View source for Given a homeomorphism all subspaces of the domain are homeomorphic to their image under the homeomorphism itself

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==Statement==
Let {{Top.|X|J}} and {{Top.|Y|K}} be [[topological spaces]] and suppose that {{M|f:X\rightarrow Y}} is a [[homeomorphism]] between them, so {{M|X\cong_f Y}}, then:
* {{M|\forall A\in\mathcal{P}(X)[A\cong_{f\vert_{A}^{\text{Im} } } f(A)]}}
** In words: For all [[topological subspace|subspaces]] of {{M|X}} {{M|f\vert_{A}^\text{Im}:A\rightarrow f(A)}} - the [[restriction onto its image]] of {{M|f}} to {{M|A}} - is a homeomorphism between {{M|A}} and {{M|f(A)\subseteq Y}}
==Proof==
{{Requires proof|grade=E|msg=I've done the proof on paper here:
<gallery>
HomeomorphismsRestrictToHomeomorphisms.JPG|Paper proof
</gallery>
It uses {{M|A\subseteq B\implies f^{-1}(A)\subseteq f^{-1}(B)}} which is somewhere under function properties, and it also expresses:
* {{M|(f\vert_A^\text{Im})^{-1}(U)\eq f^{-1}(U)\cap A}}
which is ripe for factoring out [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 13:31, 20 February 2017 (UTC)}}
==References==
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{{Requires references|grade=E|msg=I can't be the first person to have used this!}}
{{Theorem Of|Topology}}