g_0(x)}=f_0(x)\leq h(x);\quad\bot_{g_1(x)}=f_1(x)\leq h(x),
\]
whence $f_0(x)=f_1(x)=\bot_{g_0(x)}=\bot_{g_1(x)}=\bot_{h(x)}=f_W(x)$.

Therefore, $f_W$ is indeed an upper bound of $f_0$ and $f_1$.
Suppose that $f_0,f_1\leq f$ for some $f\in C(\mathbb{X},\mathbb{Y})$; we have to establish that $f_W\leq f$. Indeed, for $x\in X$, two cases are possible.

\Case{1}: $x\in\bigcup_{i\leqslant k}Z_k$. According to the considerations in \emph{Case 1} above, we have
$f_W(x)=y_j=a_{m_j}\vee b_{n_j}=f_0(x)\vee f_1(x)\leq f(x)$ in this case.

\Case{2}: $x\notin\bigcup_{i\leqslant k}Z_i$.
This implies as in \emph{Case 2} above that $f_W(x)=\bot_{g_0(x)}=\bot_{g_1(x)}=f_0(x)=f_1(x)\leq f(x)$.

Therefore, $f_W=f_0\vee f_1$ and the proof of Claim is complete.
\end{scproof}

\begin{claim}
$\mathbb{C}_0$ is a basic subspace in $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$.
\end{claim}

\begin{scproof}
Let $f\in V\in\mathcal{S}$.
According to Claim 6, $f=\bigvee\mathcal{F}_f$, where
\[
\mathcal{F}_f=\Bigl\{f_{\bigl\langle f,\{U\},\{y\}\bigr\rangle}\mid y\prec f(x)\ \text{for some}\ x\in X,\ U\prec f^{-1}(\inte\upw y)\Bigr\}.
\]
According to Claim 8, for each finite nonempty subfamily $\mathcal{G}\subseteq\mathcal{F}_f$, there is $W(\mathcal{G})\in\mathcal{W}$ such that
$f_\mathcal{G}=f_{W(\mathcal{G})}=\bigvee\mathcal{G}\leq f$. Thus, the family
\[
\mathcal{D}=\bigl\{f_\mathcal{G}\mid\mathcal{G}\subseteq\mathcal{F}_f,\ 0<|\mathcal{G}|<\omega\bigr\}
\]
is up-directed with respect to $\leq_\mathcal{P}$ and $f=\bigvee\mathcal{D}$. This implies that $f_\mathcal{G}\in V$ for some
finite nonempty family $\mathcal{G}\subseteq\mathcal{F}_f$.
Moreover, $f\in\inte_\mathcal{S}\upw g$ for all $g\in\mathcal{G}$ according to Claim 6.
Therefore,
\[
f\in\bigcap\{\inte_\mathcal{S}\upw g\mid g\in\mathcal{G}\}=\inte_\mathcal{S}\upw f_\mathcal{G},
\]
which implies that $f_\mathcal{G}\prec_\mathcal{S}f$.
%consider an arbitrary up-directed family $\{h_i\in C(\mathbb{X},\mathbb{Y})\mid i\in I\}$ such that
%$f\leq h=\bigvee_{i\in I}h_i$. Since $g\prec_\mathcal{S}f$ for all $g\in\mathcal{G}$ by Claim 5, we conclude that there is $i\in I$ such that
%$g\leq h_i$ for all $g\in\mathcal{G}$. We prove that $f_\mathcal{G}=\bigvee_{\dnw h}\mathcal{G}$.
%Indeed, we have $f_\mathcal{G}\leq f\leq h$. Therefore, $f_\mathcal{G}$ is an upper bound of $\mathcal{G}$ in $\dnw h$.
%This implies that $\bigvee_{\dnw h}\mathcal{G}\leq f_\mathcal{G}\leq f$. Hence, $\bigvee_{\dnw h}\mathcal{G}$ is an upper bound
%of $\mathcal{G}$ in $\dnw f$. Therefore, $\bigvee_{\dnw h}\mathcal{G}=f_\mathcal{G}$ which was to be proved.
%Since $h_i$ is also an upper bound of $\mathcal{G}$ in $\dnw h$, we conclude that $f_\mathcal{G}=\bigvee_{\dnw h}\mathcal{G}\leq h_i$ and thus,
%$f_\mathcal{G}\prec_\mathcal{S}f$.
\end{scproof}

\noindent
The desired statements follow from Claims 2, 8, and 9.
\end{proof}

\begin{cor}\label{C:1}
The following statements are equivalent for a $T_0$-space $\mathbb{Y}$.
\begin{enumerate}
\item
$\mathbb{Y}$ is an $A_d$-space such that $\dnw y$ has a least element for each $y\in Y$.
\item
For each core-compact space $\mathbb{X}$, $\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$ is an $A_d$-space such that $\dnw f$ has a least
element for each $f\in C(\mathbb{X},\mathbb{Y})$.
\end{enumerate}
\end{cor}

\begin{proof}
(1) implies (2) by Theorem \ref{T:FA}.
Let $\mathbb{T}$ be a trivial (one-element) topological space; $\mathbb{T}$ is obviously core-compact.
If (2) holds, then $\mathbb{Y}\cong\mathbb{C}_\mathcal{S}(\mathbb{T},\mathbb{Y})$ is an $A_d$-space such that $\dnw f$ has a least
element for each $f\in C(\mathbb{X},\mathbb{Y})$. Hence, (1) also holds.
\end{proof}

\begin{thm}\label{T:FA0}
If $\mathbb{X}$ is a core-compact space and $\mathbb{Y}$ is an $A$-space with a least element then
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is an $A$-space with a least element.
\end{thm}

\begin{proof}
The proof is very close to the one of Theorem 7.3.4 in \cite{Top}; we use certain parts of it here.

Let $\bot$ denote the least element in $\mathbb{Y}$ and let
$\mathbb{Y}_0$ be an $A$-basic subspace in $\mathbb{Y}$; in particular, $\bot\in Y_0$.
It follows from Lemma \ref{L:00} that $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ has a least element.

Similarly to the proof of \cite[Theorem 7.3.4]{Top}, we consider the set
\begin{align*}
\mathcal{W}=\bigl\{\langle\varnothing,\varnothing\rangle\bigr\}\cup
\bigl\{\bigl\langle\{&U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\mid\\
&n<\omega,\ U_0,\ldots, U_n\in\mathcal{T}(\mathbb{X}){\setminus}\{\varnothing\},\ y_0,\ldots,y_n\in Y_0\bigr\},
\end{align*}
where each $\bigl\langle\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$ possesses the following properties:
\begin{enumerate}
\item[($\ast$)]
the set $\{U_0,\ldots, U_n\}$ is closed under nonempty intersections;
\item[(${\ast}{\ast}$)]
if $U_i\subseteq U_j$ for some $i,j\leqslant n$ then $y_j\leq_\mathbb{Y}y_i$.
\end{enumerate}
For each $W=\bigl\langle\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$, we define a mapping $f_W\colon X\to Y$
by the following rule:
\[
f_W(x)=
\begin{cases}
y_i,\quad&\text{if}\ U_i=\bigcap\{U_j\mid j\leqslant n,\ x\in U_j\};\\
\bot,\quad&\text{if}\ x\notin\bigcup_{j\leqslant n}U_j.
\end{cases}
\]
We put $C_0=\{f_W\mid W\in\mathcal{W}\}$.

Claims 1--3 were established in the proof of \cite[Theorem 7.3.4]{Top}

\setcounter{claim}{0}
\begin{claim}
$f_W\in C(\mathbb{X},\mathbb{Y})$ for each $W\in\mathcal{W}$.
\end{claim}

\begin{claim}
For $f\in C(\mathbb{X},\mathbb{Y})$ and $W=\bigl\langle\{U_0,\ldots, U_n\},\{y_0,\ldots,y_n\}\bigr\rangle\in\mathcal{W}$,
we have $f_W\leq f$ if and only if $y_i\leq f(x)$ for all
$x\in\bigcup_{j\leqslant n}U_j$, where $i\leqslant n$ is such that $U_i=\bigcap\{U_j\mid j\leqslant n,\ x\in U_j\}$.
\end{claim}

\begin{claim}
If $Y_0$ is a parus in $\langle Y;\leq_\mathbb{Y}\rangle$ then
$\mathbb{C}_0$ is a parus in $\bigl\langle C(\mathbb{X},\mathbb{Y});\leq_\mathcal{P}\bigr\rangle$.
\end{claim}

\begin{claim}
$\mathbb{C}_0$ is a basic subspace in $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$.
\end{claim}

\begin{scproof}
Suppose first that $U\in\mathcal{T}(\mathbb{X})$, $W\in\mathcal{T}(\mathbb{Y})$, and $f\in C(\mathbb{X},\mathbb{Y})$ are such that
$f\in V_{U,W}\in\mathcal{I}$; then $U\prec f^{-1}(W)$.
Since $\mathbb{Y}_0$ is a basic subspace in $\mathbb{Y}$,
$f^{-1}(W)=\bigcup_{y\in Y_0\cap W}f^{-1}(\inte\upw y)$.
Therefore, $U\prec f^{-1}(\inte\upw y_0)\cup\ldots\cup f^{-1}(\inte\upw y_n)$ for some $y_0,\ldots,y_n\in Y_0\cap W$.
Since $\mathbb{X}$ is core-compact, there is $U^\prime\in\mathcal{T}(\mathbb{X})$ with
$U\prec U^\prime\prec f^{-1}(\inte\upw y_0)\cup\ldots\cup f^{-1}(\inte\upw y_n)$.
We put $\mathcal{B}=\mathcal{T}(\mathbb{X})$ and apply Proposition \ref{P:B}.
According to it and the fact that $\mathbb{X}$ is core-compact, there are $W_0,\ldots,W_n$, $S_0,\ldots,S_n\in\mathcal{T}(\mathbb{X})$ such that
\[
W_i\prec S_i\prec f^{-1}(\inte\upw y_i)\ \text{for all}\ i\leqslant n\ \text{and}\ U\subseteq W_0\cup\ldots\cup W_n\subseteq U^\prime.
\]
We claim that $\bigcap_{i\leqslant n}V_{W_i,W}\subseteq V_{U,W}$.
Indeed, if $g\in\bigcap_{i\leqslant n}V_{W_i,W}$ then $W_i\prec g^{-1}(W)$.
Hence, $U\subseteq W_0\cup\ldots\cup W_n\prec g^{-1}(W)$ which implies that $g\in V_{U,W}$.

For each $i\leqslant n$, we put $f_i=f_{\langle\{S_i\},\{y_i\}\rangle}$; then $f_i\in C_0$.
Moreover, if $x\in S_i$ then $f_i(x)=y_i\prec f(x)$ as $S_i\prec f^{-1}(\inte\upw y_i)$.
Furthermore, $y_i\in W$ whence $W_i\prec S_i\subseteq f_i^{-1}(W)$ and $f_i\in V_{W_i,W}$.
Therefore, $f_i\leq f$ for all $i\leqslant n$ by Claim 2.
According to Claim 3, there is $h=f_0\vee\ldots\vee f_n\leq f$.
As $f_i\in V_{W_i,W}$ for all $i\leqslant n$, we conclude that $h\in\bigcap_{i\leqslant n}V_{W_i,W}\subseteq V_{U,W}$.

Finally, $f\in\bigcap_{i\leqslant n}V_{S_i,\inte\upw y_i}$ as $S_i\prec f^{-1}(\inte\upw y_i)$ for each $i\leqslant n$.
Suppose that $g\in\bigcap_{i\leqslant n}V_{S_i,\inte\upw y_i}$; then
for all $i\leqslant n$ and all $x\in S_i$, one has $f_i(x)=y_i\prec g(x)$, whence $f_i\leq g$ by Claim 2.
This implies that $h\leq g$.
Summarizing, we obtain that $f\in\bigcap_{i\leqslant n}V_{S_i,\inte\upw y_i}\subseteq\upw h$; that is, $h\prec_\mathcal{I}f$.

To complete the proof, we assume that $f\in V\in\mathcal{I}$. Then there is $k<\omega$ and there are open sets
$U_0,\ldots,U_k\in\mathcal{T}(\mathbb{X})$ and $W_0,\ldots,W_k\in\mathcal{T}(\mathbb{Y})$ such that
\[
f\in V_{U_0,W_0}\cap\ldots\cap V_{U_k,W_k}\subseteq V.
\]
According to what we have just proved, there are $h_0,\ldots,h_k\in C(\mathbb{X},\mathbb{Y})$ such that
$h_i\prec_\mathcal{I}f$ and $h_i\in V_{U_i,W_i}$ for all $i\leqslant k$.
By Claim 3, there is $h=h_0\vee\ldots\vee h_n\leq f$.
In particular, $h\in V_{U_0,W_0}\cap\ldots\cap V_{U_k,W_k}\subseteq V$ and $h\prec_\mathcal{I}f$.
\end{scproof}

\noindent
The desired statement follows from Claims 3--4.
\end{proof}

\noindent
We notice that Theorem \ref{T:FA0} generalizes Theorem 7.3.4 in \cite{Top} (see Theorem \ref{T:A-top}) as $\mathcal{I}=\mathcal{P}$
on $C(\mathbb{X},\mathbb{Y})$ by Proposition \ref{P:1}(3) whenever $\mathbb{X}$ is an $\alpha^\ast$-space.

\begin{cor}\label{C:2}
The following statements are equivalent for a $T_0$-space $\mathbb{Y}$.
\begin{enumerate}
\item
$\mathbb{Y}$ is an $A$-space with a least element.
\item
For each core-compact space $\mathbb{X}$, $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is
an $A$-space with a least element.
\item
For each $\alpha^\ast$-space $\mathbb{X}$, $\mathbb{C}(\mathbb{X},\mathbb{Y})$ is
an $A$-space with a least element.
\end{enumerate}
\end{cor}

\begin{proof}
(1) implies (2) by Theorem \ref{T:FA0}. (1) implies (3) as $\mathcal{I}=\mathcal{P}$ for an $\alpha^\ast$-space $\mathbb{X}$.
A trivial (one-element) topological space $\mathbb{T}$ is obviously an $\alpha^\ast$-space and thus core-compact.
If (2) holds, then $\mathbb{Y}\cong\mathbb{C}_\mathcal{I}(\mathbb{T},\mathbb{Y})$ is
an $A$-space with a least element. Thus, (1) also holds and each of (2) and (3) implies (1).
\end{proof}




\section{Injectivity in function spaces}

\noindent
We will make use of the following results.

\begin{thm}\label{T:CharInj}\cite[Theorem 4.2.3]{Top}
A topological $T_0$-space $\mathbb{X}$ is injective if and only if the following conditions hold:
\begin{enumerate}
\item
$\mathbb{X}$ is a $d$-space;
\item
$\mathbb{X}$ is an $\alpha$-space;
\item
$\langle X;\leq_{\mathbb{X}}\rangle$ is a complete lattice.
\end{enumerate}
\end{thm}

\begin{thm}\label{T:CharRInj}\cite[Theorem 4.2.4]{Top}
A topological $T_0$-space $\mathbb{X}$ is densely injective if and only if the following conditions hold:
\begin{enumerate}
\item
$\mathbb{X}$ is a $d$-space;
\item
$\mathbb{X}$ is an $\alpha$-space;
\item
$\mathbb{X}$ is a $\mathrm{bc}$-domain.
\end{enumerate}
\end{thm}

\begin{cor}\label{C:Ad}
A topological $T_0$-space $\mathbb{X}$ is $[$densely$]$ injective if and only if $\mathbb{X}$
is a $A_d$-space with a least and a greatest element $[$with a least element$]$.
\end{cor}

\begin{prop}\label{P:FunA}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an $\alpha^\ast$-space $[$an $\alpha$-space$]$
for some space $\mathbb{X}$, then
$\mathbb{Y}$ is also an $\alpha^\ast$-space $[$an $\alpha$-space$]$.
\end{prop}

\begin{proof}
Let $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ be an $\alpha^\ast$-space.
Suppose that $y\in U\in\mathcal{T}(\mathbb{Y})$ and fix an element $x\in X$;
then $\xi_y\in V_{x,U}\in\mathcal{P}\subseteq\mathcal{T}$. According to our assumption, there are continuous functions
$f_0$, \ldots, $f_m\in V_{x,U}$ and a set $V\in\mathcal{T}$ such that
$\xi_y\in V\subseteq\upw f_0\cup\ldots\cup\upw f_m$.
Since $V\in\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$, there are
upper cones $\mathcal{H}_0$, \ldots, $\mathcal{H}_n$ in $\langle\mathcal{T}(\mathbb{X});\subseteq\rangle$
and open sets $W_0,\ldots,W_n\in\mathcal{T}(\mathbb{Y})$ such that
\[
\xi_y\in V_{\mathcal{H}_0,W_0}\cap\ldots\cap V_{\mathcal{H}_n,W_n}\subseteq V\subseteq\upw f_0\cup\ldots\cup\upw f_m\subseteq V_{x,U}.
\]
Therefore, $\xi^{-1}_i(W_j)\in\mathcal{H}_j$ for all $j\leqslant n$.
In particular, $\mathcal{H}_j\ne\varnothing$ for all $j\leqslant n$.
We put
\[
W=\bigcap\bigl\{W_j\mid j\leqslant n,\ \mathcal{H}_j\ne\mathcal{T}(\mathbb{X})\bigr\}.
\]
Then we have that $W\in\mathcal{T}(\mathbb{Y})$.
Moreover, $\xi^{-1}_y(W_j)\ne\varnothing$ whence $\xi^{-1}_y(W_j)=X$ for all $j\leqslant n$
such that $\mathcal{H}_j\ne\mathcal{T}(\mathbb{X})$.
This implies that $y\in W$.
If $y^\prime\in W$, then $\xi^{-1}_{y^\prime}(W_j)=X$ for all $j\leqslant n$
such that $\mathcal{H}_j\ne\mathcal{T}(\mathbb{X})$.
Therefore,
\[
\xi_{y^\prime}\in V_{\mathcal{H}_0,W_0}\cap\ldots\cap V_{\mathcal{H}_n,W_n}\subseteq V\subseteq\upw f_0\cup\ldots\cup\upw f_m\subseteq V_{x,U}
\]
and thus $y^\prime\in U$.
We established therefore that $y\in W\subseteq\upw f_0(x)\cup\ldots\cup\upw f_m(x)$ and $f_0(x)$, \ldots, $f_m(x)\in U$.
This implies that $\mathbb{Y}$ is an $\alpha^\ast$-space.

If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an $\alpha$-space, we use in the argument above with $m=0$.
\end{proof}

\begin{prop}\label{P:di}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a densely injective space then $\mathbb{Y}$ is also densely injective.
\end{prop}

\begin{proof}
According to Theorem \ref{T:CharRInj}, $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is an $\alpha$-space, a $d$-space, and a $\mathrm{bc}$-domain.
According to Proposition \ref{P:FunA}, $\mathbb{Y}$ is an $\alpha$-space.
According to the proof of Theorem \ref{T:FunD}, $\mathbb{Y}$ is a $d$-space.
In view of Theorem \ref{T:CharRInj}, in order to prove the dense injectivity of $\mathbb{Y}$, it suffices to show that
$\mathbb{Y}$ is a partial join-semilattice with respect to the specialization order $\leq_\mathbb{Y}$.

Let $y_0$, $y_1\leq y$ in $\mathbb{Y}$.
This means that $\xi_{y_0}$, $\xi_{y_1}\leq\xi_y$ in $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$.
As $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $\mathrm{bc}$-domain, there is a continuous function $f=\xi_{y_0}\vee\xi_{y_1}$.
We prove that $f$ is constant.
Indeed, let $f(x_0)\in U\in\mathcal{T}(\mathbb{Y})$ and let $x_1\in X$ be an arbitrary element. Then $f\in V_{x_0,U}\in\mathcal{P}\subseteq\mathcal{T}$.
By Corollary \ref{C:Ad}, $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an $A_d$-space.
By \cite[Lemma 7]{CXY}, there are open sets $V_0,V_1\in\mathcal{T}$ such that
$\xi_{y_0}\in V_0$, $\xi_{y_1}\in V_1$, and $V_0\cap V_1\subseteq V_{x_0,U}$.
Using the assumption that $\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$ and applying the same argument as in the proof of Proposition \ref{P:H}, we obtain that $f=\xi_y$ for some $y\in Y$, whence $y=y_0\vee y_1$ in $\mathbb{Y}$.

Applying Theorem \ref{T:CharRInj} again, we conclude that $\mathbb{Y}$ is densely injective.
\end{proof}

\begin{prop}\label{P:i}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an injective space then $\mathbb{Y}$ is also injective.
\end{prop}

\begin{proof}
According to Theorem \ref{T:CharInj} $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an $\alpha$-space,
a $d$-space, and a complete lattice with respect to the specialization order $\leq$.
By Proposition \ref{P:FunA}, $\mathbb{Y}$ is an $\alpha$-space.
By the proof of Theorem \ref{T:FunD}, $\mathbb{Y}$ is a $d$-space.
Since $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is an $A_d$-space having a least and a greatest element by Corollary \ref{C:Ad},
it possesses the properties
$(\mathrm{H}_0)$--$(\mathrm{H}_2)$ by \cite[Lemma 7]{CXY}.
By Lemma \ref{L:00} and Proposition \ref{P:H}, the space $\mathbb{Y}$ also possesses the properties
$(\mathrm{H}_0)$--$(\mathrm{H}_2)$. This means that
$\mathbb{Y}$ is a complete lattice with respect to the specialization order $\leq_\mathbb{Y}$. Applying Theorem
\ref{T:CharInj} again, we conclude that $\mathbb{Y}$ is an injective space.
\end{proof}

\noindent
The following statement generalizes Theorem 19 in \cite{CXY}.

\begin{thm}\label{T:CXY-4}
For a $T_0$-space $\mathbb{Y}$, the following conditions are equivalent.
\begin{enumerate}
\item
$\mathbb{Y}$ is densely injective.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is densely injective for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is densely injective for some core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is densely injective for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$
is densely injective for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is densely injective for each $\alpha^\ast$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is densely injective for some $\alpha^\ast$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is densely injective for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is densely injective for some $[$core-com\-pact$]$ $T_0$-space $\mathbb{X}$ and some topology $\mathcal{T}$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
\end{enumerate}
\end{thm}

\begin{proof}
(1) implies (2) by Proposition \ref{P:c9}(1).
(2) implies (3) implies (4), (6) implies (7) implies (8), and (3) implies (9) in a trivial way. (4) implies (1) and each of (8) and (9) implies (1) by Proposition \ref{P:di}.
Furthermore, (2) implies (5) as $\mathcal{I}=\mathcal{S}$ whenever $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is densely injective,
see \cite[Lemma 1.8.6]{Top}.

Let $\mathbb{T}$ be a trivial (one-element) topological space; $\mathbb{T}$ is obviously core-compact.
If (5) holds, then $\mathbb{Y}\cong\mathbb{C}_\mathcal{S}(\mathbb{T},\mathbb{Y})$ is a densely injective space. Hence, (1) also holds,
and (5) implies (1).

Statements (1), (6), and (7) are equivalent by \cite[Theorem 19]{CXY}.
\end{proof}


\noindent
The following statement generalizes Theorem 20 in \cite{CXY}.

\begin{thm}\label{T:CXY-3}
For a $T_0$-space $\mathbb{Y}$, the following conditions are equivalent.
\begin{enumerate}
\item
$\mathbb{Y}$ is injective.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is injective for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is injective for some core-compact space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$
is injective for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{S}(\mathbb{X},\mathbb{Y})$
is injective for each core-compact space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is injective for each $\alpha^\ast$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is injective for some $\alpha^\ast$-space $\mathbb{X}$.
\item
$\mathbb{C}(\mathbb{X},\mathbb{Y})$
is injective for some $T_0$-space $\mathbb{X}$.
\item
$\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
is injective for some $[$core-com\-pact$]$ $T_0$-space $\mathbb{X}$ and some topology $\mathcal{T}$ such that $\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
\end{enumerate}
\end{thm}

\begin{proof}
(1) implies (2) by Proposition \ref{P:c9}(1).
(2) implies (3) implies (4), (6) implies (7) implies (8), and (3) implies (9) in a trivial way.
(4) implies (1) and each of (8)and (9) implies (1) by Proposition \ref{P:i}.
Furthermore, (2) implies (5) as $\mathcal{I}=\mathcal{S}$ whenever $\mathbb{C}_\mathcal{I}(\mathbb{X},\mathbb{Y})$ is injective.

Let $\mathbb{T}$ be a trivial (one-element) topological space; $\mathbb{T}$ is obviously core-compact.
If (5) holds, then $\mathbb{Y}\cong\mathbb{C}_\mathcal{S}(\mathbb{T},\mathbb{Y})$ is a injective space, which yields (1).

Statements (1), (6), and (7) are equivalent by \cite[Theorem 20]{CXY}.
\end{proof}

\noindent
The fact that (1) implies (2) in Theorems \ref{T:CXY-4} and \ref{T:CXY-3} was established in \cite[Proposition II-4.6]{Comp}.
The fact that (3) implies (1) in Theorems \ref{T:CXY-4} and \ref{T:CXY-3} was established in \cite[Theorems 4.6, 4.7]{BQW}.

\section{$\Delta$-spaces $\mathbb{C}(\mathbb{X},\mathbb{Y})$}

\begin{definition}\label{D:D}\cite{E99-2}
A continuous function $\delta\colon\mathbb{X}\to\mathbb{X}$ from a topological space $\mathbb{X}$ into itself is a \emph{deflation}, if the set
$\delta(X)$ is finite and $\delta(x)\leq_{\mathbb{X}}x$ for all $x\in X$.

A $T_0$-space $\mathbb{X}$ is a $\Delta$-\emph{space}, if there is an up-directed family
$\{\delta_i\colon\mathbb{X}\to\mathbb{X}\mid i\in I\}$ of deflations of $\mathbb{X}$ with the property that
for every $U\in\mathcal{T}(\mathbb{X})$ and every $x\in U$, there is $i\in I$ such that $\delta_i(x)\in U$.

A space $\mathbb{X}$ is a $\Delta_d$-\emph{space}, if
$\mathbb{X}$ is a $\Delta$-space and a $d$-space simultaneously.
\end{definition}

\noindent
The following partial generalization of Theorem 23 from \cite{CXY} holds.

\begin{prop}\label{P:del}
Let $\mathbb{X}$, $\mathbb{Y}$ be $T_0$-spaces and let $\mathcal{T}$ be a topology on $C(\mathbb{X},\mathbb{Y})$ such that
$\mathcal{P}\subseteq\mathcal{T}\subseteq\mathcal{T}_A(\subseteq)^\sharp$.
If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is a $\Delta$-space $[\Delta_d$-space$]$ then $\mathbb{Y}$ is also a $\Delta$-space $[\Delta_d$-space$]$.
\end{prop}

\begin{proof}
Let $D$ be an up-directed family of deflations of the space $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$
which satisfies all the requirements of Definition \ref{D:D}.
We fix an element $a\in X$.
For each $\delta\in D$, consider the mapping
\[
\kappa_\delta\colon\mathbb{Y}\to\mathbb{Y},\quad
\kappa_\delta\colon y\mapsto\delta(\xi_y)(a).
\]
The set $\kappa_\delta(Y)$ is finite as the set $\{\delta(\xi_y)\mid y\in Y\}$ is finite.
Moreover, for every $y\in Y$, we have
$\kappa_\delta(y)=\delta(\xi_y)(a)\leq\xi_y(a)=y$, since $\delta$ is a deflation of $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$.
Let $U\in\mathcal{T}(\mathbb{X})$; then
\begin{align*}
\kappa_\delta^{-1}(U)&=\{y\in Y\mid\delta(\xi_y)(a)\in U\}=\{y\in Y\mid\delta(\xi_y)\in V_{a,U}\in\mathcal{T}\}=\\
&=\{y\in Y\mid\xi_y\in\delta^{-1}(V_{a,U})\in\mathcal{T}\}=\xi^{-1}\delta^{-1}(V_{a,U})\in\mathcal{T}(\mathbb{Y})
\end{align*}
as $\xi$ is continuous by Lemma \ref{L:xi} and $\delta$ is continuous by Definition \ref{D:D}.
Thus, $\kappa_\delta$ is continuous, whence it is a deflation of $\mathbb{Y}$.
Moreover, if $\delta,\delta^\prime\in D$ are such that $\delta\leq\delta^\prime$, then
$\kappa_\delta(y)=\delta(\xi_y)(a)\leq\delta^\prime(\xi_y)(a)=\kappa_{\delta^\prime}(y)$ for all $y\in Y$.
Therefore, $\{\kappa_\delta\mid\delta\in D\}$ is an up-directed family of deflations of $\mathbb{C}_\mathcal{T}(\mathbb{Y},\mathbb{Y})$
by Lemma \ref{L:00}.
Finally, if $y\in U\in\mathcal{T}(\mathbb{Y})$ then $\xi_y\in V_{a,U}\in\mathcal{P}\subseteq\mathcal{T}$.
In view of the choice of $D$, there is a deflation $\delta\in D$ such that
$\delta(\xi_y)\in V_{a,U}$.
Then $\kappa_\delta(y)=\delta(\xi_y)(a)\in U$, and the proof is complete.

The statement about $\Delta_d$-spaces follows with the use of Theorem \ref{T:FunD}.
\end{proof}

\noindent
Proposition \ref{P:del} was established for the case $\mathcal{T}=\mathcal{I}$ in \cite[Theorem 5.5]{BQW}.


%\noindent
%From Propositions \ref{P:1}(1) and \ref{P:del}, we obtain

%\begin{cor}\label{C:del}
%Let $\mathbb{X}$ be a core-compact space, let $\mathbb{Y}$ be a $T_0$-space, and let $\mathcal{T}$ be a proper topology on $C(\mathbb{X},\mathbb{Y})$ such that
%$\mathcal{P}\subseteq\mathcal{T}$. If $\mathbb{C}_\mathcal{T}(\mathbb{X},\mathbb{Y})$ is is a $\Delta$-space
%$[\Delta_d$-space$]$ then $\mathbb{Y}$ is also
%a $\Delta$-space $[\Delta_d$-space$]$.
%\end{cor}


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