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I'm trying to come up with an example of two topological spaces between which there exist continuous bijections in both directions, but they won't be homeomorphic. Here's my reasoning.

Let's start with the mapping: $f:X→Y$ Take $X = [0, 1]$ with the discrete topology and $Y = S^1$ with the discrete topology - a circle parameterized as $(\cos(2\pi t), \sin(2\pi t))$. Consider the identity mapping $f: [0, 1] \rightarrow (\cos(2\pi t), \sin(2\pi t))$, associating each point $t$ on the interval with a point on the circle. It is obviously injective and surjective. Moreover, the mapping will be continuous because any open arc on the circle will have a preimage as an open interval in the unit interval. This is the first continuous bijective mapping.

Now, consider the mapping in the reverse direction: $g:Y→X$ $$g(\cos(2\pi t),\sin(2\pi t)) = \frac{1}{2}+\frac{1}{2}\frac{x}{\sqrt{x^2+y^2}}$$ $g$ is injective and surjective, and it seems to be continuous...

Am I reasoning correctly?

VGT
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    $f$ is not injective ($f(0)=f(1)$), and your definition of $g$ makes no sense. +1 though (interesting question). – Anne Bauval Nov 27 '23 at 15:11
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    Any two discrete spaces with same cardinality (like yours) are homeomorphic. – Anne Bauval Nov 27 '23 at 15:12
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    Take a look here: https://mathoverflow.net/questions/30661/non-homeomorphic-spaces-that-have-continuous-bijections-between-them – Gerd Nov 27 '23 at 15:24
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    Might also be worth to note that you can't find such an example when the spaces are compact. See: https://math.stackexchange.com/questions/3042668/continuous-bijection-between-compact-and-hausdorff-spaces-is-a-homeomorphism – Keen-ameteur Nov 27 '23 at 15:30
  • how does this possible? aren't the homeomorphisms defined as the bijective, both-ways continuous functions between topological spaces ? – niobium Nov 27 '23 at 16:56

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