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

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Statement

Let $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$ be topological spaces and suppose that $f:X\rightarrow Y$ is a homeomorphism between them, so $X\cong_f Y$, then:

• $\forall A\in\mathcal{P}(X)[A\cong_{f\vert_{A}^{\text{Im} } } f(A)]$
  • In words: For all subspaces of $X$, suppose in particular $A$ is a subspace, then $f\vert_{A}^\text{Im}:A\rightarrow f(A)$ - the restriction onto its image of $f$ to $A$ - is a homeomorphism between $A$ and $f(A)\subseteq Y$
    • So $A\cong_{f\vert_A^\text{Im} }f(A)$ explicitly

Proof

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I've done the proof on paper here:

• 

  Paper proof

It uses $A\subseteq B\implies f^{-1}(A)\subseteq f^{-1}(B)$ which is somewhere under function properties, and it also expresses:

• $(f\vert_A^\text{Im})^{-1}(U)\eq f^{-1}(U)\cap A$

which is ripe for factoring out Alec (talk) 13:31, 20 February 2017 (UTC)

References

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I can't be the first person to have used this!