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Show that a continuous bijection $f : X \to Y$ with $X$ compact and $Y$ Hausdorff is a homeomorphism. Give an example to show that such a continuous bijection is not necessarily a homeomorphism if $Y$ is not assumed to be Hausdorff.

I'm having some trouble with the counterexample.

Would an $X$ interval in $\mathbb R$ and a $Y = S^1$ work?

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    $\begingroup$ $S^1$ is Hausdorff! $\endgroup$ – Cheerful Parsnip Oct 14 '11 at 1:05
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No, because $\mathbb{S}^1$ is Hausdorff (it is a subspace of the Hausdorff space $\mathbb{R}^2$).

Let $X=\{a,b\}$ with the discrete topology, i.e. the open sets of $X$ are $\varnothing$, $\{a\}$, $\{b\}$, and $\{a,b\}$, and let $Y=\{c,d\}$ with the trivial topology, i.e. the open sets of $Y$ are $\varnothing$ and $\{c,d\}$.

$X$ is compact (because it is finite), and $Y$ is not Hausdorff because the points $c$ and $d$ cannot be separated by disjoint open sets.

Let the map $f:X\to Y$ be defined by $f(a)=c$, $f(b)=d$. Then $f$ is a continuous bijection, but not a homeomorphism.

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Here is a simple example. Let $X$ be an interval with the standard topology, and let $Y$ be the same set with the coarse topology (only two open sets).

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To show that f is a homeo. is to show that it takes open sets into open sets. Let $G$ be open in $X$. Then $F = X - G$ is closed and therefore compact (closed subspace of a compact space.) Since f is continuous, $f(F)$ is compact in $Y$. Since $Y$ is $T_2$ we have that $f(F)$ is closed in $Y$.

Then $Y - f(F) = f(G)\ $ is open in Y. therefore $f^{-1}$ is continuous

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