# Homeomorphism restricted to an open subset

Theorem: Let $$f:X\rightarrow Y$$ be a homemorphism and $$U\subseteq X$$ an open subset. Show that $$f(U)$$ is open in $$Y$$ and that $$f|_{U} \rightarrow f(U)$$ is a homeomorphism.

My Proof:

As $$f$$ is a homemorphism, it maps open sets to open sets, so if $$U$$ is open in $$X$$ then $$f(U)$$ is indeed open in $$Y$$. Observe that the map $$f|_{U}$$ is clearly bijective. So it does have an inverse $$g : f(U)\rightarrow U$$. It suffices to show that $$f|_{U}$$ and $$g$$ are both continuous. Let $$A$$ be open in $$f(U)$$ so $$A= A' \cap f(U)$$ for some open set $$A'$$ in $$Y$$. Then, since $$f|_{U}$$ is bijective, it follows that $$f|_{U}^{-1}(A'\cap f(U))= f^{-1}(A')\cap U\cap f^{-1}(f(U)).$$ Similarly, by bijectivity of $$f$$, the latter expression is equal to $$f^{-1}(A') \cap U$$, which is open in $$U$$, as $$f^{-1}$$ is a homeomorphism. Similarly, let $$g:f(U)\rightarrow U$$ denote the inverse. Let $$B$$ be open in $$U$$ so $$B=B' \cap U$$ for some open $$B'$$ in $$X$$. Then $$g^{-1}(B)=g^{-1}(B'\cap U)= g^{-1}(B') \cap f(U)= f(B')=f(U)$$ (since the inverse is bijective) and, as $$f$$ is a homeomorphism, $$f(B')$$ is open in $$Y$$ and so $$g^{-1}(B)$$ is open in $$f(U)$$.

Is the proof correct?

The proof of your claim now goes through very quickly and cleanly: Let $$f: X \rightarrow Y$$ be a homeomorphism. By the Lemma, both $$f|_U$$ and $$g = f^{-1}|_{f(U)}$$ are continuous. So $$f|_U$$ is a homeomorphism.