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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?

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1 Answer 1

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I am pretty sure your proof is correct, congratulations! However, it could be written down more concisely by first proving the following Lemma.

The restriction of a continuous map to any subspace is continuous.

(note that you basically prove this lemma twice in your solution.)

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.

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