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			{\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]  %Headings in Russian
			{\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
			{\large Siberian Electronic Mathematical Reports} \\[1mm]
			{\LARGE\tt{http://semr.math.nsc.ru}}\\[0.5mm]
			%     {\small Институт математики им. С.Л.Соболева СО РАН}\\[-1mm]
			%     {\small Sobolev Institute of Mathematics SB RAS}
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		\noindent
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		{\flushleft\it Том 20, № 2 стр. 814--826 (2023) \hspace{54mm}{\rm\small УДК \udcs}} %Volume, pages in Russian,
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		{\rm\small DOI~10.33048/semi.2023.20.000}\hphantom{aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa}{\rm\small MSC\ \ \mscs }
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	\title[Multivalued quasim\"{o}bius property and bounded turning]{Multivalued quasim\"{o}bius property \\and bounded turning}
	\author{{N.V. ABROSIMOV, V.V. ASEEV}}%
	\address{Nikolay Abrosimov \smallskip % Контактные данные всех авторов и место работы указываются только на английском языке
		\newline\hphantom{iii} Sobolev Institute of Mathematics,
		\newline\hphantom{iii} pr. Koptyuga, 4,
		\newline\hphantom{iii} 630090, Novosibirsk, Russia\smallskip}%
	\email{abrosimov@math.nsc.ru}%
	
	
	\address{Vladislav Aseev \smallskip % Контактные данные всех авторов и место работы указываются только на английском языке
		\newline\hphantom{iii} Sobolev Institute of Mathematics,
		\newline\hphantom{iii} pr. Koptyuga, 4,
		\newline\hphantom{iii} 630090, Novosibirsk, Russia\smallskip}%
	\email{aseevvv@yandex.ru}%
	
	\thanks{\sc Abrosimov, N.V., Aseev, V.V.,
		Multivalued quasim\"{o}bius property and bounded turning}
	\thanks{\copyright \ 2022 Abrosimov N.V., Aseev V.V}
	\thanks{\rm The study was carried out within the framework of the state contract of the Sobolev Institute of Mathematics (project no. FWNF-2022-0005).}
	\thanks{\it Received April, 17, 2022, published ???, ??,  ????.}%
	
	
	\semrtop \vspace{1cm}
	\maketitle {\small
		\begin{quote}
			\noindent{\sc Abstract. } The class of multivalued mappings with bounded angular distortion (BAD) property in metric spaces can be considered as a multivalued analog for quasim\"{o}bius mappings. We study the connections between quasimeromorphic self-mappings of $X= \bar{R}^n$ and multivalued mappings $F: X\to 2^X$ with BAD property. The main result of the paper concerns the multivalued mappings $F: D\to 2^{\bar{\bf C}}$ with BAD property of a domain $D\subset \bar{\bf C}$. If the image $F(x)$ of each point $x\in D$ is either a point or a continuum with bounded turning then $F$ is proved to be a single-valued quasim\"{o}bius mapping. The crucial point in the proof of this result is the local connectedness of the set $F(X)$ for the multivalued continuous mapping $F: X\to 2^Y$ with BAD property. We obtain sufficient conditions providing $F(X)$ to have local connectedness or bounded turning property in the most general case.\medskip
			
			\noindent{\bf Keywords:} multivalued quasim\"{o}bius mapping, multivalued hyperinjective mapping, Ptolemaic characteristic of tetrad, generalized angle, bounded angular distortion, local connectedness.
		\end{quote}
	}
	
	
	\section{Introduction}
	
	The definition of quasisymmetric and quasim\"{o}bius mappings was given by P.~Tukia and J.~V\"{a}is\"{a}l\"{a} \cite{1TV1980}, \cite{2V1981}, \cite{3V1984} in order to extend the notion of quasiconformality to the case of arbitrary (and in particular, discrete) metric spaces. The quasim\"{o}bius property was defined as the existence of some estimate for the distortion of absolute cross-ratio of a quadruple of points (i.e. tetrad) under the mapping. It is also equivalent to the existence of two-sided bounds for the distortion of Ptolemaic characteristics of tetrads (see \cite{4AST2005}, proposition~3.4). Moreover, it was shown in \cite{5A2018} under some assumptions on metric spaces that the lower estimate alone (for the distortion of Ptolemaic characteristic) is sufficient for the quasim\"{o}bius property of the mapping (see \cite{5A2018}, theorem~6.2).
	
	A multivalued mapping is called {\em hyperinjective} if the images of distinct points do not intersect each other. Thus, the multivalued hyperinjective mapping $F: X\to 2^Y$ transforms each tetrad from $X$ into a four of pairwise disjoint sets in $Y$ (so-called {\em generalized angle}, or {\em generalized tetrad}). The value of generalized angle was introduced in \cite{4AST2005} (see definition~4.1) as a counterpart for the Ptolemaic characteristic of tetrads. Then the condition of lower boundedness of the distortion of this characteristic distinguishes the class BAD (Bounded Angular Distortion) of multivalued mappings in metric spaces which may be considered as a multivalued analog for quasim\"{o}bius mappings. Some topological and metric properties of such mappings were investigated in \cite{5A2018}, \cite{6A2018}, \cite{7A2019} and \cite{8A2018}. In one special case, the metric spaces $X$ and $Y$ being unit circles in $\bar{\bf C}$, the multivalued quasim\"{o}bius mappings $F: X\to 2^{Y}$ with BAD property have been thoroughly studied in \cite{9A2021}. Namely, it was proved that the left inverse to $F$ mapping $f= F^{-1}: F(X) \to X$ is of the form $f(z) = [\varphi(z)]^N$ where $\varphi$ is a quasim\"{o}bius homeomorphism of unit circle onto itself.
	
	The purpose of our further research is to study the connection between quasimeromorphic self-mappings of $X= \bar{R}^n$ and multivalued mappings $F: X\to 2^X$ with BAD property. The main result presented below as Theorem~4.1. of section~\ref{th3.1} concerns the multivalued mappings $F: D\to 2^{\bar{\bf C}}$ with BAD property of a domain $D\subset \bar{\bf C}$. If the image $F(x)$ of each point $x\in D$ is either a point or a continuum with bounded turning then $F$ is proved to be a single-valued quasim\"{o}bius mapping. The crucial point in the proof of this result is the local connectedness of the set $F(X)$ for the multivalued continuous mapping $F: X\to 2^Y$ with BAD property. The sufficient conditions providing $F(X)$ to have local connectedness or bounded turning property in the most general case are obtained in Theorem~3.1. of section~\ref{th2.1} which will be also useful in our further studies.
	
	%\centerline{\bf \S 1. Definitions and notations.}
	\section{Definitions and notations}
	
	%\begin{defi}[see \cite{1TV1980}, Definition 2.7, p.100 or \cite{10T1996}, \S 1, p. 559]
	%	A metric space $(X, \rho)$ has {\it bounded turning} property $c$-BT with a constant $c\geq 1$ if any two points $x'$ and $x''$ of $X$ can be joined by a continuum $\tau\subset X$ such that ${\rm diam}_{\rho}(\tau) \leq c\cdot \rho(x',x'')$.
	%\end{defi}
	
	\noindent{\bf Definition} (see \cite{1TV1980}, Definition 2.7, p.100 or \cite{10T1996}, \S 1, p. 559). A metric space $(X, \rho)$ has {\it bounded turning} property $c$-BT with a constant $c\geq 1$ if any two points $x'$ and $x''$ of $X$ can be joined by a continuum $\tau\subset X$ such that ${\rm diam}_{\rho}(\tau) \leq c\cdot \rho(x',x'')$. \medskip
	
	In this definition, the space $X$ is supposed to be conncted. In more general case, we use the following 
	
	%\begin{definition}
	%	A metric space $(X, \rho)$ has {\it bounded turning} property $(c, \delta)$-BT with constants $c\geq 1$ and $\delta >0$ if any two points $x', x'' \in X$ with the distance $\rho(x',x'')< \delta$ can be joined by a continuum $\tau\subset X$ such that ${\rm diam}_{\rho}(\tau) \leq c\cdot \rho(x',x'')$.
	%\end{definition}
	
	\medskip
	\noindent{\bf 2.1. Definition.} A metric space $(X, \rho)$ has {\it bounded turning} property $(c, \delta)$-BT with constants $c\geq 1$ and $\delta >0$ if any two points $x', x'' \in X$ with the distance $\rho(x',x'')< \delta$ can be joined by a continuum $\tau\subset X$ such that ${\rm diam}_{\rho}(\tau) \leq c\cdot \rho(x',x'')$.\medskip

	
	If $\delta \geq {\rm diam}_{\rho}(X)$, then the property $(c, \delta)$-BT is equivalent to $c$-BT and provides the connectedness of $X$. If $X$ is a continuum of bounded diameter ${\rm diam}_{\rho}(X)= D< \infty$, then the property $(c, \delta)$-BT implies $c'$-BT with the constant $c' = \max\{ c, D/\delta\}$.
	
	%\begin{defi}[see \cite{11K1966}, \S 49.I]
	%	A topological space $X$ is {\it locally connected} if for every point $x\in X$ and it's open neighbourhood $U$ there exists a connected neighbourhood $V$ of $x$ such that $V\subset U$.
	%\end{defi}

	\medskip
	\noindent{\bf Definition} (see \cite{11K1966}, \S 49.I). A topological space $X$ is {\it locally connected} if for every point $x\in X$ and it's open neighbourhood $U$ there exists a connected neighbourhood $V$ of $x$ such that $V\subset U$.
	\medskip
	
	\noindent{\bf 2.2. Lemma.} {\it A metric space with bounded turning $(c, \delta)$-BT is locally connected.} \medskip
	
	See paragraph 5.1. for the proof. \medskip 
	
	\noindent{\bf Definitions.} Throughout this paper $2^Y$ denotes the set of all nonempty subsets of $Y$. The multivalued mapping $F: X\to 2^Y$ is said to be {\it hyperinjective} if $x_1\neq x_2$ implies $F(x_1)\cap F(x_2) = \emptyset$. The images $F(x)\subset Y$ of points $x\in X$ will be called {\it thick points}. In metric spaces $(X, \rho)$ and $(Y,\sigma)$ the multivalued mapping $F: X\to 2^Y$ with closed thick points is said to be {\it continuous} if the convergence $x_n \to x$ in $X$ implies the existence of the topological limit of the sequence $\{ F(x_n)\}$ of closed sets in $Y$ and the equality  ${\rm Lim}_{n\to \infty} F(x_n) = F(x_0)$. For more detailed definition see paragraph~5.2.  \medskip
	
	\noindent{\bf 2.3. Proposition.} {\it Let $X, Y$ be topological spaces and $F: X\to 2^Y$ be a multivalued hyperinjective continuous mapping. 
		
		(i) If a set $K\subset X$ is compact and each of thick points $F(x)$ for $x\in K$ is compact then the set $F(K): = \cup\{ F(x): x\in K\}$ is compact as well. Moreover, since $Y$ is a Hausdorff space, the left inverse to $F$ mapping $f= F^{-1}: F(K) \to K$ is continuous.
		
		(ii) If $\gamma \subset X$ is a connected set and $F(x_0)$ is connected for some point $x_0\in \gamma$, then $F(\gamma)$ is connected as well. In particular, if $\gamma\subset X$ is a continuum and all thick points $F(x)$ for $x\in \gamma$ are continuums, then $F(\gamma)$ is a continuum in $Y$. } \medskip 
	
	The proof of Proposition~2.3. will be given in paragraph~5.3.\medskip
	
	\noindent{\bf Definition} (see \cite{12B1953}, section~32.1). A metric (or semimetric) space $(X, \rho)$ is said to be {\it Ptolemaic} if the Ptolemy's inequality
	$$
	\rho(x_1, x_2)\cdot \rho(x_3, x_4) + \rho(x_1, x_4)\cdot \rho(x_2, x_3) \geq 
	\rho(x_1, x_3)\cdot \rho(x_2, x_4)
	$$ 
	holds for every four of distinct points $x_1, x_2, x_3, x_4 \in X$. \medskip
	
	In particular, the space $\bar{R}^n$ with chordal metric and the space $R^n$ with Euclidean or chordal metric both are Ptolemaic spaces. Any subspace of a Ptolemaic space remains to be Ptolemaic. \medskip
	
	\noindent{\bf Definition} (see \cite{4AST2005}, definition 4.1). A {\it general angle} in a metric (or semimetric) space $(Y, \sigma)$ is a four of non-empty pairvise disjoint sets $\Psi = (A_1,A_2; B_1, B_2)$, and it's {\it value} $\alpha(\Psi)$ is defined as
	$$
	\alpha(\Psi): = \inf_{u_1\in A_1; u_2\in A_2}\left( 
	\sup_{v_1\in B_1; v_2\in B_2} \frac{\sigma(u_1, u_2)\cdot \sigma(v_1, v_2)}
	{\sigma(u_1, v_1)\cdot \sigma(u_2, v_2) + \sigma(u_1, v_2)\cdot \sigma(u_2, v_1)}\right).
	$$
	It is clear, that $\alpha(\Psi)\leq 1$ for every generalized angle $\Psi$ in Ptolemaic metric space.\medskip
	
	\noindent{\bf 2.4. Definition} (see \cite{6A2018}, section 4.1). Let $(X, \rho)$ and $(Y, \sigma)$ be Ptolemaic metric (or semimetric) spaces. Let a real non-decreasing function $\omega: [0,1] \to [0,1]$ be positive on $(0,1]$, and $\omega(0) = 0 = \lim_{t\searrow 0}\omega(t)$. Then a multivalued hyperinjective mapping $F: X \to 2^Y$ is said to have $\omega$-BAD property (Bounded Angular Distortion) with {\it control function} $\omega$ if the estimate $\alpha(F(\Psi)) \geq \omega(\alpha(\Psi))$ is valid for every generalized angle $\Psi = (A_1, A_2; B_1, B_2)$ in $X$ and it's image $F(\Psi) = (F(A_1),F(A_2); F(B_1), F(B_2))$ in $Y$.\medskip
	
	For every control function $\omega$ in $\omega$-BAD property the real function $\omega^{(-1)}: [0,1]\to [0,1]$ is defined by formula $\omega^{(-1)}(s):= \sup\{ t\in [0,1]: \omega(t)\leq s\}$. In general, the function $\omega^{(-1)}$ is not inverse to $\omega$ in the strict sense, but it has all the properties of control function, and $\omega^{(-1)}(\omega(t)) \geq t$ for all $t\in [0,1]$.
	
	It was shown in \cite{6A2018} (section~4.2) that it is sufficient to check the inequality  $\alpha(F(\Psi))\geq\omega(\alpha(\Psi))$ in $\omega$-BAD property only for general angles $\Psi$ of the form $\Psi = (\{a_1\}, \{a_2\}; \{b_1\}, \{b_2\})$. In this case, the value  
	$$
	\alpha(\Psi)= \alpha(a_1, a_2; b_1, b_2)= \frac{\rho(a_1, a_2)\cdot \rho(b_1, b_2)}
	{\rho(a_1, b_1)\cdot \rho(a_2, b_2) + \rho(a_1, b_2)\cdot \rho(a_2, b_1)}
	$$
	is called the {\it Ptolemaic characteristic of tetrad $(a_1, a_2; b_1, b_2)$}.\medskip
	
	\noindent{\bf 2.5. Definition.} Let $(X,\rho)$ and $(Y, \sigma)$ be metric spaces, and $F: X \to 2^Y$ be a multivalued hyperinjective mapping. A pair of positive numbers $(\delta, \Delta)$ is called the {\it normalizator} of $F$ if there exists a set $\{a_1, a_2, a_3\}$ of points in $X$ such that $\rho(a_i, a_j)\geq \delta$ and ${\rm dist}_{\sigma}(F(a_i), F(a_j))\geq \Delta$ for all $i,j \in \{1,2,3\}$ with $i\neq j$.\medskip
	
	It should be noted that a multivalued hyperinjective mapping $F: X \to 2^Y$ with compact thick points has many normalizators. It suffices to take an arbitrary triple of distinct points $(a_1, a_2, a_3)$ in $X$ and put $\delta:= \min\{ \rho(a_i, a_j): i\neq j\}$ and $\Delta: = \min\{ {\rm dist}_{\sigma}(F(a_i), F(a_j)): i\neq j\}$.
	
	%\centerline{\bf \S 2. Local connectedness and bounded turning}
	\section{Local connectedness and bounded turning}
	
	In this section, we prove the following theorem. \medskip
	
	\noindent{\bf 3.1. Theorem.}\label{th2.1} {\it Let $(X, \rho)$ and $(Y, \sigma)$ be Ptolemaic metric spaces and a multivalued hyperinjective continuous mapping $F: X\to 2^{Y}$ has $\omega$-{\rm BAD} property. If all thick points $F(x)$ with $x\in X$ are compact sets with bounded turning property $(C_0, \delta_0)$-{\rm BT}, then the following statements hold.
		
		(i) If $X$ is compact and locally connected, then $F(X)$ is also compact and locally connected.
		
		(ii) If $X$ has a finite diameter ${\rm diam}_{\rho}(X) = d < \infty$, normalizator $(\delta, \Delta)$, and bounded turning property $(C, \delta')$-{\rm BT}, then $F(X)$ has bounded turning property $(C^*, \delta^*)$-{\rm BT}. Here, the constants $C^*$ and $\delta^*$ depend only on 
		the normalizator $(\delta, \Delta)$, constants $(C_0, \delta_0)$, $(C, \delta')$ in {\rm BT} properties, the diameter $d$ of $X$, and the control function $\omega$ in {\rm BAD} property.} \medskip
	
	The proof of the theorem is divided into several consecutive steps.\medskip
	
	\noindent{\bf 3.2. Proposition.}\label{pr2.2} {\it Under the assumptions in Theorem~3.1., if the mapping $F$ has a normalizator $(\delta, \Delta)$, then for every $x_0\in X$ and $y_0\in F(x_0)$, there exists a point $a\in X$ such that}
	$$
	\rho(x_0, a)> \delta/2\,,\quad {\rm dist}_{\sigma}(y_0, F(a)) \geq \Delta/2\,. \eqno(3.2.1)
	$$
	
	\medskip\noindent{\it Proof.} Let $A = \{ a_1, a_2, a_3\}$ be the triple of points in $X$ which realizes the normalizator $(\delta, \Delta)$. Then there is a pair of distinct points $\{a_i, a_j\}\subset A$ such that $\rho(x_0, a_i) > \delta/2$ and $\rho(x_0, a_j) > \delta/2$. Indeed, if $\rho(x_0, a') > \delta/2$ for all $a'\in A$, then we can take an arbitrary pair of distinct points in $A$ as $\{ a_i, a_j\}$. If $\rho(x_0, a')\leq \delta/2$, then $\rho(a'', x_0) \geq \rho(a'', a')- \rho(a', x_0)> \delta - \delta/2= \delta/2$ for every $a''\in A\setminus \{a'\}$. Then $\{a_i, a_j\} = A\setminus \{a'\}$ is the required pair of points.
	
	Next, there exists a pair of distinct points $\{a_k, a_s\}$ such that 
	$$
	\min\{ {\rm dist}_{\sigma}(y_0, F(a_k)) \geq \Delta/2\,,\quad  
	{\rm dist}_{\sigma}(y_0, F(a_s)) \} \geq \Delta/2\,.
	$$ 
	Indeed, if ${\rm dist}_{\sigma}(y_0, F(a')) \geq \Delta/2$ for all $a'\in A$, then we can take an arbitrary pair of distinct points in $A$ as the pair $\{ a_k, a_s\}$. If ${\rm dist}_{\sigma}(y_0, F(a')) < \Delta/2$ for a point $a'\in A$, then there exists a point $y'\in F(a')$ such that $\sigma(y_0, y') < \Delta/2$. Therefore for any point $a''\in A$ and every $z\in F(a'')$, the inequality $\sigma(y_0, z)\geq \sigma(z, y') - \sigma(y_0,y')> \Delta - \Delta/2 = \Delta/2$ holds. Thus $\{ a_k, a_s\} = A\setminus \{a'\}$ is the required pair of points.
	
	Since $\{i,j\} \cap \{ k, s\} \neq \emptyset$, there exists a point $a\in \{ a_i, a_j\}\cap \{a_k, a_s\}$ for which (3.2.1) is valid. 
	
	Proposition~3.2. is proved.\medskip 
	
	\noindent{\bf 3.3. Proposition.}\label{pr2.3} {\it Let $(X, \rho)$ be a metric space with a pair of distinct points $x_0$ and $a$. If $C\geq 1$ and points $x_1, x $ in $X$ are such that 
		$$
		\rho(x_0, x') \leq C \cdot \rho(x_0, x_1) < \rho(x_0, a)/2\,, \eqno(3.3.1) 
		$$  
		then the generalized angle $\Psi = (x_0, x_1; x', a)$ has the value $\alpha(\Psi) > (3+4C)^{-1}$.}\medskip
	
	\noindent{\it Proof.} Since $\rho(x_0, x') \leq C \rho(x_0, x_1)$ and $\rho(x_1, x') \leq (1+C)\rho(x_0, x_1)$, then
	$$
	\alpha(\Psi) = \frac{\rho(x_0, x_1) \rho(x', a)}{\rho(x_0, x')\rho(x_1, a) + 
		\rho(x_0, a)\rho(x_1, x')} \geq \frac{\rho(x', a)}{C\rho(x_1, a) + (1+C)\rho(x_0, a)}\,.
	$$
	Then in the numerator, we use the inequalities $\rho(x', a)\geq \rho(x_0,a)- \rho(x_0, x') > \rho(x_0,a)/2$, and in the denominator, we note that $\rho(x_1, a)\leq \rho(x_0, a)+ \rho(x_0, x_1)<(1 + 1/(2C))\rho(x_0, a)$. Continuing the inequality, we have
	$$
	\alpha(\Psi)  > \frac{1/2}{(1_ 2C)/2 + (1+C)} = \frac{1}{3+4C}\,.
	$$ 
	The Proposition 3.3. is proved. \medskip
	
	\noindent{\bf 3.4. Proposition.} {\it Under the assumptions in Theorem~3.1., let the mapping $F$ has the normalizator $(\delta, \Delta)$. Let $C\geq 1$ and the points $y_0\in F(x_0)$ and $y_1 \in F(x_1)$ be such that $x_0\neq x_1$ and
		$$
		\rho(x_0, x_1) \leq \delta/(4C)\,,\quad \sigma(y_0, y_1) \leq (\Delta/4)\omega_C\,, 
		\eqno(3.4.1)
		$$
		where $\omega_C: = \omega(1/(3+4C))$. Then for every point $x'\in X$ the condition $\rho(x_0, x')\leq  C\cdot \rho(x_0, x_1)$ implies that }
	$$
	F(x')\cap B(y_0, \sigma(y_0,y_1)\cdot 2/\omega_C) \neq \emptyset \,. \eqno(3.4.2)
	$$ 
	
	\medskip\noindent{\it Proof.} By Proposition~3.2., the estimates (3.2.1) are valid for some point $a\in X$. Let $x'\in X\setminus \{x_0, x_1, a\}$ be such that $\rho(x_0, x') \leq C\cdot \rho(x_0, x_1)$. Estimate (3.3.1) in Proposition~3.3. follows from inequalities (3.2.1) and (3.4.1), indeed $\rho(x_0, x')\leq C\cdot \rho(x_0, x_1)\leq \delta/4 < \rho(x_0,a)/2$. Thus we have the estimate $\alpha(\Psi)> (3+4C)^{-1}$ for the value of the generalized angle $\Psi = (x_0, x_1; x', a)$. For the value of the generalized angle $F(\Psi) = (F(x_0, F(x_1); F(x'), F(a))$, we have the following inequalities
	$$
	\alpha(F(\Psi)) \leq \sup_{v'\in F(x');\ v\in F(a)}\frac{\sigma(y_0, y_1)\cdot \sigma(v',v)}
	{\sigma(y_0, v')\cdot\sigma(y_1, v) + \sigma(y_0, v)\cdot\sigma(y_1, v')}\,.    
	$$  
	Since $\sigma(y_0, v') \geq {\rm dist}_{\sigma}(y_0, F(x'))$ and $\sigma(y_0, v) \geq {\rm dist}_{\sigma}(y_0, F(a)) \geq \Delta/2$) then we have
	\begin{multline*}
	\alpha(F(\Psi)) \leq \\
	\frac{\sigma(y_0, y_1)}{\min\{ \Delta/2,\ {\rm dist}_{\sigma}(y_0, F(x'))\}}
	\left( \sup_{v'\in F(x');\ v\in F(a)}\frac{\sigma(v',v)}
	{\sigma(v',y_0) + \sigma(v, y_0)} \right) \leq  \\
	\frac{\sigma(y_0, y_1)}{\min\{\Delta/2,\ {\rm dist}_{\sigma}(y_0, F(x') \} }\,.
	\end{multline*}  
	
	From the $\omega$-BAD property and the estimate $\alpha(\Psi)> (3+4C)^{-1}$ we get the inequality
	$$
	\frac{\sigma(y_0, y_1)}{\min\{\Delta/2,\ {\rm dist}_{\sigma}(y_0, F(x'))\}} \geq 
	\alpha(F(\Psi))\geq \omega((3+4C)^{-1}) = \omega_C\,.
	$$  
	Hence, $\min\{ \Delta/2,\ {\rm dist}_{\sigma}(y_0, F(x'))  \} \leq \sigma(y_0, y_1)/\omega_C$.
	
	Since $\Delta/2 > \Delta/4 \geq \sigma(y_0, y_1)/ \omega_C$ by (3.4.1), then
	$$
	{\rm dist}_{\sigma}(y_0, F(x')) \leq \sigma(y_0, y_1)/\omega_C\,.
	$$
	This means that the thick point $F(x')$ intersect the open ball $B(y_0, R)$ with radius $R = \sigma(y_0, y_1)\cdot 2/\omega_C$. 
	
	If $x'= x_0$, then (3.4.2) is obvious. If $x'= x_1$, then $y_1 \in B(y_0, \sigma(y_0, y_1)\cdot 2/\omega_C)$, and (3.4.2) is true. Under the condition $\rho(x_0, x') \leq C\cdot \rho(x_0, x_1)$, the point $x'$ cannot coincide with $a$ since $\rho(x_0, a) > \delta/2 = C\cdot \delta/(4C)\geq C\cdot \rho(x_0, x_1)$.
	
	Thus (3.4.2) holds for every point $x'$ such that $\rho(x_0, x')\leq C\cdot \rho(x_0, x_1)$. 
	
	Proposition 3.4. is proved.\medskip
	
	\noindent{\bf 3.5. Proposition.} {\it Let the mapping $F: X \to 2^{Y}$ satisfy the conditions in Theorem~3.1. Let $y_0 \in F(x_0)$ and $\varepsilon \in (0, \delta_0/2)$ be given. Then for every thick point $F(x')$ the set $F(x')\cap B(y_0, \varepsilon)$ (it may be empty) contained in a component of the set $F(x')\cap B(y_0, (1+ 2C_0)\varepsilon)$.}
	
	{\it Proof.} Clearly, it suffices to consider the case of non-empty set $F(x')\cap B(y_0, \varepsilon)$. Let $y', y'' \in F(x')\cap B(y_0, \varepsilon)$. Because of bounded turning property $(C_0, \delta_0)$-BT and the inequality $\sigma(y', y'')\leq \sigma(y', y_0) + \sigma(y'', y_0)< \delta_0$, there exists a continuum $\tau \subset F(x')$ such that $y',y'' \in \tau$ and $\mbox{diam}_{\sigma}(\tau) \leq C_0\cdot \sigma(y', y'') < 2C_0\varepsilon$. Since $\sigma(y_0, y') + \mbox{ diam}_{\sigma}(\tau) < \varepsilon + 2C_0\varepsilon$, $\tau \subset B(y_0. (1+2C_0)\varepsilon)$. It means that $y'$ and $y''$ enter into the same component $U$ of the set $F(x')\cap B(y_0, \varepsilon)$. Since $y', y''$ have been arbitrarily chosen points in $F(x')\cap B(y_0, \varepsilon)$, the desired result $F(x')\cap B(y_0, \varepsilon)\subset U$ follows. 
	
	Proposition 3.5 is proved.\medskip
	
	\noindent{\bf 3.6. Proposition.} {\it Let the mapping $F: X \to 2^Y$ satisfy the conditions in Theorem~3.1. and have the normalizator $(\delta, \Delta)$. Let $C\geq 1$ and the points 
		$y_0\in F(x_0), y_1\in F(x_1)$ be such that 
		$$
		0< \rho(x_0, x_1)\leq \delta/(4C)\,;\quad r = \sigma(y_0, y_1)\leq \min\{ \Delta/4, 
		\delta_0/8\}\cdot \omega_C\,. \eqno(3.6.1) 
		$$
		Let the connected set $\gamma\subset X$ be such that $x_0, x_1\in \gamma \subset 
		B(x_0, C\cdot \rho(x_0, x_1))$. Then the set 
		$$
		F(\gamma)\cap B(y_0, 2r(1+2C_0)/\omega_C)
		$$
		has a component $W$\,such that $y_0, y_1\in W$\,and }
	$$ 
	F(\gamma)\cap B(y_0,  2r/\omega_C) \subset W.
	$$
	
	\medskip\noindent{\it Proof.} Let $\mathcal{U}$ be the family of all those components of the set 
	$F(\gamma)\cap B(y_0, 2r(1+2C_0)/\omega_C)$, that intersect  
	the ball $B(y_0, 2r/\omega_C)$. Then $\rho(x_0, x_1)\leq \delta/(4C)$, 
	$\sigma(y_0, y_1)\leq (\Delta/4)\omega_C$ by (3.6.1) and $\rho(x_0, x')< C\cdot 
	\rho(x_0, x_1)$ for all $x'\in \gamma$. Hence, the set 
	$F(x')\cap B(y_0, 2r/\omega_C)$ is not empty (by Proposition~3.4.). It means that 
	for every point $x'\in \gamma$ there exists an element $U(x')$ 
	in the family $\mathcal{U}$.
	
	Since $\varepsilon: = 2r/\omega_C < (\delta_0/4)\cdot 2/\omega_C = \delta_0/2$, 
	by Proposition~3.5. the set $F(x')\cap B(y_0, 2r/\omega_C)$ is contained in a 
	component of the set $F(x')\cap B(y_0, 2r(1+ 2C_0)/\omega_C)$. It follows that the 
	element $U(x')$ corresponding to the point $x'\in \gamma$ contains non-empty set 
	$F(x')\cap B(y_0, 2r/\omega_C)$. 
	
	Let us show that the set $W: = \cup\{U: U\in \mathcal{U}\}$ has required properties. It 
	is contained in the ball $B(y_0, 2r(1+ 2C_0)/\omega_C)$ (because every 
	element of $\mathcal{U}$ is contained 
	in that ball) and contains the set $F(\gamma) \cap B(y_0, 2r/\omega_C)$. In particular, 
	$y_0,y_1 \in W$. Therefore, we have to prove that $W$ is connected and then 
	$\mathcal{U} = \{W\}$. 
	
	Suppose $W$ to be non-connected. Then there exist open sets $S_1, S_2 \subset Y$ such that 
	$W\subset S_1\cup S_2$, $S_1\cap S_2 = \emptyset$ and  
	$W\cap S_1 \neq \emptyset \neq W\cap S_2$. Replacing $S_j$ ($j=1,2$) by open sets 
	$S_j\cap B(y_0, 2r(1+ 2C_0)/\omega_C)$ we may assume that $S_1\cup S_2 \subset B(y_0, 
	2r(1+2C_0)/\omega_C)$. Since every $U\in \mathcal{U}$ is a connected set, for every 
	$U\in \mathcal{U}$ either $U\subset S_1$ or $U\subset S_2$ holds.  The restriction 
	$F|\gamma$ is a lower semicontinuous mapping, so the sets 
	$V_j = \{ x\in \gamma: F(x) \cap U_j \neq \emptyset \}$ ($j=1,2$) are open in $\gamma$. 
	For every point $x\in \gamma $ the set $F(x)\cap B(y_0, 2r/\omega_C)$ is non-empty and 
	is contained in an element $U(x)\in \mathcal{U}$ which is contained either in $S_1$ or 
	in $S_2$. It means that for every one point $x\in \gamma$ we have either $x\in V_1$ or 
	$x\in V_2$ but not both $x\in V_1$ and $x\in V_2$. Thus $V_1 \neq \emptyset \neq 
	V_2$, $\gamma = V_1\cup V_2$ and $V_1\cap V_2 = \emptyset$ for open sets $V_1, V_2$,  
	and this contradicts the connectedness of $\gamma$. 
	
	Proposition~3.6. is proved. \medskip
	
	\noindent{\bf 3.7. Proposition.}  {\it Let the mapping $F: X \to 2^Y$ satisfy the conditions in Theorem~3.1. and have the normalizator $(\delta, \Delta)$. 
		Let $d = {\rm diam}_{\rho}(X)< \infty$. Then for each pair of points $x', x'' $ in $X$ 
		the following inequality holds
		$$
		{\rm dist}_{\sigma}(F(x'), F(x'')) \geq \Delta\cdot \omega\left( \frac{\delta^2}
		{4d^2}\right)\cdot \omega\left( \rho(x', x'')\frac{\delta}{4d^2}\right). \eqno(3.7.1)
		$$
		Furthermore, $\rho(f(y'), f(y''))\leq \zeta(\sigma(y',y''))$ for all $y',y''\in F(X)$, 
		where}
	$$
	\zeta(t): = \frac{4d^2}{\delta}\cdot \omega^{(-1)}\left( \frac{t}{\Delta \cdot 
		\omega(\delta^2/(4d^2))}\right).  \eqno(3.7.2)
	$$  
	
	\medskip\noindent{\it Proof}. Let the normalizator $(\delta, \Delta)$ be realized by 3-point set 
	$A = \{ a_0, a_1, a_2\}$. For a point $x\in X\setminus A$ let us consider the generalized 
	angle $\Psi = (x, a_j; a_i, a_k)$ where $\{j, i, k \} = \{ 0, 1, 2\}$. 
	From $\rho(a_i, a_k)\geq \delta$ and $\rho(\,.\,,\,.\,)\leq d$ 
	it follows that
	$$
	\alpha(\Psi)= \frac{\rho(x, a_j)\rho(a_i,a_k)}{\rho(x, a_i)\rho(a_j,a_k)+ \rho(x, a_k)
		\rho(a_j, a_i)} \geq \frac{\delta}{2d^2}\rho(x, a_j)\,. \eqno(3.7.3)
	$$
	The value of generalized angle $F(\Psi) = (F(x), A_j; A_i, A_k)$ (here $A_j: = F(a_j)$, 
	$A_i: = F(a_i)$, $A_k: = F(a_k)$) is
	$$
	\alpha(F(\Psi)) = \inf_{u_1\in F(x);\ u_2\in A_j}\left( \sup_{v_1\in A_i;\ v_2\in A_k}
	\frac{\sigma(u_1, u_2)\sigma(v_1, v_2)}{\sigma(u_1, v_1)\sigma(u_2, v_2) + 
		\sigma(u_1, v_2)\sigma(u_2, v_1)}  \right).
	$$
	For an arbitrary pair of points $u_1^0\in F(x)$, $u_2^0 \in A_j$ it follows from estimates 
	$\sigma(u_2^0, v_2)\geq \Delta$ and $\sigma(u_2^0, v_1)\geq \Delta$ that
	$$
	\alpha(F(\Psi))\leq \frac{\sigma(u_1^0, u_2^0)}{\Delta} \sup_{v_1\in A_i;\ v_2\in A_k}
	\frac{\sigma(v_1, v_2)}{\sigma(v_1, u_1^0) + \sigma(u_1^0, v_2)} \leq 
	\frac{\sigma(u_1^0, u_2^0)}{\Delta}\,.
	$$
	It means that $\alpha(F(\Psi)) \leq {\rm dist}_{\sigma}(F(x), A_j)/\Delta$. Then 
	$\omega$-BAD property and (3.7.3) give us the inequality 
	$$
	{\rm dist}_{\sigma}(F(x), A_j)\geq 
	\Delta\cdot \omega\left(\rho(x, a_j)\frac{\delta}{2d^2}  \right)  \eqno(3.7.4) 
	$$
	which holds for every point $x\in X\setminus A$ and every $j\in \{0,1,2\}$. 
	If $i\neq j$ we have  
	${\rm dist}_{\sigma}(A_i, A_j) \geq \Delta \geq \Delta\cdot 
	\omega(\rho(a_i, a_j))\delta/(2d^2) $ and if $x\in A_i$ where $i=j$ the inequality (3.7.4) 
	turns to the equality $0 = 0$. Thus (3.7.4) is true for every $x\in X$ and for each  
	$j\in {0,1,2}$.
	
	For an arbitrary point $x'\in X$ there exists a pair of points $\{ b_1, b_2\}$ such that 
	$\rho(x', b_1)\geq \delta'$ and $\rho(x', b_2)\geq \delta'$ where $\delta' = \delta/2$. 
	Then for $B_1= F(b_1)$ and $B_2= F(b_2)$ the estimate (3.7.4) gives us the inequality 
	$$
	{\rm dist}_{\sigma}(F(x'), B_j) \geq 
	\Delta \cdot \omega\left( \rho(x', b_j)\frac{\delta}{2d^2}  \right)
	\geq \Delta\cdot \omega\left( \frac{\delta^2}{4d^2} \right)
	$$ 
	where $j\in \{1,2\}$. The set $B=\{ x', b_1, b_2\}$ defines for the mapping $F$ the
	normalizator 
	$(\delta', \Delta')$ with $\Delta': = \Delta\cdot \omega(\delta^2/(4d^2)) \leq \Delta$. 
	The estimate (3.7.4) being applied to an arbitrary point $x''\in X$ and 
	the normalizator $(\delta', \Delta')$ gives us the desired inequality (3.7.1):
	$$
	{\rm dist}_{\sigma}(F(x''), F(x')) \geq \Delta'\cdot \omega\left( \rho(x'', x')
	\frac{\delta'}{2d^2}\right) = \Delta\cdot \omega\left( \frac{\delta^2}{4d^2}\right)
	\cdot \omega\left(\rho(x', x'')\frac{\delta}{4d^2}  \right).
	$$
	Now, for every pair of points $y', y'' \in F(X)$ and the corresponding 
	points $f(y'), f(y'')\in X$ 
	we get the inequality
	$$
	\frac{\sigma(y', y'')}{\Delta\cdot \omega(\delta^2/ (4d^2))}\geq 
	\omega\left( \rho(f(y'), f(y'')) \frac{\delta}{4d^2}  \right).
	$$
	The increasing function $\omega^{(-1)}$ with $\omega^{(-1)}(\omega(p))\geq p$ (see 
	section~2.4.) being applied to both sides of this inequality gives the desired 
	estimate
	$$
	\rho(f(y'), f(y'')) \leq \frac{4d^2}{\delta}\cdot \omega^{(-1)}
	\left( \frac{\sigma(y', y'')}{\Delta\cdot \omega(\delta^2/(4d^2))}   \right) = 
	\zeta(\sigma(y',y''))\,.
	$$
	Proposition 3.7 is proved.\medskip
	
	\noindent{\bf 3.8. Proof of Theorem~3.1(i).} We may assume without loss of generality that 
	$Y= F(X)$. The compact metric space $X$ has a finite diameter ${\rm diam}_{\rho}(X) = d < 
	\infty$. Let $(\delta, \Delta)$ be a normalizator for $F$ and the function $\zeta(t)$ 
	be difined by (3.7.2).
	
	Let $y_0$ be an arbitrary point in $F(X)$ and $y_0\in F(x_0)$ for $x_0\in X$. 
	We are to show that every open 
	ball $B(y_0, R)$ contains a connected neighbourhood of the point $y_{0}$. Let 
	$C:= 2$ and $\omega_0:= \Omega_C = \omega(1/11)$.\\ 
	
	Since the set $\{ t>0: \zeta(t) < \delta/(4C)\}$ is not empty and $\zeta(t)\to 0$ as 
	$t\to 0$, there exists $t_0>0$ such that $\zeta(t) < \delta/(4C)$ for all $t\leq t_0$.
	Let
	$$
	R_0: = \min\left\{ \frac{R\cdot \omega_0}{2(1+C_0)}\ ;\ 
	\frac{\Delta}{4}\omega_0\ ;\ \frac{\delta_0}{8}\omega_0\ ;\ t_0  \right\}.
	$$ 
	The mapping $F$ being lower semicontinuous, the set 
	$V_0: = \{ x\in X: F(x)\cap B(y_0, R_0)\neq \emptyset\}$ is an open neighbourhood of 
	the point $x_0$.  The space $X$ is locally connected, so there exists a 
	connected open neighbourhood $V_0'\subset V_0$ of the point $x_0$. Consider a point 
	$x_1\in V_0'$ such that $V_0'\subset B(x_0, C\rho(x_0,x_1))$. Since $x_1\in V_0$, 
	the set $F(x_1)\cap B(y_0, R_0)$ is not empty, so there exists a point $y_1$ in this 
	set. Then Proposition~3.7. gives the inequalities 
	$\rho(x_0, x_1)\leq \zeta(\sigma(y_0, y_1))\leq \zeta(R_0)\leq \zeta(t_0) < \delta/(4C)$.
	
	Thus the points $x_1$ and $y_1\in F(x_1)$ satisfy the conditions (3.6.1) in Proposition~3.6.:
	$$
	\rho(x_0, x_1) < \delta/(4C)\ ;\ \sigma(y_0, y_1)< R_0 \leq \min\left\{ 
	\frac{\Delta}{4}, \frac{\delta_0}{8}  \right\}\omega_0\,.
	$$
	The connected set $\gamma = V_0'$ contains the points $x_0, x_1$ and is contained in 
	the open ball $B(x_0, C\cdot \rho(x_0, x_1))$. Then by Proposition~3.6. the nonempty 
	set $F(V_0')\cap B(y_0, \sigma(y_0, y_1)\cdot 2(1+2C_0)/\omega_0)$ has a component 
	$W$ such that $y_0, y_1\in W$ and $W_0\subset W$ where 
	$W_0: = F(V_0')\cap B(y_0, \sigma(y_0, y_1)\cdot 2/\omega_0)$. Besides 
	$W\subset B(y_0, \sigma(y_0, y_1) \cdot 2(1+2C_0)/\omega_0) \subset B(y_0, R_0)$.
	
	The set $V_0'$ is open, the set $X\setminus V_0'$ is compact, all thick points of the 
	mapping $F$ are compact. Then by Proposition~2.3(i) the set $F(X)\setminus V_0'$ is 
	compact. Hence the set $F(V_0') = F(X)\setminus F(X\setminus V_0')$ is open. Then 
	$W_0$ is open and the connected set $W$ is a neighbourhood of the point $y_0$ which 
	is contained in the given open neighbourhood $B(y_0, R)$ of the point $y_0$. 
	
	It means that $F(X)$ is locally connected at every point $y_0\in F(X)$. 
	
	The proposition (i) in Theorem~3.1. is proved.\medskip
	
	\noindent{\bf 3.9. Proof of Theorem~3.1(ii).} For the given normalizator $(\delta, \Delta)$ and 
	the constant $d= {\rm diam}_{\rho}(X)$ let the function $\zeta(t)$ be defined by (3.7.2).
	Since $\zeta(t)\searrow 0$ as $t\to 0$, there exists $t_0>0$ such that 
	$\zeta(t) < \min\{ \delta', \delta/(4C)  \}$ for all $t\leq t_0$. Let 
	$\omega_C: = \omega(1/(3+4C))$ and 
	$\delta^*: = \min \{ (\Delta/4)\omega_C, (\delta_0/8)\omega_C, t_0\}$.
	
	Let points $y_0\in F(x_0)$ and $y_1 \in F(x_1)$ be such that $\sigma(y_0, y_1)\leq 
	\delta^*$. 
	
	It follows from $\delta^*\leq t_0$ that $\rho(x_0, x_1) \leq \zeta(\delta^*)\leq 
	\zeta(t_0) \leq \delta'$. The bounded turning property $(C, \delta')$-BT of the space $X$ 
	gives the existence of a continuum $\gamma\subset X$ such that $x_0, x_1\in \gamma$ 
	and ${\rm diam}_{\rho}(\gamma) \leq C\cdot \rho(x_0, x_1)$. It means that 
	$\gamma \subset \bar{B}(x_0, C\cdot \rho(x_0, x_1))$. Since $\rho(x_0, x_1)< \delta/(4C)$ 
	and $\sigma(y_0, y_1)\leq \min\{ \Delta/4, \delta_0/8\}\cdot \omega_C$, we may 
	apply Proposition~3.6. to $\gamma$ and obtain a connected set $W\subset F(\gamma)$ 
	such that $y_0, y_1\in W$ and 
	$W\subset \bar{B}(y_0, \sigma(y_0, y_1)\cdot 2(1+2C_0)/\omega_C))$. The set $F(\gamma)$ 
	being compact (by Proposition~2.3.), the set $\bar{W}\subset F(\gamma)$ is a 
	continuum such that $y_0, y_1\in \bar{W}$ and ${\rm diam}_{\sigma}(\bar{W}) \leq 
	\sigma(y_0, y_1)\cdot C^*$ where $C^*: = 4(1+2C_0)/\omega_C$. 
	
	Due to the arbitrary choice of points $y_0, y_1\in F(X)$ with $\sigma(y_0, y_1)\leq 
	\delta^*$ the set $F(X)$ has bounded turning property $(C^*, \delta^*)$-BT. 
	
	Theorem~3.1. is proved.
	
	%\centerline{\bf \S 3. Main result.}
	\section{Main result}
	
	In this section the following result will be posed.\medskip
	
	\noindent{\bf 4.1. Theorem.}\label{th3.1} {\it Let $D$ be a domain in the extended complex plane $\bar{\bf C}$ and 
		$F: D \to 2^{\bar{\bf C}}$ be a hyperinjective mapping with $\omega$-BAD property. 
		If all thick points of $F$ are continuums with bonded turning property $(C, \delta)$-BT, 
		then $F$ is a single-valued $\eta$-quasim\"{o}bius mapping with the distortion function 
		$\eta$ depending only on the control function $\omega$.}  \medskip
	
	The proof of the theorem is divided into several consecutive steps. 
	
	Recall that a point $p$ in a connected space $X$ is called 
	{\it dividing point} if  the set $X\setminus \{p\}$ is not connected (see \cite{11K1966}, \S 46.VIII).\medskip
	
	\noindent{\bf 4.2. Proposition.} {\it Let $X, Y$ be metric spaces and $F: X\to 2^Y$ be a hyperinjective 
		continuous mapping with all thick points being continuums. If $X$ has no dividing points 
		then $F(X)$ has no dividing points as well.}\medskip
	
	The proof will be given in paragraph~5.4.\medskip
	
	\noindent{\bf 4.3. Proposition.}  {\it Under the conditions in Theorem~4.1. let the mapping $F$ be 
		continuous. If a circle $\gamma \subset D$ bounds an open ball $B_0\subset D$, 
		then $F(\gamma)$ is a Jordan curve of bounded turning BT.} \medskip
	
	\noindent{\it Proof.} By Proposition~2.3. $F(\gamma)$ is a continuum. By Theorem~3.1(i) $F(\gamma)$ 
	is locally connected, and by Proposition~4.2. it has no dividing points.
	
	The set $B_1: = D\setminus \bar{B}_0$ is a domain (see \cite{13N1939}, Theorem 10-2). By Proposition~2.3. 
	the sets $F(B_0)$, $F(B_1)$ are connected and therefore each of them is contained in a 
	component of $\bar{\bf C}\setminus F(\gamma)$. 
	
	Let us assume that a point $y_0\in F(\gamma)$  
	is not a limit point for $F(B_j)$ where 
	$j\in \{0,1\}$. Then $y_0 \in F(x_0)$ for $x_0\in \gamma$ and 
	there exists an open ball $U$ centered at $y_0$ such that 
	$U\cap F(B_j) = \emptyset$. Since $F$ is continuous, it follows that the set 
	$V: = \{ x\in D: F(x)\cap U \neq \emptyset \}$ is an open neighbourhood of $x_0$. Hence 
	there exists a point $x_1\in V\cap B_j$ such that $F(x_1)\cap U \neq \emptyset$ and 
	$F(x_1)\cap F(B_j) \neq \emptyset$. It contradicts to $U\cap F(B_j) = \emptyset$. 
	
	Thus every point $y_0\in F(\gamma)$ is a limit point both for $F(B_0)$ and for $F(B_1)$.
	
	Recall (see \cite{11K1966}, \S 52.IV) that $\theta$-{\it curve} is a join 
	$\tau_1\cup \tau_2\cup \tau_3$ of 
	three Jordan arcs with common ends $a,b$ such that $\tau_i\cap \tau_j = \{a,b\}$ for all 
	$i, j \in \{ 1,2,3\}$, $i\neq j$. Let us show that $F(\gamma)$ contains no $\theta$-curves.
	
	If a $\theta$-curve $\tau = \tau_1\cup \tau_2\cup \tau_3$ contains in $F(\gamma)$, then 
	it divides $\bar{\bf C}$ into three non-intersecting domains $G_{12}, G_{13}, G_{23}$ 
	with boundaries $\partial G_{ij} = \tau_i\cup \tau_j$ (see \cite{11K1966}, \S 61.II, Theorem~2). 
	Then for some indices $j,k \in \{i,j,k\} = \{1,2,3\}$ the set $F(B_0)\cup F(B_1)$ 
	intersects with $G_{jk}$. Since every point of the arc $\tau_i$ is a limit point for 
	each of connected sets $F(B_0)$ and $F(B_1)$, we obtain the inclusion 
	$F(B_0)\cup F(B_1) \subset G_{ij}\cup G_{ik}$. But $G_{ij}\cup G_{ik}$ does not 
	intersect $G_{jk}$, and it contradicts to $F(B_0)\cup F(B_1) \cap G_{jk} \neq \emptyset$. 
	Thus there is no $\theta$-curves in $F(\gamma)$.
	
	It is known (see \cite{11K1966}, \S 52.IV, Theorem~1), that if a nondegenerated locally 
	connected continuum in $\bar{\bf C}$ has no dividing points and does not 
	contain $\theta$ curves, then it is Jordan curve. It means that $F(\gamma)$ is 
	a Jordan curve, and it has bounded turning BT by Theorem~3.1(ii).
	
	Proposition~4.3. is proved.\medskip   
	
	\noindent{\bf 4.4. Proposition.} {\it If under the assumptions in Theorem~4.1. the mapping 
		$F(\gamma)$ is continuous, then $F(x)$ is a singleton for every $x\in D$.}\medskip
	
	\noindent{\it Proof.} For an arbitrary given point $x_0\in D$ there exists an open disk 
	$B\subset D$ with the boundary $\partial B=\gamma \subset D$.  
	According to Proposition 4.3. the set $F(\gamma)\subset \bar{\bf C}$ 
	is a Jordan curve. Then there exists a homeomorphism $g: \bar{\bf C} \to \bar{\bf C}$ such that $g(F(\gamma))=\gamma$ (see, e.g., \cite{11K1966}, \S 61.V, Corollary~2). 
	The multivalued mapping $G=g \circ F: \gamma \to 2^{\gamma}$ is continuous and all it's 
	thick points are compact. According to Lemma~5.5. there exists an integer $N>0$ 
	such that $\# G(z) =N$ for each of thick points $G(z)$ with $z\in \gamma$. Since every thick 
	point $G(z)$ is a continuum, $N=1$. It means that all thick points $G(z)$ and hence 
	all thick points $F(y)$, $y\in \gamma$, 
	are singletons. In particular, $\# F(x_0) =1$. Thus the mapping 
	$F: D\to 2^{\bar{\bf C}}$ is actually a single-valued mapping $F: D\to \bar{\bf C}$ 
	since $y_0$ was an arbitrary point in $D$. 
	
	Proposition~4.4. is proved.\medskip
	
	\noindent{\bf 4.5. Proof of Theorem~4.1}. There is no 
	assumption on $F$ to be continuous in Theorem~4.1. However, the mapping 
	$F^0: D \to 2^{\bar{\bf C}}$ in Lemma~5.6. is continuous and possesses 
	$\omega$-BAD property with the same control function $\omega$. Since the topological 
	limit of continuums with bounded turning $(C, \delta)$-BT is either a single point or 
	a continuum with bounded turning $(C, \delta)$-BT, the mapping $F^0$ satisfies the 
	conditions in Proposition~4.4. Thus $F^0$ is a single-valued mapping 
	$F^0: D \to \bar{\bf C}$ with $\omega$-BAD property. Then the mapping $F^0$ is $\eta$-quasim\"{o}bius mapping where the control function $\eta$ 
	depends only on $\omega$ (see \cite{5A2018}, Theorem~6.2). In particular, $F^0$ is $K$-quasiconformal mapping whith  the coefficient of quasiconformality $K$ depending only on $\omega$.
	
	For every point $z_0\in D$ the continuum $F(z_0)$ contains the point $w_0= F^0(z_0)$. 
	If there exists a point $w\in F(z_0)\setminus \{ w_0\}$, then $w=F^0(z)$ where $z\neq z_0$. 
	This contradicts the hyperinjectivity of $F$. It means that $F(z_0)= \{ F^0(z_0)\}$ 
	and that $F(z) \equiv F^0(z)$ at every point $z\in D$. 
	
	Hence $F$ is a single-valued mapping and it is $\eta$-quasim\"{o}bius with $\eta$ 
	depending only on $\omega$. 
	
	The proof of Theorem~4.1. is complete. 
	
	
	%\centerline{\bf \S 4. Appendix}
	\section{Appendix}
	
	\noindent{\bf 5.1.} {\it Proof for Lemma~2.2}. For connected spaces it was noticed in \cite{10T1996}, \S 2, p.562. 
	Let the metric space $(X, \rho)$ has the bounded turning property $(c, \delta)$-BT and 
	$U\subset X$ be an open neighbourhood of an arbitrary given point $p\in X$. 
	Consider a closed ball $\bar{B}(p,r)$ 
	with radius $r< \delta$. Let $U_0$ be the component of $U$ such that $p\in U_0$. 
	Because of $(c, \delta)$-BT property every point $x\in B(p, r/c)$ may be connected to $p$ by 
	a continuum $\tau \subset \bar{B}(p,r)$. Since $U_0$ is the maximal connected set in 
	$U$ that contains the point $p$, we have $x \in \tau \subset U_0$. Thus 
	$B(p, r/c)\subset U_0$, and $p$ is the inner point in $U_0$. It means that $U_0 \subset U$ is  
	the desired connected neibourhood of $p$. \medskip    
	
	\noindent{\bf 5.2.} (see \cite{11K1966}, \S 17.III). In topological spaces $X$ and $Y$, 
	the miltivalued mapping $F: X\to 2^Y$ with closed thick points is called to 
	be {\it below semicontinuous} if the set $\{ x\in X: F(x)\cap U\neq \emptyset\}$ is open 
	for an arbitrary open set $U\subset Y$. This mapping is said to be {\it upper semicontinuous} 
	if the set $\{ x\in X: F(x) \subset U\}$ is open for an arbitrary open set $U\subset Y$. 
	The mapping $F: X\to 2^Y$ is said to be {\it continuous} if it is both upper and lower 
	semicontinuous. 
	
	If $Y$ is a compact metric space, then the exponential topology (i.e. Vietoris topology) in 
	$2^Y$ (see \cite{11K1966}, \S 17.I) coinsides with the metric topology defined by the Hausdorff 
	distance (see \cite{11K1966}, \S 21.7 and \S 42.II) as well as with topology generated in 
	$2^Y$ by the operation of topological limit ${\rm Lim}$ (see \cite{11K1966}, \S 42.II, note~1).\medskip  
	
	
	\noindent{\bf 5.3.} {\it Proof for Proposition~2.3(i)}. Let an open covering $\mathcal{U}$ of the set $F(K)$ be given. 
	Then for each point $x_0\in K$ the thick point $F(x_0)$ being compact has a finite 
	subcovering $\mathcal{U}(x_0)$. The set $U(x_0) = \cup \{ U\in \mathcal{U}(x_0)\}$ is an open 
	neibourhood of thick point $F(x_0)$. Then the upper semicontinuous property of the mapping $F$ 
	implies the existence of an open neibourhood $V(x_0)$ of $x_0$ such that 
	$F(V(x_0)) \subset U(x_0)$. Thus we obtain the open covering $\{ V(x_0): x_0\in K\}$ of the 
	compact $K$, and then there exists a finite subcovering $\{V(x_1),..., V(x_k)\}$ of $K$.
	Then $\{ \mathcal{U}(x_1), ..., \mathcal{U}(x_k)\} \subset \mathcal{U}$ is a desired finite 
	subcovering of $F(K)$. Thus the compactness of $F(K)$ is proved.
	
	Let the space $Y$ be Hausdorff, and let us consider the left inverse mapping $f: F(K)\to K$. 
	An arbitrary  
	given close subset $E\subset K$ is compact (see \cite{11K1966}, \S 41.II, Theorem~2]). 
	Then $F(E)$ is compact in $Y$ by above proof. 
	Since every 
	compact set in a Hausdorff space is closed (see \cite{11K1966}, \S 41.II, Theorem~1]), 
	then the set $f^{-1}(E) = F(E)$ is closed. 
	It means (see \cite{11K1966}, \S 13.IV) that the mapping $f$ is continuous.
	
	\noindent{\it Proof for Proposition~2.3(ii)}. The restriction $F|\gamma$ remains to be 
	hyperinjective and continuous. Suppose on contrary that $F(\gamma)$ is not connected. Then 
	$F(\gamma) = U_1\sqcup U_2$ for nonempty 
	open subsets $U_1,U_2\subset F(\gamma)$ with $U_1\cap U_2 = \emptyset$. Since $F|\gamma$ 
	is continuous, we get three pairvise non-intersecting open sets 
	$V_1= \{ x\in \gamma: F(x) \subset U_1\}$, $V_2 = \{x\in \gamma: F(x)\subset U_2\}$, and 
	$V_{12} = \{ x\in \gamma: F(x)\cap U_1 \neq \emptyset \neq F(x)\cap U_2\}$. Since the 
	set $\gamma$ is connected and $\gamma = V_1\sqcup V_2\sqcup V_{12}$, just only one of the sets 
	$V_1, V_2, V_{12}$ is not empty. There exists at least one thick point $F(x_0)$ which is 
	connected and therefore it cannot intersect both $U_1$ and $U_2$. It means that 
	$F(x_0)\subset U_j$ for some $j\in \{1,2\}$. Clearly, we may assume that $F(x_0)\subset U_1$. 
	Then $x_0 \in V_1 \neq \emptyset$ and $V_2 = V_{12} = \emptyset$. Thus $\gamma = V_1$ and 
	$F(\gamma) \cap U_2 = \emptyset$. It means that $U_2 = \emptyset$, and it contradicts our 
	assumption. It follows that $F(\gamma)$ is a connected set.\medskip      
	
	\noindent{\bf 5.4.} {\it Proof for Proposition~4.2.} The set $F(X)$ is compact (by Proposition~2.3.), 
	so we can assume that $Y= F(X)$. Let us assume (on the contrary) that  
	a point $y_0 \in F(x_0)$  is dividing point in $Y$. 
	
	The point $x_0$ is not dividing in $X$, so the set $X\setminus \{x_0\}$ is connected. 
	Then (by Proposition~2.3.) the set $F(X\setminus \{x_0\}) = Y\setminus F(x_0)$ is also 
	connected and is contained in a component $U_0$ of the set $Y\setminus \{y_0\}$. 
	Being a component the set $U_0$ is closed with respect to $Y\setminus \{y_0\}$ 
	(see \cite{11K1966}, \S 46.III, Theorem~1). Then the set $V_0: = (Y\setminus \{y_0\})\setminus U_0$ is 
	open with respect to $Y\setminus \{y_0\}$ 
	and non-empty according to our assumption. Since $Y\setminus \{y_0\}$ is open in $Y$, then 
	$V_0$ is open in $Y$ as well. Besides, $V_0 = (Y\setminus \{y_0\})\setminus U_0 \subset 
	(Y\setminus \{y_0\})\setminus (Y\setminus F(x_0)) = F(x_0)\setminus \{y_0\}$. 
	Since the mapping $F$ is lower semicontinuous, then the set 
	$\{x\in X: F(x)\cap V_0 \neq \emptyset\} = \{x_0\}$ must be open in $X$. But the connected 
	set $X$ does not have isolated points. This contradiction completes the proof of Proposition~4.2.\medskip 
	
	\noindent{\bf 5.5. Lemma} (see \cite{9A2021}, Lemma 1.3). {\it Let $X = Y = \{ z\in \bar{\bf C}: |z|=1\}$ and
		$F: X\to 2^Y$ be a hyperinjective mapping from a 
		circle $X\subset \bar{\bf C}$ into a circle $Y\subset \bar{\bf C}$. If $F$ is continuous and  
		all thick points $F(x)$ are compact, then there exist an integer $N>0$ and a homeomorphism 
		$\varphi: Y \to Y$ such that the left inverse mapping 
		$f: Y\to X$ is $f(y) = (\varphi(y))^N$.}\medskip 
	
	\noindent{\bf 5.6. Lemma} (see \cite{7A2019}, Theorems~3.2 and 3.5). {\it Let $X$, $Y$ be Ptolemaic M\"{o}bius spaces. 
		Assume that a hyperinjective  mapping $F: X \to 2^Y$ possesses 
		the $\omega$-BAD property and all 
		it's thick points $F(x)$ are closed in Y.  Then at every limit point $x_0\in X$ there 
		exists a topological limit 
		$$
		L(x_0): = \mbox{\rm Lim }_{x\to x_0;\ x\neq x_0} F(x) \subset F(x_0).
		$$ 
		Moreover, the mapping 
		\begin{equation*}\label{pp}
		F^{0}(x): =
		\begin{cases}
		& F(x), \mbox{ if } x \mbox{ is an isolated point in } X; \\[5pt]
		& L(x), \mbox{ if } x \mbox{ is a limit point}
		\end{cases}
		\end{equation*}
		%$$
		%F^{0}(x): = \{ F(x) \mbox{ if } x \mbox{ is an isolated point in } X;\ L(x) \mbox{ if } x 
		%\mbox{ is 
		%	a limit point} \}
		%$$
		is continuous and possess the $\omega$-BAD property witn the same 
		control function $\omega$.}
	
	
	
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