Does Hausdorffication preserve finite limits?

Question

Does the left adjoint to the inclusion of T$_2$ spaces into the category of topological spaces preserve equalizers and finite products?

(Apologies if this question has already been posted, but I did't find it)

Answer 1

It does not preserve equalizers. The example in Stefan Hamcke's comment to Does weak Hausdorffication preserve equalizers and finite products? applies also in the Hausdorff case.

Concerning products we can say the following.

For $x,x' \in X$ define $x \sim x'$ if any two open neighborhoods $U$ of $x$ and $U'$ of $x'$ intersect. Then the Hausdorffication of $X$ is given by $X/E$, where $E$ is the equivalence relation generated by $\sim$. Now let $X_\alpha$, $\alpha \in A$, be a family of spaces and $P = \prod_{\alpha \in A} X_\alpha$.

1) If $x_\alpha,x'_\alpha \in X_\alpha$ such that $x_\alpha \sim x'_\alpha$, then $(x_\alpha) \sim (x'_\alpha)$ in $P$. Let $V, V'$ be open neigborhoods of $(x_\alpha), (x'_\alpha)$ in $P$. There exist $W \subset V, W' \subset V'$ having the form $W = \prod_{\alpha \in A} W_\alpha$, $W' = \prod_{\alpha \in A} W'_\alpha$ with open neighborhoods $W_\alpha$ of $x_\alpha$ and $W'_\alpha$ of $x'_\alpha$ in $X_\alpha$, such that almost all of them are $= X_\alpha$. But now $W_\alpha \cap W'_\alpha \ne \emptyset$ for all $\alpha$, therefore $V \cap V' \ne \emptyset$.

2) Let $(x_\alpha) \sim (x'_\alpha)$ in $P$. Consider any fixed $\alpha$ and any two open neighborhoods $W_\alpha$ of $x_\alpha$ and $W'_\alpha$ of $x'_\alpha$ in $X_\alpha$. Then $W = p_\alpha^{-1}(W_\alpha), W' = p_\alpha^{-1}(W'_\alpha)$ are open neigborhoods of$(x_\alpha), (x'_\alpha)$ in $P$. They must intersect, and this implies that $W_\alpha, W'_\alpha$ intersect. Therefore $x_\alpha \sim x'_\alpha$ for all $\alpha$.

This shows that there is a canonical continuous bijection $(\prod_{\alpha \in A} X_\alpha)/\sim \phantom{.} \to \prod_{\alpha \in A} (X_\alpha/\sim)$. Whether this map is a homeomorphism is not known to me.