Showing that a Cofibration has closed image

Question

I am learning cofibrations and have seen the answer here: Cofibration necessarily has closed image?. I have been able to show that if $i: A \to X$ is a cofibration then it is an embedding. If in the case that $X,A$ is Hausdorff, I now want to show that $i(A)$ is closed. I have shown that $i$ is injective and that it is a homeomorphism onto its image. I was thinking that this would imply that $i(A)$ is closed since $A $ is closed in the subspace topology. But I think this is wrong since I never use the Hausdorff condition.

Answer 1

You are right, each cofibration is an embedding (i.e. we have a homeomorphism $i' : A \to i(A)$, where $i(A)$ has the subspace topology inherited from $X$). However, in general $i(A)$ is not closed in $X$ as shown by the example in the above link.

Let us show that if $X$ is Hausdorff, then $A' = i(A)$ is closed. We know that there exists a retraction $r : X \times [0,1] \to X \times \{ 0 \} \cup A' \times [0,1]$. Then $r(\overline{A'} \times \{ 1\}) = r(\overline{A' \times \{ 1\}}^{X \times [0,1]}) \subset \overline{r(A' \times \{ 1\})}^{X \times \{ 0 \} \cup A' \times [0,1]} = \overline{A' \times \{ 1\}}^{X \times \{ 0 \} \cup A' \times [0,1]} = A' \times \{ 1\}$ which means that for $x \in \overline{A'}$ we can write $r(x,1) = (\rho(x),1)$ with a continuous map $\rho : \overline{A'} \to \overline{A'}$ such that $\rho(x) = x$ for $x \in A'$ and $\rho(\overline{A'}) = A'$. The identity on $\overline{A'}$ and $\rho$ agree on $A'$. Since $\overline{A'}$ is Hausdorff, we conclude $\rho = id$. This implies $\overline{A'} = id(\overline{A'}) = \rho(\overline{A'}) = A'$.