$f:X\to Y$ be a continuous bijection between $X$ a compact Hausdorff space and $Y$ is just a topological space, is it true that $f$ has to be open or closed map?
I know if $Y$ were Hausdorff then $f$ will be a homeomorphism.
$f:X\to Y$ be a continuous bijection between $X$ a compact Hausdorff space and $Y$ is just a topological space, is it true that $f$ has to be open or closed map?
I know if $Y$ were Hausdorff then $f$ will be a homeomorphism.
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Let $Y$ have the same underlying set as $X$, but give $Y$ the cofinite topology. Then $Y$ is a compact $T_1$ space, and the identity map from $X$ to $Y$ is a continuous bijection that is open or closed if and only if $X$ is finite. This is because if $X$ is infinite, compact, and Hausdorff, its topology must be strictly finer than the cofinite topology. In that case there is an open set $U$ in $X$ whose complement is not finite, and the identity map takes the open set $U$ to a non-open set in $Y$ and the closed set $X\setminus U$ to a non-closed set in $Y$.
Added: I just saw Daniel Fischer’s comment. Of course he’s absolutely right: you can do much the same thing by giving $Y$ the indiscrete topology. $Y$ will still be compact, though it won’t have any nice separation properties unless $|X|=1$, so that it has them vacuously. And now the identity map will fail to be either closed or open as long as $X$ has at least two points: you no longer need infinitely many.