Prove $X/A$ is Hausdorff if $A↪X$ is a cofibration

Question

Let $X$ be a Hausdorff space and $A$ a closed subspace. Suppose the inclusion $A↪X$ is a cofibration. Prove $X/A$ is Hausdorff.

I have no idea that how Hausdorff is related to the cofibration.

I can understand the unswer given by Stefan Hamcke, but I am wonder if there is an easier answer not using adjunction space.

Answer 1

If $A\hookrightarrow X$ is a cofibration, then there exists a retraction $ r: X\times I \to X\times\{0\} \cup A\times I$ with coordinates $(r_1,r_2)$. Now define $$ \textstyle u: X \to I, x\mapsto \max_{t\in I}(t-r_2(x,t)) $$ and $$ h: X\times I \to X, (x,t)\mapsto r_1(x,t) $$ Then $u$ is continuous and $A=u^{-1}(0)$ is a zero set. When $u(x)>0$, then disjoint neighborhoods of $0$ and $u(x)$ pull back to disjoint neighborhoods of $A$ and $x$ in $X$.
It is easy to show that $h$ is a homotopy rel $A$ starting at $1_X$ and ending at a map sending $x$ to $A$ whenever $u(x)<1$, so $A$ is a retract of a neighborhood.

Now there is a proposition saying that an adjunction space $X \cup_f B$ is Hausdorff if

• $B$ and $X$ are Hausdorff,
• each $x \in X \backslash A$ can be separated from $A$ by neighborhoods, and
• $A$ is a neighborhood retract.

This is proven in Brown's Topology and Groupoids, page 128.

Since $X/A$ is the same as $X \cup_f \{*\}$ with $f:A\to\{*\}$ the constant map, this shows that $X/A$ is Hausdorff.