TITLE: Topology: Show restriction of continuous function is continuous, and restriction of a homeomorphism is a homeomorphism
QUESTION [15 upvotes]: 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.

REPLY [2 votes]: 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 $f|_A(A\cap U)=f(A\cap U)=f(A)\cap f(U).$