Lack of homeomorphism between compact space and non-Hausdorff space 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? 
 A: 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.
A: 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). 
A: 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
