Let $X$$X$XX$X$ be a nonvold set and $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ a multivalued mapping. Let

denote the fixed point set, respectively the strict fixed point set of T.

For the multivalued mapping $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ we consider the graph

which may be regarded as a relation on $X$$X$XX$X$. The properties of the relations are exposed, for example, in [1].

It is obvious that any function $f:X\to X$$f:X\to X$f:X rarr Xf: X \rightarrow X$f:X\to X$ satisfies the equality $f\left(F_f\right)=F_f$$f\left(F_f\right)=F_f$f(F_(f))=F_(f)f\left(F_{f}\right)=F_{f}$f\left(F_f\right)=F_f$. In the paper [2], I.A. Rus shows that there are multivalued mappings which have not this property, but any multivalued mapping having only strict fixed points ( ${\mathrm{F}}_T=(SF{)}_T$${\mathrm{F}}_T=(SF{)}_T$F_(T)=(SF)_(T)\mathrm{F}_{T}=(S F)_{T}${\mathrm{F}}_T=(SF{)}_T$ ) verifies the condition $T\left(F_T\right)=F_T$$T\left(F_T\right)=F_T$T(F_(T))=F_(T)T\left(F_{T}\right)=F_{T}$T\left(F_T\right)=F_T$ ( Lemma 4.1). The Problem 4.1[2] asks what are the conditions under which the set $F_T$$F_T$F_(T)F_{T}$F_T$ is fixed for a multivalued mapping T.

In the sequel we give sufficient conditions for the equality $T\left(F_T\right)=F_T$$T\left(F_T\right)=F_T$T(F_(T))=F_(T)T\left(F_{T}\right)=F_{T}$T\left(F_T\right)=F_T$ holds and then we study some properties of a multivalued mapping $\tilde{T}$$\tilde{T}$tilde(T)\tilde{T}$\tilde{T}$ induced by the given $T$$T$TT$T$.

For the set $X\ne \varnothing$$X\ne \varnothing$X!=O/X \neq \varnothing$X\ne \varnothing$ we denote

${\mathrm{\Delta }}_X=G\left(I_X\right)=\{(x,x)\in X\times X:x\in X\}$${\mathrm{\Delta }}_X=G\left(I_X\right)=\{(x,x)\in X\times X:x\in X\}$Delta_(X)=G(I_(X))={(x,x)in X xx X:x in X}\Delta_{X}=G\left(I_{X}\right)=\{(x, x) \in X \times X: x \in X\}${\mathrm{\Delta }}_X=G\left(I_X\right)=\{(x,x)\in X\times X:x\in X\}$,
and for the multivalued mapping $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$
$\mathrm{dom}T=\{x\in X:T(x)\ne \varnothing \}$$\mathrm{dom}T=\{x\in X:T(x)\ne \varnothing \}$dom T={x in X:T(x)!=O/}\operatorname{dom} T=\{x \in X: T(x) \neq \varnothing\}$\mathrm{dom}T=\{x\in X:T(x)\ne \varnothing \}$
Im $T=\{y\in X$$T=\{y\in X$T={y in XT=\{y \in X$T=\{y\in X$ : there is $x\in X$$x\in X$x in Xx \in X$x\in X$ such that $y\in T(x)\}$$y\in T(x)\}$y in T(x)}y \in T(x)\}$y\in T(x)\}$
$T^{-1}:X\to X,T^{-1}(y)=\{x\in X:y\in T(x)\}$$T^{-1}:X\to X,T^{-1}(y)=\{x\in X:y\in T(x)\}$T^(-1):X rarr X,T^(-1)(y)={x in X:y in T(x)}T^{-1}: X \rightarrow X, T^{-1}(y)=\{x \in X: y \in T(x)\}$T^{-1}:X\to X,T^{-1}(y)=\{x\in X:y\in T(x)\}$.
For a multivalued mapping $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ the following theorem holds.

THHOREM 1. Let the below conditions be given
(a) $T\left(F_T\right)\subseteq F_T$$T\left(F_T\right)\subseteq F_T$T(F_(T))subeF_(T)T\left(F_{T}\right) \subseteq F_{T}$T\left(F_T\right)\subseteq F_T$
(a') $T\left(F_T\right)=F_T$$T\left(F_T\right)=F_T$T(F_(T))=F_(T)T\left(F_{T}\right)=F_{T}$T\left(F_T\right)=F_T$
(a") $T\left(\mathrm{dom}(I_X\cap T)\right)\subseteq \mathrm{dom}(I_X\cap T)$$T\left(\mathrm{dom}\left(I_X\cap T\right)\right)\subseteq \mathrm{dom}\left(I_X\cap T\right)$T(dom(I_(X)nn T))sube dom(I_(X)nn T)T\left(\operatorname{dom}\left(I_{X} \cap T\right)\right) \subseteq \operatorname{dom}\left(I_{X} \cap T\right)$T\left(\mathrm{dom}(I_X\cap T)\right)\subseteq \mathrm{dom}(I_X\cap T)$
(b) for any $x\in {\mathbb{F}}_{T}T(x)\subseteq \bigcap _{y\in T(x)}T(y)$$x\in {\mathbb{F}}_{T}T(x)\subseteq \bigcap _{y\in T(x)} T(y)$x inF_(T)T(x)subennn_(y in T(x))T(y)x \in \mathbb{F}_{T} T(x) \subseteq \bigcap_{y \in T(x)} T(y)$x\in {\mathbb{F}}_{T}T(x)\subseteq \bigcap _{y\in T(x)}T(y)$
(b') for any $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ and $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$, it follows $T(x)\subseteq T(z)$$T(x)\subseteq T(z)$T(x)sube T(z)T(x) \subseteq T(z)$T(x)\subseteq T(z)$
(c) $G(T)$$G(T)$G(T)G(T)$G(T)$ is a symmetrical and transitive relation
(d) $G(T)$$G(T)$G(T)G(T)$G(T)$ is a reflexive relation
(e) $F_T=(SF{)}_T$$F_T=(SF{)}_T$F_(T)=(SF)_(T)F_{T}=(S F)_{T}$F_T=(SF{)}_T$

The following implications are true
$(\mathrm{a})\Rightarrow (\mathrm{a})\iff \left({\mathrm{a}}^{\prime }\right)\iff \left({\mathrm{a}}^{\prime \prime }\right)$$(\mathrm{a})\Rightarrow (\mathrm{a})\iff \left({\mathrm{a}}^{\prime }\right)\iff \left({\mathrm{a}}^{\prime \prime }\right)$(a)=>(a)<=>(a^('))<=>(a^(''))(\mathrm{a}) \Rightarrow(\mathrm{a}) \Leftrightarrow\left(\mathrm{a}^{\prime}\right) \Leftrightarrow\left(\mathrm{a}^{\prime \prime}\right)$(\mathrm{a})\Rightarrow (\mathrm{a})\iff \left({\mathrm{a}}^{\prime }\right)\iff \left({\mathrm{a}}^{\prime \prime }\right)$
$\hat{n}$$\widehat{n}$hat(n)\hat{n}$\hat{n}$
(e) $\Rightarrow (b)⟺\left(b^{\prime }\right)$$\Rightarrow (b)⟺\left(b^{\prime }\right)$=>(b)Longleftrightarrow(b^('))\Rightarrow(b) \Longleftrightarrow\left(b^{\prime}\right)$\Rightarrow (b)⟺\left(b^{\prime }\right)$
$\hat{n}$$\widehat{n}$hat(n)\hat{n}$\hat{n}$
(c)

Proof. (a) $\Rightarrow \left(a^{\prime }\right)$$\Rightarrow \left(a^{\prime }\right)$=>(a^('))\Rightarrow\left(a^{\prime}\right)$\Rightarrow \left(a^{\prime }\right)$. It is obvious that $\left(a^{\prime }\right)\Rightarrow (a)$$\left(a^{\prime }\right)\Rightarrow (a)$(a^('))=>(a)\left(a^{\prime}\right) \Rightarrow(a)$\left(a^{\prime }\right)\Rightarrow (a)$; if $(a)$$(a)$(a)(a)$(a)$ is true, we obtain ( $a^{\prime }$$a^{\prime }$a^(')a^{\prime}$a^{\prime }$ ) because from $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ we deduce $x\in T(x)$$x\in T(x)$x in T(x)x \in T(x)$x\in T(x)$, so $x\in \in T\left(F_T\right)$$x\in \in T\left(F_T\right)$x∈∈T(F_(T))x \in \in T\left(F_{T}\right)$x\in \in T\left(F_T\right)$.
$(\mathrm{a})\iff \left({\mathrm{a}}^{\prime \prime }\right)\cdot \mathrm{dom}(I_{\mathrm{X}}\cap T)=\{x\in X:I_{\mathrm{X}}(x)\cap T(x)\ne \mathrm{\varnothing }\}==\{x\in X:x\in T(x)\}=F_T$$(\mathrm{a})\iff \left({\mathrm{a}}^{\prime \prime }\right)\cdot \mathrm{dom}\left(I_{\mathrm{X}}\cap T\right)=\left\{x\in X:I_{\mathrm{X}}(x)\cap T(x)\ne \mathrm{\varnothing }\right\}==\{x\in X:x\in T(x)\}=F_T$(a)<=>(a^(''))*dom(I_(X)nn T)={x in X:I_(X)(x)nn T(x)!=O/}=={x in X:x in T(x)}=F_(T)(\mathrm{a}) \Leftrightarrow\left(\mathrm{a}^{\prime \prime}\right) \cdot \operatorname{dom}\left(I_{\mathrm{X}} \cap T\right)=\left\{x \in X: I_{\mathrm{X}}(x) \cap T(x) \neq \emptyset\right\}= =\{x \in X: x \in T(x)\}=F_{T}$(\mathrm{a})\iff \left({\mathrm{a}}^{\prime \prime }\right)\cdot \mathrm{dom}(I_{\mathrm{X}}\cap T)=\{x\in X:I_{\mathrm{X}}(x)\cap T(x)\ne \mathrm{\varnothing }\}==\{x\in X:x\in T(x)\}=F_T$ and the equivalence holds.
(b) $\Rightarrow$$\Rightarrow$=>\Rightarrow$\Rightarrow$ (b'). Let $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ and $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$. We prove that $T(x)\subseteq T(z)$$T(x)\subseteq T(z)$T(x)sube T(z)T(x) \subseteq T(z)$T(x)\subseteq T(z)$. For any $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$, we obtain $y\in T(x)\subseteq \bigcap _{y\in T(x)}T(y)\subseteq {\mathrm{TV}}_3(\notin \mathrm{so}T(x)$$y\in T(x)\subseteq \bigcap _{y\in T(x)} T(y)\subseteq {\mathrm{TV}}_3(\notin \mathrm{so}T(x)$y in T(x)subennn_(y in T(x))T(y)subeTV_(3)(!in so T(x)y \in T(x) \subseteq \bigcap_{y \in T(x)} T(y) \subseteq \operatorname{TV}_{3}(\notin \operatorname{so} T(x)$y\in T(x)\subseteq \bigcap _{y\in T(x)}T(y)\subseteq {\mathrm{TV}}_3(\notin \mathrm{so}T(x)$
$\subseteq T(z)$$\subseteq T(z)$sube T(z)\subseteq T(z)$\subseteq T(z)$.
(b') $\Rightarrow$$\Rightarrow$=>\Rightarrow$\Rightarrow$ (B). Let $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ and $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$; by (b') we have $T(x)\subseteq T(\mathbf{B})$$T(x)\subseteq T(\mathbf{B})$T(x)sube T(B)T(x) \subseteq T(\mathbf{B})$T(x)\subseteq T(\mathbf{B})$, hence $T(x)\subseteq \bigcap _{z\in T(x)}T(z)$$T(x)\subseteq \bigcap _{z\in T(x)} T(z)$T(x)subennn_(z in T(x))T(z)T(x) \subseteq \bigcap_{z \in T(x)} T(z)$T(x)\subseteq \bigcap _{z\in T(x)}T(z)$.
(b) $\Rightarrow$$\Rightarrow$=>\Rightarrow$\Rightarrow$ (a). Let $y\in T\left(F_T\right)$$y\in T\left(F_T\right)$y in T(F_(T))y \in T\left(F_{T}\right)$y\in T\left(F_T\right)$, i.e. there exists $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ such that $y\in \in T(x)$$y\in \in T(x)$y∈∈T(x)y \in \in T(x)$y\in \in T(x)$. The condition (b) implies $T(x)\subseteq \bigcap _{z\in T(x)}T(z)\subseteq T(y)$$T(x)\subseteq \bigcap _{z\in T(x)} T(z)\subseteq T(y)$T(x)subennn_(z in T(x))T(z)sube T(y)T(x) \subseteq \bigcap_{z \in T(x)} T(z) \subseteq T(y)$T(x)\subseteq \bigcap _{z\in T(x)}T(z)\subseteq T(y)$; but $y\in \in T(x)$$y\in \in T(x)$y∈∈T(x)y \in \in T(x)$y\in \in T(x)$ and it follows $y\in T(y)$$y\in T(y)$y in T(y)y \in T(y)$y\in T(y)$, hence $y\in F_T$$y\in F_T$y inF_(T)y \in F_{T}$y\in F_T$.
(c) $\Rightarrow$$\Rightarrow$=>\Rightarrow$\Rightarrow$ (B). Let $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$ and $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$. We prove that for any $y\in X$$y\in X$y in Xy \in X$y\in X$ such that $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$ we have $y\in T(z)$$y\in T(z)$y in T(z)y \in T(z)$y\in T(z)$. The symmetry of $G(T)$$G(T)$G(T)G(T)$G(T)$ implies $x\in T(y)$$x\in T(y)$x in T(y)x \in T(y)$x\in T(y)$; but $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$ and from the transitivity of $G(T)$$G(T)$G(T)G(T)$G(T)$ we obtain $z\in \mathit{f}(y)$$z\in \mathit{f}(y)$z in f(y)z \in \boldsymbol{f}(y)$z\in \mathit{f}(y)$. Applying again the symmetry, we have $y\in T(z)$$y\in T(z)$y in T(z)y \in T(z)$y\in T(z)$ and $T(x)\subseteq \subseteq \bigcap _{z\in T(x)}T(z)$$T(x)\subseteq \subseteq \bigcap _{z\in T(x)} T(z)$T(x)⊆⊆nnn_(z in T(x))T(z)T(x) \subseteq \subseteq \bigcap_{z \in T(x)} T(z)$T(x)\subseteq \subseteq \bigcap _{z\in T(x)}T(z)$.
$(e)\Rightarrow (b)$$(e)\Rightarrow (b)$(e)=>(b)(e) \Rightarrow(b)$(e)\Rightarrow (b)$. Let $x\in F_T=(SF{)}_T$$x\in F_T=(SF{)}_T$x inF_(T)=(SF)_(T)x \in F_{T}=(S F)_{T}$x\in F_T=(SF{)}_T$ so $T(x)=\{x\}=\bigcap _{y\in T(x)}T(y)$$T(x)=\{x\}=\bigcap _{y\in T(x)} T(y)$T(x)={x}=nnn_(y in T(x))T(y)T(x)=\{x\}=\bigcap_{y \in T(x)} T(y)$T(x)=\{x\}=\bigcap _{y\in T(x)}T(y)$ and (b) holds.
$(d)\Rightarrow (a).G(T)$$(d)\Rightarrow (a).G(T)$(d)=>(a).G(T)(d) \Rightarrow(a) . G(T)$(d)\Rightarrow (a).G(T)$ being reflexive, we have ${\mathrm{\Delta }}_X\subseteq G(T)$${\mathrm{\Delta }}_X\subseteq G(T)$Delta_(X)sube G(T)\Delta_{X} \subseteq G(T)${\mathrm{\Delta }}_X\subseteq G(T)$ and it follows $F_T=X$$F_T=X$F_(T)=XF_{T}=X$F_T=X$; it is obvious that $T\left(F_T\right)\subseteq F_T$$T\left(F_T\right)\subseteq F_T$T(F_(T))subeF_(T)T\left(F_{T}\right) \subseteq F_{T}$T\left(F_T\right)\subseteq F_T$.

We mention now some connections between the classes of multivalued mappings satisfying the conditions in Theorem. 1.

THEOREM 2. If the graph of the multivalued mapping $T$$T$TT$T$ satisfying (b) is a reflexive relation, it is also symmetrical and transitive.

Proof. $G(T)$$G(T)$G(T)G(T)$G(T)$ being reflexive, the condition (b) is satisfied for any $x\in X$$x\in X$x in Xx \in X$x\in X$. Let $\overline{x}\in X$$\overline{x}\in X$bar(x)in X\bar{x} \in X$\overline{x}\in X$ be arbitrary and $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$. Using $(b),x\in T(x)\subseteq$$(b),x\in T(x)\subseteq$(b),x in T(x)sube(b), x \in T(x) \subseteq$(b),x\in T(x)\subseteq$ 里 $\subseteq T(y)$$\subseteq T(y)$sube T(y)\subseteq T(y)$\subseteq T(y)$, so $x\in T(y)$$x\in T(y)$x in T(y)x \in T(y)$x\in T(y)$ and the symmetry is proved.

For the transitivity, we consider $x\in X,y\in T(x)$$x\in X,y\in T(x)$x in X,y in T(x)x \in X, y \in T(x)$x\in X,y\in T(x)$ and $z\in T(y)$$z\in T(y)$z in T(y)z \in T(y)$z\in T(y)$. It follows by (b) that $x\in T(x)\subseteq T(y)\subseteq T(z)$$x\in T(x)\subseteq T(y)\subseteq T(z)$x in T(x)sube T(y)sube T(z)x \in T(x) \subseteq T(y) \subseteq T(z)$x\in T(x)\subseteq T(y)\subseteq T(z)$, so $x\in T(z)$$x\in T(z)$x in T(z)x \in T(z)$x\in T(z)$. Applying the symmetry of $G(T)$$G(T)$G(T)G(T)$G(T)$ we obtain $z\in T(x)$$z\in T(x)$z in T(x)z \in T(x)$z\in T(x)$ and the proof is over.

REMARK 1. There is only one mailhytrod mapping which has only strict fixed points and a reflexive graph, namely $I_X$$I_X$I_(X)I_{X}$I_X$, whose graph ${\mathrm{\Delta }}_X$${\mathrm{\Delta }}_X$Delta_(X)\Delta_{X}${\mathrm{\Delta }}_X$ is also symmetrical and transitive.

We are able now to present the relative position of the classes
of multivalued mappings involved in Theorem 1 using the diagram in Fig. 1; rectangles having the basas on the same line and the top vertexes marked with a letter stand for the classes denoted by that letter. All the regions marked by a number are nonvoid, as the following examples show.

EXAMPLIES.

By Remark 1, the only multivalued mapping satisfying these conditions is $I_X$$I_X$I_(X)I_{X}$I_X$.
6. I satisfies (c) and (d), but it does not satisfy (e).

REMARK 2. If we consider only multivalued mappings $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ such that $X=\mathrm{dom}T$$X=\mathrm{dom}T$X=dom TX=\operatorname{dom} T$X=\mathrm{dom}T$ (all the values of $T$$T$TT$T$ are nonvidid), the condition (c) implies (d.).

Indeed, for any $x\in X$$x\in X$x in Xx \in X$x\in X$ we have $T(x)\ne \varphi$$T(x)\ne \varphi$T(x)!=phiT(x) \neq \phi$T(x)\ne \varphi$ and we can choose $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$; from the symmetry we obtain $x\in T(y)$$x\in T(y)$x in T(y)x \in T(y)$x\in T(y)$ and the transitivity of $G(x)$$G(x)$G(x)G(x)$G(x)$ imp plies $x\in T(x)$$x\in T(x)$x in T(x)x \in T(x)$x\in T(x)$, i.e. $G(T)$$G(T)$G(T)G(T)$G(T)$ is reflexive.

If this is the case, the regions denoted by 7 and 8 in Fig. 1 an are void and the diagram looks like this

It follows that a multivalued mapping $\mathrm{\Psi }:X\to X$$\mathrm{\Psi }:X\to X$Psi:X rarr X\Psi: X \rightarrow X$\mathrm{\Psi }:X\to X$ with $X=\mathrm{dom}X$$X=\mathrm{dom}X$X=dom XX=\operatorname{dom} X$X=\mathrm{dom}X$ has a symmetrical and transitive graph ( satisfies the condition (c) ) if and only if $T$$T$TT$T$ satisfies the condition (b) and $G(T)$$G(T)$G(T)G(T)$G(T)$ is a reflexive relation.

The condition (b) leeds us to the definition of a multivalued mapping attached to $T$$T$TT$T$.

Let $X$$X$XX$X$ be a nonvoid set and $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ a multivalued mapping. We define $\tilde{T}:X\to X$$\tilde{T}:X\to X$tilde(T):X rarr X\tilde{T}: X \rightarrow X$\tilde{T}:X\to X$ given by

The the termes of $\tilde{T}$$\tilde{T}$tilde(T)\tilde{T}$\tilde{T}$, the conditione (6) becomes ${{T|}_{F_T}\subseteq T|}_{F_T}$${\left.{\left.T\right|}_{F_T}\subseteq T\right|}_{F_T}$T|_(F_(T))sube T|_(F_(T))\left.\left.T\right|_{F_{T}} \subseteq T\right|_{F_{T}}${{T|}_{F_T}\subseteq T|}_{F_T}$.

If we consider a new condition
(P) $T\subseteq \tilde{T}$$T\subseteq \tilde{T}$T sube tilde(T)T \subseteq \tilde{T}$T\subseteq \tilde{T}$
we obtain obviously that (f) implies (b).
If $T=G:X\to X$$T=G:X\to X$T=G:X rarr XT=G: X \rightarrow X$T=G:X\to X$ is a function, we have $g^N(x)=(g\circ g)(x)$$g^N(x)=(g\circ g)(x)$g^(N)(x)=(g@g)(x)g^{N}(x)=(g \circ g)(x)$g^N(x)=(g\circ g)(x)$ for any $x\in X$$x\in X$x in Xx \in X$x\in X$; the condition $(f)$$(f)$(f)(f)$(f)$ is equivalent to $g(x)=(g\circ g)(x)$$g(x)=(g\circ g)(x)$g(x)=(g@g)(x)g(x)=(g \circ g)(x)$g(x)=(g\circ g)(x)$ for an any $x\in X$$x\in X$x in Xx \in X$x\in X$, i.e. to $\mathrm{Im}g=F_g$$\mathrm{Im}g=F_g$Im g=F_(g)\operatorname{Im} g=F_{g}$\mathrm{Im}g=F_g$.

RMARK 3. If $T$$T$TT$T$ is a multivalued mapping, we have only $T\subseteq T\circ T$$T\subseteq T\circ T$T sube T@TT \subseteq T \circ T$T\subseteq T\circ T$ on dom $T$$T$TT$T$, the inclusion being generally strict, as the following example shows. Let $T:R\to R$$T:R\to R$T:R rarr RT: R \rightarrow R$T:R\to R$ be given by $T(x)=\{0,x\}$$T(x)=\{0,x\}$T(x)={0,x}T(x)=\{0, x\}$T(x)=\{0,x\}$; then $T\cdot T(x)==\{0,x\}\rightleftharpoons \tilde{T}(x)=\{0\}$$T\cdot T(x)==\{0,x\}\rightleftharpoons \tilde{T}(x)=\{0\}$T*T(x)=={0,x}⇌ tilde(T)(x)={0}T \cdot T(x)= =\{0, x\} \rightleftharpoons \tilde{T}(x)=\{0\}$T\cdot T(x)==\{0,x\}\rightleftharpoons \tilde{T}(x)=\{0\}$, for any $x\ne 0$$x\ne 0$x!=0x \neq 0$x\ne 0$.

THEOREM 3. If $T:X\to X$$T:X\to X$T:X rarr XT: X \rightarrow X$T:X\to X$ satisfles the condition $(f)$$(f)$(f)(f)$(f)$ we have Im $T=F_T{}^{\ast }$$T=F_T{}^{\ast }$T=F_(T)^(**)T=F_{T}{ }^{*}$T=F_T{}^{\ast }$

Procif. We have obviously $F_T\subseteq \mathrm{Im}T$$F_T\subseteq \mathrm{Im}T$F_(T)sube Im TF_{T} \subseteq \operatorname{Im} T$F_T\subseteq \mathrm{Im}T$. Let $y\in \mathrm{Im}T$$y\in \mathrm{Im}T$y in Im Ty \in \operatorname{Im} T$y\in \mathrm{Im}T$ and $x\in X$$x\in X$x in Xx \in X$x\in X$ such that $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$. Applying $(f)$$(f)$(f)(f)$(f)$, we get $y\in T(x)\subseteq \tilde{T}(x)\le T(y)$$y\in T(x)\subseteq \tilde{T}(x)\le T(y)$y in T(x)sube tilde(T)(x) <= T(y)y \in T(x) \subseteq \tilde{T}(x) \leq T(y)$y\in T(x)\subseteq \tilde{T}(x)\le T(y)$ and $y\in F_T$$y\in F_T$y inF_(T)y \in F_{T}$y\in F_T$. It follows that In $T\subseteq {\mathbb{F}}_T$$T\subseteq {\mathbb{F}}_T$T subeF_(T)T \subseteq \mathbb{F}_{T}$T\subseteq {\mathbb{F}}_T$, so $\mathrm{Im}T={\mathbb{F}}_T$$\mathrm{Im}T={\mathbb{F}}_T$Im T=F_(T)\operatorname{Im} T=\mathbb{F}_{T}$\mathrm{Im}T={\mathbb{F}}_T$ holds.

REMARK 4. The reverse implication is not true. For the multivalued mapping $T$$T$TT$T$ from Example 1 we have $\mathrm{Im}T=F_T=[-1,1]$$\mathrm{Im}T=F_T=[-1,1]$Im T=F_(T)=[-1,1]\operatorname{Im} T=F_{T}=[-1,1]$\mathrm{Im}T=F_T=[-1,1]$, but $T(x)=\{\begin{array}{l}\{0\},x\in [-1,1]\\ \{0,1\},x\in R\mathrm{\setminus }[-1,1]\end{array}$$T(x)=\left\{\begin{array}{l}\{0\},x\in [-1,1]\\ \{0,1\},x\in R\mathrm{\setminus }[-1,1]\end{array}\right.$T(x)={[{0}","x in[-1","1]],[{0","1}","x in R\\[-1","1]]:}T(x)=\left\{\begin{array}{l}\{0\}, x \in[-1,1] \\ \{0,1\}, x \in R \backslash[-1,1]\end{array}\right.$T(x)=\{\begin{array}{l}\{0\},x\in [-1,1]\\ \{0,1\},x\in R\mathrm{\setminus }[-1,1]\end{array}$, so $T⊈\tilde{T}$$T⊈\tilde{T}$T⊈ tilde(T)T \nsubseteq \tilde{T}$T⊈\tilde{T}$.

The next theorem gives a condition for a point $\mathbf{x}\in \mathbf{X}$$\mathbf{x}\in \mathbf{X}$xinX\mathbf{x} \in \mathbf{X}$\mathbf{x}\in \mathbf{X}$ be a fixed point for T.

THEOREM 4. The element $x\in X$$x\in X$x in Xx \in X$x\in X$ is a fixed point for $T$$T$TT$T$ if and only if $T(x)\subseteq T^{-1}(x)$$T(x)\subseteq T^{-1}(x)$T(x)subeT^(-1)(x)T(x) \subseteq T^{-1}(x)$T(x)\subseteq T^{-1}(x)$.

Proof. Let $x\in X$$x\in X$x in Xx \in X$x\in X$ be a fixed point for $\tilde{T}$$\tilde{T}$tilde(T)\tilde{T}$\tilde{T}$; if $x\in \mathrm{dom}T$$x\in \mathrm{dom}T$x in dom Tx \in \operatorname{dom} T$x\in \mathrm{dom}T$, we have $T(x)=\varphi \subseteq T^{-1}(x)$$T(x)=\varphi \subseteq T^{-1}(x)$T(x)=phi subeT^(-1)(x)T(x)=\phi \subseteq T^{-1}(x)$T(x)=\varphi \subseteq T^{-1}(x)$. In the case that $x\in$$x\in$x inx \in$x\in$ dom $T,T(x)$$T,T(x)$T,T(x)T, T(x)$T,T(x)$ is a nonvoid set; let $y\in T(x)$$y\in T(x)$y in T(x)y \in T(x)$y\in T(x)$. It follows that $x\in \tilde{T}(x)\subseteq T(y)$$x\in \tilde{T}(x)\subseteq T(y)$x in tilde(T)(x)sube T(y)x \in \tilde{T}(x) \subseteq T(y)$x\in \tilde{T}(x)\subseteq T(y)$, i.e. $y\in T^{-1}(x)$$y\in T^{-1}(x)$y inT^(-1)(x)y \in T^{-1}(x)$y\in T^{-1}(x)$ and we obtain again $T(x)\subseteq T^{-1}(x)$$T(x)\subseteq T^{-1}(x)$T(x)subeT^(-1)(x)T(x) \subseteq T^{-1}(x)$T(x)\subseteq T^{-1}(x)$.

Let now $x\in X$$x\in X$x in Xx \in X$x\in X$ be a point such that $T(x)\subseteq T^{-1}(x)$$T(x)\subseteq T^{-1}(x)$T(x)subeT^(-1)(x)T(x) \subseteq T^{-1}(x)$T(x)\subseteq T^{-1}(x)$. If $T(x)=\mathrm{\varnothing }$$T(x)=\mathrm{\varnothing }$T(x)=O/T(x)=\emptyset$T(x)=\mathrm{\varnothing }$, we have $T(x)=X$$T(x)=X$T(x)=XT(x)=X$T(x)=X$ and $x$$x$xx$x$ is obviously a fixed point for $\tilde{T}$$\tilde{T}$tilde(T)\tilde{T}$\tilde{T}$. But $y$$y$yy$y$ was
arbitrary in $T(x)$$T(x)$T(x)T(x)$T(x)$, so $x\in \bigcap _{y\in T(x)}T(y)=T(x)$$x\in \bigcap _{y\in T(x)} T(y)=T(x)$x innnn_(y in T(x))T(y)=T(x)x \in \underset{y \in T(x)}{\bigcap} T(y)=T(x)$x\in \bigcap _{y\in T(x)}T(y)=T(x)$, hence $x\in F_T$$x\in F_T$x inF_(T)x \in F_{T}$x\in F_T$.
COROILARY. The fixed point set for $\tilde{T}$$\tilde{T}$tilde(T)\tilde{T}$\tilde{T}$ is the largest subset of $X$$X$XX$X$ on which $G(T)$$G(T)$G(T)G(T)$G(T)$ is symmetrical.