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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.

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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.

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  • $\begingroup$ 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
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    $\begingroup$ Yes, see for example math.stackexchange.com/questions/359693/… $\endgroup$ – Nigel Overmars Jun 19 '17 at 14:50
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    $\begingroup$ 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
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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.

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