I am trying to prove a topology statement.

Let $X,Y$ be topological spaces, and let $f: X \to Y$ be a bijection. Prove that $f$ is a homeomorphism if and only if $f$ is continuous and open.

My Attempt:

Suppose that $f$ is a homeomorphism. Then by definition, $f$ is continuous. And by definition of continuity, for every open $U \subset Y$, $f^{-1}(U)$ is open in X. Let $f^{-1}(U) =V$. Then, since $f$ is bijective, we have that $f(V)=f(f^{-1}(U))$. Since $f$ is continuous, all the open subsets of $X$ can be obtained as $f^{-1}(U)$ and since $U$ is open, $f(V)$ is open for all $V \subset X$.

(Other direction) Assume that $f$ is continuous and open. Then Since $f$ is bijective, it has an inverse $f^{-1}$. Since $f$ is open, for any open set $U$ of $X$, $f(U)$ is open in $Y$. We need to prove that $f^{-1}$ is continuous. Since $f$ is open, for all open sets $U \in X$, $(f^{-1})^{-1}(U)=f(U)$ is open in $Y$. Thus $f^{-1}$ is continuous and $f$ is a homeomorphism

Any help would be appreciated.


Continuing where you stopped: Since $f$ is bijective, we have that $f(f^{-1}(U)) = U$ and since $U$ is open, we have that $f(V)$ is open (in $Y$)

To prove that $f^{-1}$ is continuous, we should prove that for all open sets $U \in X$, $(f^{-1})^{-1}(U)$ is open in $Y$, since $(f^{-1})^{-1}(U) = f(U)$. And since $f$ is open, this follows directly.

  • $\begingroup$ Thank you for the answer. Is it always true that $f(f^{-1}(U)) = U$ when $f$ is bijective? $\endgroup$ – Cruso James Jun 19 '17 at 14:48
  • 1
    $\begingroup$ Yes, see for example math.stackexchange.com/questions/359693/… $\endgroup$ – Nigel Overmars Jun 19 '17 at 14:50
  • 2
    $\begingroup$ Except that you need to prove that $f(V)$ is open for all open $V \subset X$, not just for $V = f^{-1}(U)$. Of course since $f$ is a homeomorphism then all the open subsets of $X$ can be obtained as $f^{-1}(U)$ for some open $U \subset Y$, but you have not proved this anywhere. $\endgroup$ – Najib Idrissi Jun 19 '17 at 14:54
  • $\begingroup$ @NajibIdrissi you are correct, I will edit my answer. $\endgroup$ – Nigel Overmars Jun 19 '17 at 14:56
  • $\begingroup$ Thank you guys. Could you check my edited answer please? $\endgroup$ – Cruso James Jun 19 '17 at 15:59

Homeomorphism means a continuous bijection whose inverse is continuos too. Now use the fact that f is continuous iff for every open set $U$ of Y , $f^{-1}(U)$ is open in X. The bijection is needed for the other direction, when you have to prove f is homeomorphism. $f^{-1}$ exists since it is a bijection and continuos as f is an open map.


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.