# Restriction of domain and codomain of homeomorphism is still a homeomorphism

Let $$f:X \rightarrow Y$$ be a homeomorphism and let $$A \subset X$$. Then $$f$$ restricts to a homeomorphism $$f|_A:A \rightarrow f(A)$$ between the subspaces $$A$$ and $$f(A)$$.

I am going to omit the restriction symbol when showing continuity? Any harm doing this?

Let $$U$$ be an open set in $$f(A)$$ as a subspace of $$Y$$. Then $$U=f(A) \cap V$$ for some $$V$$ open in $$Y$$. So $$f^{-1}(f(A) \cap V)=f^{-1}(f(A)) \cap f^{-1}(V)=A \cap f^{-1}(V)$$ which is open in $$A$$.

Would $$f^{-1}(f(A))=A$$ since $$f$$ is bijective? I believe it would.So $$f|_A$$ is continuous.

Let $$W$$ be open in $$A$$ as a subspace of $$X$$. Then $$W=A \cap Z$$ for some $$Z$$ open in $$X$$. Then $$(f^{-1})^{-1}(A \cap Z)=(f^{-1})^{-1}(A) \cap (f^{-1})^{-1}(Z)=f(A) \cap f(Z)$$ which is open in $$f(A)$$ since $$f$$ is bijective. So $$(f|_A)^{-1}$$ is continuous.

Now I'm attempting to prove bijectivity of $$f|_A$$.Suppose $$f|_A(x)= f|_A(y)$$. Then since $$f|_A(x)=f(x)=f(y)=f|_A(y)$$ and $$f$$ is injective $$x=y$$.

Onto surjective. I don't know I figure it would be automatic. Or would it? Any help here? Wait isn't any injective function with codomain restricted to its range necessarily a bijection?

Thanks

• Your proof is fine, but it might be easier to observe that $f|_A=f\circ i$ where $i$ is inclusion of $A$ into $X$. Apr 4, 2021 at 2:01

To simplify notation call $$f\restriction_A$$ just $$f_A$$, once you've fixed $$A$$. Then $$f_A: A \to Y$$ and $$f_A = f \circ i_A$$ where $$i_A:A \to X: i_A(x)=x$$ is the canonical injection. The subspace topology on $$A$$ is chosen such that $$i_A$$ is continuous (and minimally so), so $$f$$ continuous implies $$f_A$$ continuous is immediate. Also $$i_B: B \to Y$$ where $$B = f[A]$$ is likewise a continuous map and using the universal property for initial topologies and $$i_B \circ f' = f_A = f \circ i_A$$
$$f': A \to f[A]$$ being the map we have to prove continuous, we see that $$f'$$ is continuous iff $$f_A$$ is continuous and we know the latter to be true.
So continuity of $$f$$ is no problem. You could do a siimailr argument for $$g: Y \to X$$ which is the inverse of $$f$$. Being bijective, $$g$$ maps $$f[A]$$ back to $$A$$ etc. No topology, just basic set theory.
So $$g': f[B] \to A$$ is also continuous when $$g$$ is (which it is), so $$f'$$ is a homeomorphism, as $$g'$$ is the inverse of $$f'$$ etc.