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I'd like to prove that if the topological pair $(X,A)$ has the homotopy extension property, then so does $(X\cup CA,CA)$, where $CA$ is the cone over $A$ and where $A\times\{0\}$ is identified with $A$. My idea is the following:

If $f:CA\times I\to Y$ is a homotopy and $g:X\cup CA\to Y$ a given continuous function with $g|_{CA}=f(\cdot,0)$, then in particular $g|_{A} = f|_{A\times I}(\cdot,0)$, and thus by hypothesis there exists a homotopy $\tilde{F}:X\times I\to Y$ such that $\tilde{F}|_{A\times I}=f|_{A\times I}$ and $g|_{X} = \tilde{F}(\cdot,0)$. Then intuitively the function $F:(X\cup CA)\times I\to Y$ defined by $F|_{CA\times I} = f$ and $F|_{X\times I} = \tilde{F}$ should work. However, I don't see how to prove that $F$ is continuous. If we suppose additionally that $A$ is closed in $X$, then $F$ is continuous, because if $B\subseteq Y$ is closed, then so is $F^{-1}(B)=(\tilde{F}^{-1}(B)\cap(X\times I))\cup (f^{-1}(B)\cap(CA\times I))$, because $X\times I$ and $CA\times I$ are closed in $(X\cup CA)\times I$.

But how to prove that $F$ is continuous when we don't suppose $A$ to be closed in $X$?

I don't consider this question a duplicate of if $(X,A)$ has homotopy extension, so does $(X \cup CA,CA)$, because I'm asking about whether the sketch above can be completed, and furthermore I don't understand the answers given over there.

Redundant Aunt
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  • In a nice category of spaces, $A$ will always be closed. If you do not want to assume that $A$ is closed, see Hatcher: prop A.18 math.cornell.edu/~hatcher/AT/ATapp.pdf –  May 24 '18 at 15:06

1 Answers1

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Let $p :A \times [0,1] \to CA = A \times [0,1]/A \times \{ 1 \}$ denote the quotient map and $j : A \to CA$ be the embedding defined by $j(a) = p(a,0)$. It is well-known and easy to verify that $j$ is a (closed) cofibration.

The space $X \cup CA$ is defined as the pushout of the two maps $i :A \hookrightarrow X$ and $j$. It comes with maps $i' : CA \to X \cup CA$ and $j' : X \to X \cup CA$ such that $j' \circ i = i' \circ j$ and such that the usual universal property is satisfied.

It is well-known that pushouts preserve cofibrations, i.e. if one of the base maps is a cofibration, then the map on the opposite side of the pushout diagram is also one. This applies to $i'$.

Since cofibrations are always embeddings, we see that $CA$ can be regarded as subspace of $X \cup CA$ which justifies the notation $(X \cup CA, CA)$.

Moreover we see that $j' : X \to X \cup CA$ is always a cofibration. Since $j$ is closed, the same holds for $j'$.

Paul Frost
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