Show that a Hausdorff space $X$ is regular if and only if $X/A$ is Hausdorff for all $A$ closed in $X$.

Question

Show that a Hausdorff space $X$ is regular if and only if $X/A$ is Hausdorff for all $A$ closed in $X$.

Assume that $X$ is regular. Let $x, y \in X/A$ such that $x \ne y$. If $x \ne A$ and $y \ne A$, then $q^{-1}(\{x\})=x$ and $q^{-1}(\{y\})=y$. As $X$ is Hausdorff there exists open neighborhoods $U_1$ and $U_2$ of $x$ and $y$ respectively. Now $q$ being surjective implies that $q^{-1}(q(U_1))=U_1$ and $q^{-1}(q(U_2))=U_2$. Since each $U_i$ is open the quotient topology that $X/A$ has implies that $q(U_1)$ and $q(U_2)$ are both open. We also have that $$q^{-1}(q(U_1 \cap U_2))= \emptyset$$ but I don't know if we can derive the result that $q(U_1) \cap q(U_2) = \emptyset$ which would imply that $X/A$ is Hausdorff?

Answer 1

As $A$ is closed, we may assume that $U_i \cap A = \emptyset$, $i = 1$, $2$ (otherwise replace $U_i$ by $U_i \cap (X \setminus A)$). Then on both $U_i$ $q$ is a bijection, therefore $$ q[U_1] \cap q[U_2] = q[U_1 \cap U_2] = q[\emptyset] = \emptyset$$ Note that this does not imply $X/A$ being Hausdorff, you still have to consider the case $x = A$, $y \ne A$. Use regularity here.

Answer 2

First of all recall the following:

Lemma. If $f:X\to Y$ is a surjective function and $A,B\subseteq Y$ are such that $f^{-1}(A)$ is disjoint from $f^{-1}(B)$ then $A,B$ have to be disjoint as well.

Proof. Assume otherwise that $x\in A\cap B$. Then $f^{-1}(x)\subseteq f^{-1}(A)$ and $f^{-1}(x)\subseteq f^{-1}(B)$. Since $f$ is surjective then $f^{-1}(x)$ is nonempty showing that $f^{-1}(A)$ and $f^{-1}(B)$ are not disjoint. Contradiction. $\Box$

This property reduces the problem "$q(U)$ is disjoint from $q(V)$" to "$q^{-1}(q(U))$ is disjoint from $q^{-1}(q(V))$", which is immediate if $q^{-1}(q(U))=U$ and $q^{-1}(q(V))=V$. However:

Now $q$ being surjective implies that $q^{-1}(q(U_1))=U_1$

That is unfortunately wrong. The real equality is $q(q^{-1}(F))=F$ for surjective functions. The equality you wrote is for injective functions. In fact consider what happens when $U_1$ intersects $A$ (nothing you wrote stops that from happening). Then $q^{-1}(q(U_1))=U_1\cup A$ which no longer has to be open. And if the same happens for $U_2$ then additionally their images are no longer disjoint, even though they still can be disjoint in $X$ (obviously $A$ has to have more than $1$ element in such scenario).

What you can show is that for a subset $B\subseteq X$ we have

$$q^{-1}(q(B))=\begin{cases} B&\text{if }B\cap A=\emptyset\\ B\cup A&\text{otherwise} \end{cases}$$

So how do we utilize all of that?

"$\Rightarrow$" Assume $[x],[y]\in X/A$, $[x]\neq[y]$. There are two cases:

1. $[x]\neq A$ and $[y]\neq A$. Then as you wrote $q^{-1}([x])=\{x\}$ and $q^{-1}([y])=\{y\}$. Since $X$ is Hausdorff then so is $X-A$ (which is open) and thus $x$ and $y$ can be be separated by open subsets $U$, $V$ of $X-A$. Therefore $U$, $V$ are open subsets of $X$ that do not intersect $A$. Thus $q^{-1}(q(U))=U$ and $q^{-1}(q(V))=V$ showing that $q(U)$ and $q(V)$ are open and (by our lemma) disjoint in $X/A$.

So my reasoning is almost the same as yours, except I had this one additonal step that my open neighbourhoods are taken in $X-A$ instead of whole $X$. In order to guarantee the $q^{-1}(q(B))=B$ property, which then gives us that images are open and disjoint.

2. One of them is $A$, e.g. $[y]=A$. Then $q^{-1}([x])=\{x\}$ but $q^{-1}([y])=A$. Since $X$ is regular than these can be separated by open neighbourhoods $U$, $V$ of $x$ and $A$ respectively. Note that $U$ doesn't intersect $A$ and thus $q^{-1}(q(U))=U$, while $q^{-1}(q(V))=V\cup A=V$ as well (because $A\subseteq V$). This shows that $q(U)$ and $q(V)$ are open and (by our lemma) disjoint in $X/A$.

"$\Leftarrow$" Let $A\subseteq X$ be a closed subset and $x\in X-A$. Since $X/A$ is Hausdorff then $A$ and $[x]$ can be separated by open subsets $U$, $V$ in $X/A$. It follows that $q^{-1}(U)$ and $q^{-1}(V)$ are open subsets separating $A$ from $x$ in $X$.