In Willard's text, the way he shows that if a product $\Pi_{\alpha}X_{\alpha}$ is locally compact, then each $X_{\alpha}$ is locally compact, is by using the fact that if $X$ is locally compact and $f:X\to Y$ is continuous, onto, and open, then $Y$ is locally compact.
I was thinking you could argue that each $X_{\alpha}$ is homeomorphic to a closed subspace of $\Pi_{\alpha}X_{\alpha}$ (local compactness is not preserved by general subspaces, so it is not enough to know that $X_{\alpha}$ is homeomorphic to a subspace of $\Pi_{\alpha}X_{\alpha}$. But local compactness is preserved by closed subspaces). However, I think it's only true that each $X_{\alpha}$ is homemorphic to a closed subspace if each $X_{\alpha}$ is Hausdorff. Is this correct, and hence the reason for the argument in Willard?
Best Answer
Each $X_\alpha$ is homeomorphic to a closed subset of the product if each $X_\alpha$ has a closed singleton, e.g. So it often will be, but you cannot count on it. For example, if all $X_\alpha$ are indiscrete it will not be. But the open projection argument always works and is simple enough. In practice many locally compact spaces are assumed to be Hausdorff too and the closed subspace argument works too (because then all singletons are closed in all components).
But the open-map fact is also interesting in its own right, so from a general theory perspective the Willard proof is better, IMHO.