# Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism

I need to prove two trivial results but I don't know how to work with restricted function and its inverse

Consider the topological spaces $(X, \mathcal{T}), (Y, \mathcal{J})$

Claim 1: Let $f:X \to Y$ be continuous function, $A \subset X$ equipped with subspace topology, then $f|_{A}:A \to Y$ is continuous

Proof: Take some $V \in \mathcal{J}$, then $f^{-1}|_A(V) = f^{-1}(V) \cap A$, where $f^{-1}(V)$ is open, therefore $f^{-1}(V) \cap A$ is open in the subspace topology.

Claim 2: Let $f:X \to Y$ be homeomorphism, $A \subset X$ equipped with subspace topology, then $f(A)$ is a subspace of $Y$ and $f|_{A}:A \to f(A)$ is continuous

Proof: We proceed by showing $f|_A$ is continuous and open.

First show $f|_A$ is continuous, take some open set $W$ in the subspace topology on $f(A)$, $W = f(A) \cap V, V \in \mathcal{J}$ then $f^{-1}|_A(V \cap f(A) ) = f^{-1}|_A(V) \cap f^{-1}|_A(f(A)) = $$(f^{-1}(V) \cap A) \cap (f^{-1}(f(A)) \cap A) = f^{-1}(V) \cap A is open. Next show f|_A is open. Take some open set M in the subspace topology on A, then M = A \cap U, U \in \mathcal{T}. Then f|_A(A \cap U) = f|_A(A) \cap f|_A(U) = f(A) \cap f|_A(U) = f(A) \cap f(U \cap A) =$$ f(A) \cap f(U) \cap f(A) = f(A) \cap f(U)$. Note $f(U)$ is open since $f$ is open, therefore $f(A) \cap f(U)$ is open in the subspace topology of $f(A)$

This shows all homeomorphisms are local homeomorphism

Can someone check the two proofs? The second one is a bit messy.

• Please be careful, there is a difference between $f^{-1}\vert_A$ and $f\vert_A^{-1}$. Sep 1, 2021 at 3:47

Your proof of Claim 1 is correct, and your proof of Claim 2 is correct but unclear.

First, the last line of your proof This shows all homeomorphisms...'' can be omitted. The fact that all homeomorphisms are local homeomorphisms has nothing to do with your claim that $f|_A$ is a homeomorphism.

Second, some of your reasoning in the third paragraph of Claim 2 depends on the fact that the restricted mapping $f|_A$ is a bijection. You could start out by noting that $f|_A$ is a bijection since $f$ is a homeomorphism. Then it suffices to prove that $f|_A$ is continuous and open (since being open in this case is the same thing as the inverse map being continuous).

Now consider the second line in your third paragraph, which starts with $M = A\cap U \ldots$ I don't understand the presence of $Y$ in this line. $f|_A(A) = f(A),$ not $Y$. So you can replace $Y$ with $f(A)$ and then delete the extraneous references to $Y$. Finally, as I alluded to above, your assertion that the image of the intersection of two sets is the intersection of the images is only true because $f|_A$ is a bijection-- this fact isn't true in general. So it might be good to mention that in your proof.

• Thanks for the comment, I was setting up a proof involving local homeomorphism then I decided not to include it, but the result holds noneteless! Jun 15, 2016 at 8:05

Correct.

It would be better in #2 to say " It suffices to show that $f|_A\to f(A)$ is continuous and open because it is a bijection." First, because, although it is necessary, it matters that the proof is done when it has been shown, because it is sufficient. Second, because you state the properties "continuous" and "open" in the same order that you prove them.

To show that $f|_A\to f(A)$ is open , it is briefer to observe that, because $f$ is a bijection and $A\cap U\subset A,$ we have $f|_A(A\cap U)=f(A\cap U)=f(A)\cap f(U).$