The double mapping track is an absolute neighborhood retract

Question

I have a question that is likely true and can probably be easily resolved, but I'm unsure where to begin. It pertains to a result that can be found without proof, for instance, in G. Warner's book "Topics in Topology and Homotopy Theory" on page 6-26.

The result states the following:

Consider continuous maps $X\stackrel{f}{\longrightarrow}Z\stackrel{g}{\longleftarrow}Y$, where $X$, $Y$, and $Z$ are ANRs (Absolute Neighborhood Retracts). Then, the double mapping track $W_{f,g}$ of $f$ and $g$ is also an ANR. Here, the double mapping track is understood to be the standard (or canonical) homotopy pullback of $f$ and $g$: $$W_{f,g}:=\{(x,y,\alpha)\in X\times Y\times Z^I:\alpha(0)=f(x), \alpha(1)=g(y)\}$$

Is there any reference where the complete proof of this result can be found? If not, are there any hints about its proof? Thank you.

Answer 1

Since $I$ is compact, $Z^I$ is metrisable in the compact-open topology. In fact, $Z^I$ is also an ANR, although we will not use this directly. What we need is that the product $X\times Y\times Z^I$ is metrisable, which implies that so is its subspace $W_{f,g}$.

A metrisable space is an ANR for metric spaces if and only if it is absolute neighbourhood extensor (ANE) for metric spaces.

Thus to show that $W_{f,g}$ is an ANR it will suffice to show that it is an ANE.

For this we realise $W_{f,g}$ as the categorical pullback of the cospan $$X\times Y\xrightarrow{f\times g}Z\times Z\xleftarrow{e_{0,1}}Z^I,$$ where $e_{0,1}(\ell)=(\ell(0),\ell(1))$. Given a metrisable space $M$ and a map $\phi:A\rightarrow W_{f,g}$ from its closed subspace $A\subseteq M$, we will show how to extend $\phi$ over a neighbourhood $V\subseteq M$ of $A$ in $M$.

To this end, let $\phi_X:A\rightarrow X$, $\phi_Y:A\rightarrow Y$, and $\phi_Z:A\rightarrow Z^I$ be the maps obtained by projection. Since $X,Y$ are ANEs, the first two maps have extensions $$\widetilde\phi_X:U_X\rightarrow X,\qquad\qquad \widetilde\phi_Y:U_Y\rightarrow Y,$$ where $U_X,U_Y\subseteq M$ are open neighbourhoods of $A$. Restricting to $A$ we have $$f\widetilde\phi_X|_A=f\phi_X\simeq g\phi_Y=g\widetilde\phi_Y|_A.$$ Indeed, $\phi_Z:A\rightarrow Z^I$ is exactly such a homotopy.

Suppose that $N$ is metrisable and $B\subseteq N$ is closed. If $Z$ is an ANE and $\alpha,\beta:N\rightarrow Z$ satisfy $\alpha|_B\simeq\beta|_B$, then there is a neighbourhood $V\subseteq N$ of $B$ and a homotopy $\alpha|_V\simeq\beta|_V$.

We apply this with $N=U_X\cap U_Y$ and $B=A\subseteq U_X\cap U_Y$. We obtain an neighbourhood $V\subseteq U_X\cap U_Y$ of $A$ and a homotopy $$\lambda:V\rightarrow Z^I,\qquad f\widetilde\phi_X|_V\simeq g\widetilde\phi_Y|_V.$$ Since $U_X\cap U_Y$ is open in $M$, $V$ is also a neighbourhood of $A$ in $M$.

Now, by construction $$e_{0,1}\circ\lambda|_V=(f\widetilde\phi_X|_V)\times (g\widetilde\phi_Y|_V).$$ Therefore, by the universal property of the pullback we obtain $\widetilde\phi:V\rightarrow W_{f,g}$ satisfying $$\pi_X\widetilde\phi=\widetilde\phi_X|_V,\qquad \pi_Y\widetilde\phi=\widetilde\phi_Y|_V,\qquad \pi_{Z^I}\widetilde\phi=\widetilde\lambda|_V,$$ where $\pi_X,\pi_Y,\pi_{Z^I}$ are the three projections.

Checking the universal property we obtain $$\widetilde\phi|_A=\phi,$$ which was what was needed to be shown.