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

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Statement

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

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

Proof

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I've done the proof on paper here:

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  Paper proof

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

• [ilmath](f\vert_A^\text{Im})^{-1}(U)\eq f^{-1}(U)\cap A[/ilmath]

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

References

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