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I know that you can prove that a bijective $f:X\to Y$ is a homeomorphism if and only if $A$ is open in $X$ if and only if $f(A)$ is open in $Y$, and that a bijective $f:X\to Y$ is a homeomorphism if and only if $f$ and $f^{-1}$ both map open sets to open sets, meaning that you have to either assume $f$ to be continuous or that $f^{-1}$ also maps open sets to open sets.

Could it hold that a bijective $f:X\to Y$ is a homeomorphism if and only $f$ maps open sets to open sets? I can prove the direction that $f$ being a homeomorphism implies that $f$ maps open sets to open sets, and to the other direction i can show that if $f$ maps open sets to open sets then $f^{-1}$ is continuous, but i can't seem to get the continuity of $f$ from that. Is it possible?

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  • $\begingroup$ Technically, you aren’t assuming that $f$ or $f^{-1}$ is continuous. You can just prove both are continuous based on the condition that $A$ is open if and only if $f(A)$ is open. But no, the weaker condition is not enough. $\endgroup$ Jul 17, 2021 at 21:47
  • $\begingroup$ The correct statement is “a continuous bijection is a homeomorphism iff it maps open sets to open sets. $\endgroup$ Jul 18, 2021 at 11:07

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Let $\tau_1$ be the usual topology on $\Bbb R$ and let $\tau_2$ be the discrete topology. Then the identity from $(\Bbb R,\tau_1)$ into $(\Bbb R,\tau_2)$ is a bijection which maps open sets into open sets. But it is not a homeomorphism.

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Given any bijective continuous function $g$ which is not a homeomorphism has $f=g^{-1}$ a counterexample to your claim.

The standard example of a continuous bijection which is not a homeomorphism is:

$$g:[0,1)\to S^1, t\mapsto e^{2\pi i t}.$$ $g$ is a continuous bijection.

So $f=g^{-1}$ has the property that $f$ is a bijection, and for any open $A\subseteq S^1,$ $f(A)$ open. But $f$ is not continuous.

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