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This is a problem from the book "modern classical homotopy theory" which I can't solve.

Let $i : A \rightarrow X$ be a cofibration and $Y$ any space. Show that $i : A\times Y \rightarrow X\times Y$ is also a cofibration.

I am supposed to use the following result:

$i : A \rightarrow X$ is a cofibration if and only if the canonical map $T \rightarrow X\times I$ has a retraction, where $T$ is the pushout of the diagram $A\times I \leftarrow A \rightarrow X$.

I'm not able to construct a map $X\times Y\times I \rightarrow T_2$ without using projections (and I don't think that is the way), and even if that is ok I have not been able to show that the map is the retraction of $T_2 \rightarrow X\times Y \times I$. By $T_2$ I mean the pushout of $A\times Y\times I \leftarrow A\times Y \rightarrow X\times Y$

Ben
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2 Answers2

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Without loss of generality $i$ can be taken to be the inclusion of $A\subset X$, since cofibrations are embeddings. Let $T$ be the pushout of diagram $A\times\mathbb{I}\leftarrow A\stackrel{i}{\rightarrow}X$ and let $r:X\times\mathbb{I}\rightarrow T$ be a retraction. Under the extra condition that $Y$ is locally compact $T\times Y$ is the pushout of diagram $A\times\mathbb{I}\times Y\leftarrow A\times Y\stackrel{i\times1}{\rightarrow}X\times Y$. Then the existence of retraction $r\times1:X\times\mathbb{I}\times Y\rightarrow T\times Y$ tells us that $i\times1$ is a cofibration. However, you were supposed to prove this for an arbitrary $Y$ so the condition of 'being locally compact' is bothering. There is a solution for that. In fact the inclusion $i:A\rightarrow X$ is a cofibration if and only if $X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is a retract of $X\times\mathbb{I}$. A proof of this can be found in Algebraic Topology of Allen Hatcher on page 532. Here $X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is equipped with the subspace topology inherited from $X\times\mathbb{I}$, and if it is a retract of $X\times\mathbb{I}$ then it follows directly that $X'\times\left\{ 0\right\} \cup A'\times\mathbb{I}$ is a retract of $X'\times\mathbb{I}$ where $X':=X\times Y$ and $A':=A\times Y$. So inclusion $i\times1$ is a cofibration! We do not need pushouts here. Note that space $T$ can be seen as topological space having the same underlying set, but it is equipped with the topology final with respect to $\bar{i}$ and the embedding $X\rightarrow X\times\left\{ 0\right\} $. This topology can be properly finer than the subspace topology. In my view they sent you into the wrong direction by pointing you to the pushouts.

drhab
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  1. proof that $T\times Y\cong T_2$, or say that $T\times Y$ is the pushout of $A\times Y\times I\leftarrow A\times Y\to X\times Y$

    • by the universal property, we have two arrow $i_1: A\times Y\times I\to T\times Y$ and $i_2:X\times Y\to T\times Y$. and the pushout diagram in (1) commute
    • to proof universal arrow, suppose $K$ making the diagram commute, then by adjoint of $\times$, we have $K^Y$ commute in the pushout $A\times I\leftarrow A\to X$, so there is unique $T\to K^Y$ so there is unique $T\times Y\to K$

  2. so the retraction you wanted is $r\times \mathrm{id}_Y: X\times I\times Y\to T\times Y\cong T_2$

Minghao Liu
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