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(a) Let X be a topological space. Prove that the set $Homeo(X)$ of homeomorphisms $f:X \to X$ becomes a group when endowed with the binary operation $f \circ g$ .

(b) Let $G$ be a subgroup of $Homeo(X)$. Prove that the relation "$xR_G y \iff \exists g \in G$ such that $g(x)=y$" is an equivalence relation.

I have done part (a) but I am unsure what part (b) is really even saying, I don't understand what the proposed equivalence relation means so am not sure how to prove it is an equivalence relation.

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    $\begingroup$ For (2): the relation is on $\,X\,$ . This is the first thing we have to understand. $\endgroup$ – DonAntonio Apr 15 '13 at 11:36
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HINT: Let $x,y\in X$; if there is at least one auto-homeomorphism $g:X\to X$ such that $g(x)=y$, then we write $xR_Gy$. You’re asked to prove that this binary relation $R_G$ on $X$ is an equivalence relation: it’s reflexive, symmetric, and transitive.

  • Reflexivity: If $x\in X$, is there an auto-homeomorphism of $X$ that takes $x$ to $x$?

  • Symmetry: If $x,y\in X$, and there is some auto-homeomorphism $f:X\to X$ such that $f(x)=y$, is there an auto-homeomorphism $g$ of $X$ such that $g(y)=x$?

  • Transitivity: If $x,y,z\in X$, and there are auto-homeomorphisms $f,g:X\to X$ such that $y=f(x)$ and $z=g(y)$, is there an an auto-homeomorphism $h$ of $X$ such that $z=h(x)$?

Note: $\operatorname{Homeo}(X)$ must be defined as the set of surjective homeomorphisms $h:X\to X$, or the result is false.

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Hints: be sure you understand and can explain and justify the following

1)Reflexivity: clear since the identity map $\,Id(x)=x\,\,,\,\forall\,x\in X\,$ is the unit element in the group $\,Homeo(X)\,$ and thus is in $\,G\,$

2) Symmetry: $\,xR_Gy\iff \exists f\in G\,\,s.t.\,\,f(x)=y\iff x=f^{-1}y\,$ , and it must be clear that also $\,f^{-1}\in G\,$

3) Transitivity: $\,xR_Gy\,\,\wedge\,\,yR_Gz\implies\;\;\exists\, f\,,\,g\in G\;\;s.t.\,\,f(x)=g\;,\;\;g(y)=z$ . Now just check what happens with $\,g\circ f\,$ to get $\,xR_Gz\,$ and why $\,g\circ f\in G\,$ ...

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$x\, R_G\, y$ means that the ordered pair $(x,y)$ is in the $R_G$ relation, that is, as written, there is an element $g\in G\subseteq Homeo(X)$, such that $g(x)=y$. We can call this property: "$x$ and $y$ are on the same orbit", where the orbit of a point $y$ w.r.t. $G$ is defined as $Gy:=\{g(y)\,\mid\,g\in G\}$.

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