# A bijection f is a homeomorphism iff f maps open sets to open sets

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?

• 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. Jul 17, 2021 at 21:47
• The correct statement is “a continuous bijection is a homeomorphism iff it maps open sets to open sets. Jul 18, 2021 at 11:07

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.
Given any bijective continuous function $$g$$ which is not a homeomorphism has $$f=g^{-1}$$ a counterexample to your claim.
$$g:[0,1)\to S^1, t\mapsto e^{2\pi i t}.$$ $$g$$ 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.