Is the adjunction space of two Hausdorff spaces also Hausdorff

Question

I was reading the definition of CW-complex in terms of pushouts given by Lück's Algebraische Topologie: Homologie und Mannigfaltigkeiten (Chapter 3). It is stated (though not proven) that such a topological space is always Hausdorff. I guess this follows readily by induction if we can prove that the adjunction space $Z$ formed by attaching $Y$ to $X$ along $f$, where $f:A\rightarrow X$ is some continuous map and $A$ is a closed subset of $Y$, is Hausdorff whenever $X$ and $Y$ are (alternatively, $Z$ is just the pushout of $(\iota, f)$, with $\iota: A\hookrightarrow Y$ being the inclusion).

However, the quotient of a Hausdorff space is not Hausdorff in general, so I was wondering how to prove it in this particular case. If I choose two distinct $z,z'\in Z$, it is easy to separate them if they both lie in $q(Y\setminus A)$, with $q:X\amalg Y\rightarrow Z$ being the quotient map, as $q(Y\setminus A)$ is open in $Z$ and $q\vert_{Y\setminus A}$ is a topological embedding. Now, if both $z,z'$ lie in $q(X)$, I've seen people claiming that the same argument may be used, as $q\vert_X$ is also a topological embedding. What I don't see is how we can ignore the fact that $q(X)$ need not be open in $Z$, in general.

Moreover, in the case that one point lies in $q(X)$ and the other in $q(Y\setminus A)$, I have no clue on how to proceed. Could anyone help?

Answer 1

Here's a counterexample

Let $Q=\Bbb{Q}\cap (0,1)$.

Let $Y$ be $\Bbb{R}$ with the following topology

• A set $U\subseteq Y$ is open if it's open in the usual topology on $\Bbb{R}$ or $U=V\setminus Q$ where $V$ is open in the usual topology on $\Bbb{R}$. (This is due to Munkres - a Hausdorff space that's not regular.)

Since $Y$'s topology contains the usual topology on $\Bbb{R}$, it's Hausdorff.

Let $A=Q\cup\{0\}$. $A$ as a subspace of $Y$ which is Hausdorff, is also Hausdorff. $Q$ is closed in $Y$ as the complement of $Y\setminus Q$. $\{0\}$ is also closed in $Y$. So $A$ is a closed subset of $Y$.

Let $X=\{0,1\}$. In the subspace topology of $A$, $Q$ is closed and $\{0\}$ is closed, so there is a continuous map $f\colon A\to X$ sending $Q$ to $0$ and $f(0)=1$.

In the quotient topology $X\sqcup Y/f$, the points $0,1$ from $X$ don't have disjoint open neighborhoods. One such neighborhood must be an open neighborhood of $0$ in $Y$ - containing all irrationals in some interval $(0,b)$ with $b>0$. The other an open neighborhood of $Q$ - which must also contain all irrationals in some interval $(0,c)$ with $c>0$. So they can't be disjoint.