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Let $f:X\rightarrow X$ a bijective map between topological spaces (the same space X). A priori not known to be continuous. If we know $f\circ f=id$ does it mean that $f$ has to be continuous map and hence a homeomorphism ?

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No. For instance, consider the topological space $X = \{a, b\}$ with the topology $$ \tau = \{\varnothing, \{a\}, X\} $$ and the bijection $f:X\to X$ with $f(a) = b, f(b) = a$.

Then $f^{-1}(\{a\}) = \{b\}$ is not open, so $f$ is not continuous.

For a more "interesting" example (i.e. closer to something you may reasonably encounter more often), consider $X = \Bbb R$ with the standard topology, and $$ f(x) = \cases{\frac1x & if $x<0$\\x&otherwise} $$ which is a discontinuous, self-inverse bijection.

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If $X$ is the Sorgenfrey line $\Bbb S$, i.e. $\Bbb R$ in the topology generated by the sets $[a,b)$ (aka as the lower limit topology) and $f(x)=-x$ then $f$ is self inverse but continuous at no point, while $X$ is hereditarily normal etc., so quite nice.

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