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?
Best Answer
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.