What are the sufficient conditions for a function to take a Hausdorff space to another Hausdorff space.

Question

Assume that $f: X \rightarrow Y$ and $X$ is Hausdorff. I'm thinking that we have to have $f$ being an onto, open and injective map to guarantee that $Y$ is also Hausdorff.

Proof: Assume that $f: X \rightarrow Y$ is onto, open and injective and $X$ is Hausdorff. For any $r, s \in Y$ such that $r \ne s$, there exits $p \ne q$ such that $f(p) = r$ and $f(q) = s$. Since $X$ is Hausdorff, there exists disjoint $U, V \in F_X$ such that $p \in U, q \in V$. Since $f$ is injective, $f(U) \cap f(V) = \varnothing$ and $r \in f(U), s \in f(V)$. Hence, $Y$ is Hausdorff.

But this seems to be a pretty strong assumption on $f$. Are there looser conditions that suffice to make $Y$ Hausdorff?

Plenty of thanks!!!

Answer 1

As mentioned in the comments by Pedro, your assumption is simply that $f$ is a homeomorphism. In general, there are very few topological properties that are transferred from a space $X$ to a space $Y$ under a continuous mapping, the most notable one being compactness. You're right that in this case the only way we can guarantee that $Y$ is Hausdorff under the assumption that $X$ is Hausdorff is if the mapping is a homeomorphism. It is interesting to note however that if $f: X \to Y$ is a continuous injection and $Y$ is Hausdorff, then so is $X$.

Answer 2

It works better with $T_1$ and normality, then with Hausdorffness. But if $X$ is compact, you can relax to $f$ being a closed quotient map. More precisely:

• If $f$ is a quotient map, then $Y$ is $T_1$ if and only if the fibers are closed.
• If $f$ is a closed quotient map and $X$ is normal, then $Y$ is normal.
• If $f$ is a closed quotient map, $X$ is normal, and the fibers are closed, then $Y$ is $T_4$ and hence $T_2$.
• If $f$ is a closed quotient map, $X$ is compact, and the fibers are closed, then $Y$ is $T_2$ if $X$ is $T_2$.