2
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.