Let $X$ be a Hausdorff topological space, $A$ subset of $X$. Let $I$ be an interval $[0,1] \subset {\mathbb R}$. Suppose that $X \times \{0\} \cup A \times I$ is closed in $X \times I$. Then, is $A$ closed in $X$?
It seems that it is , since that is the explanation in Algebraic Topology by Allen Hatcher of the claim:
If $(X,A)$ has the homotopy extension property, then $A$ must be a closed subspace of $X$, at least when $X$ is Hausdorff.
But, I don't understand why it is.
If $Y=A\times I \cup X\times\{0\}$ is closed in $X\times I$, then $Y\cap (X\times\{1\})$ is closed in $X\times\{1\}$. But this set is just $A\times\{1\}$, and if $A\times\{1\}$ is closed in $X\times\{1\}$, then $A$ must be closed in $X$.