Recently I read a book about linear algebraic group written by Ian Macdonald. There is a conclusion which I can't prove.
It says that if $X$ is locally compact Hausdorff space, then $X$ is compact if and only if, for all locally compact spaces $Y$, the projection $X\times Y \to Y$ is a closed map. Is it a fact for all topology spaces?
Thank you in advance
Yes, you can find this in Engelking's General Topology as Theorem 3.1.16 (for Hausdorff spaces but it works in general).
See also Compactness and product spaces, Coll. Math., 7 (1959), 19--22 by S. Mrowka.
See Problem 3.12.1 in Engelking's book for the following: if $X$ is not compact then there is a sequence $\langle F_\alpha:\alpha<\kappa\rangle$ of nonempty closed sets that is decreasing ($\alpha<\beta$ imlies $F_\alpha\supseteq F_\beta$) and has an empty intersection. In addition we assume that when $\alpha$ is a limit then $F_\alpha=\bigcap_{\beta<\alpha}F_\beta$. Then $\{\langle\alpha,x\rangle:x\in F_\alpha\}$ is a closed subset of $(\kappa+1)\times X$ (where $\kappa+1$ has the order topology and so is compact), whose projection along $X$ is the set $\kappa$ and hance not closed.
If $X$ isn't compact, let $Y$ be its one point compactification (or any compactification, really) and consider the set $\{(x,x): x\in X\}$. It's closed in $X\times Y$, but its projection into $Y$, namely $X\subset Y$, is not.
