$(i_0,i_1):A\vee A\rightarrow A\wedge I_+$ is a cofibration

Question

This is May & Ponto's More concise algebraic topology. They claimed that $(i_0,i_1):A\vee A\rightarrow A\wedge I_+$ is a cofibration. But I don't know why.

Question: Why the map $(i_0,i_1):A\vee A\rightarrow A\wedge I_+$ is a cofibration.

Could anybody give a proof?

Answer 1

Consider the folding map $\nabla:A\vee A\rightarrow A$, which is defined to be the identity on each of the wedge summands, and the map $j$ which results by replacing $\nabla$ by a cofibration using the pointed mapping cylinder. Thus we let $M_\nabla$ be the pushout of $$(A\vee A)\wedge I_+\xleftarrow{in_0}A\vee A\xrightarrow\nabla A$$ and have the cofibration $j:A\vee A\hookrightarrow M_\nabla$ which is induced by the inclusion of $A\vee A$ into the top end of the cylinder.

To proceed we identify $$(A\vee A)\wedge I_+\cong (A\wedge I_+)\vee (A\wedge I_+)$$ and so have that $M_\nabla$ is the quotient $$M_\nabla\cong \frac{(A\wedge I_+)\vee (A\wedge I_+)\vee A}{[((a,0),\ast)\sim (\ast,(a,0))\sim a]}.$$ The identifications glue the two cylinders together along the $A$ summand. Up to homeomorphism what is left is $$M_\nabla\cong A\wedge I_+.$$ Under this identification, one of the cylinders is turned upside down. The map $j$ now includes one summand of $A$ in at the bottom of $A\wedge I_+$ and the other in at the top.

Of course a homemorphism is a cofibration and a composite of cofibrations is a cofibration, so we conclude that $$j=(in_0,in_1):A\hookrightarrow A\wedge I_+$$ is a cofibration.

Answer 2

In general it is not true. However, May and Ponto do not consider arbitrary pointed spaces. On p. 7 they make the following general assumption

We nearly always work with based spaces. To avoid pathology, we assume once and for all that basepoints are nondegenerate, meaning that the inclusion $∗ \to X$ is a cofibration.

Note the May and Ponto define free and based cofibrations and clarify their relationship in Lemma 1.3.4. On p. 20 they say

One of the many motivations for our standing assumption that basepoints are nondegenerate is that it ensures that based maps are cofibrations or fibrations in the free sense if and only if they are cofibrations or fibrations in the based sense.

Let us first observe that for a based space $A = (A,a_0)$ we can identify $A \wedge I_+$ with the quotient space $A \times I / \{a_0\} \times I$.

$A \vee A$ can be regarded as the quotient space obtained from $A \times \{0,1\} \cup \{a_0\} \times I$ by identifying $ \{a_0\} \times I$ to a point. Then $(i_0,i_1) : A \vee A \to A \times I / \{a_0\} \times I$ is given as the map induced on the quotients by the inclusion $i : A \times \{0,1\} \cup \{a_0\} \times I \to A \times I$. The following diagram is a pushout diagram:

$\require{AMScd}$ \begin{CD} A \times \{0,1\} \cup \{a_0\} \times I @>{i}>> A \times I \\ @V{}VV @V{}VV \\ A \vee A @>{(i_0,i_1)}>> A \times I / \{a_0\} \times I \end{CD}

Therefore, if $i$ is a cofibration, then so is $(i_0,i_1)$.

But if $A$ is nondegenerately based, then it is well-known that $i$ is a cofibration. Thus, under May and Ponto's general assumption $(i_0,i_1)$ is in fact a cofibration.

For degenerately based $A$ is in general not true. As an example take $A = \{0\} \cup \{ 1/n \mid n \in \mathbb N \}$ with basepoint $0$. It is not hard to show that $(i_0,i_1)$ is no based cofibration (hence no free cofibration); you can do that as a nice exercise.