This problem is bugging me for some time.
$f:(X,\mathcal{T}) \to (Y,\mathcal{T}')$ is a continuous bijection where $(X,\mathcal{T})$ is compact and $(Y,\mathcal{T}')$ is $T_2$ (i.e. Hausdorff and $T_1$). Then show that $f$ must be a homeomorphism.
So far I have failed to understand any relation between the compactness of domain and $T_2$-ness of the range. I tried in vain to prove that $f(U)$ is open in $(Y,\mathcal{T}')$ whenever $U$ is so in $(X,\mathcal{T})$. I know $f$ makes $X$ and $Y$ respectively $T_2$ and compact and hence makes both of them $T_4$ spaces. But that is not coming to any use. I would be glad if someone helps me with a solution. Thank you.
Best Answer
Hint: You only have to prove that it's a closed map.
Now, if you take a closed subspace $A\subseteq X$ you know it is (BLANK) and so its image is (BLANK). But, by Hausdorffness, (BLANK) subspaces are closed.