In our definition of homeomorphic topological spaces there must exist a bijective function $f:X\mapsto X'$ such that $f$ is continuous on $X$ and $f^{-1}$ is continuous on $X'$.
Is it necessary to assure that $f^{-1}$ is continuous on $X'$?
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1Sure. Take the identity on $\Bbb R$ with the discrete topology onto $\Bbb R$ with the usual topology. – David Mitra Jan 20 '19 at 10:04
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The requirement that
$f^{-1}$ is continuous, is essential. Consider for instance the function
$f:[0,2 \pi)\rightarrow S_1$ (the unit circle in $\mathbb{R}^2$ defined by
$f(\phi )=(\cos \phi ,\sin \phi ))$.
This function is bijective and continuous, but not a homeomorphism ($S^{1}$ is compact but $[0,2\pi )$ is not).
The function $f^{-1}$ is not continuous at the point $(1,0)$, because although $f^{-1}$ maps $(1,0)$ to $0$, any neighbourhood of this point also includes points that the function maps close to $2\pi $, but the points it maps to numbers in between lie outside the neighbourhood.
user289143
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You have to assume this, it is not automatic:
If $X=\{1,2\}$ in the discrete topology and $Y=X$ in the indiscrete/trivial topology, then $f(x)=x$ is continuous from $X$ to $Y$ but its inverse (the same identity) is not continuous from $Y$ to $X$.
Henno Brandsma
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