I'm having trouble with the proof of Proposition 2.4.9 in Hovey's Model Categories.

Proposition. Deformation retracts are closed under pushouts.

Proof. Suppose we have a pushout diagram $$\require{AMScd} \begin{CD} A @>{f}>> C\\ @V{i}VV @VV{j}V\\ B @>>{g}> D\end{CD}$$ where $i$ is the inclusion of a deformation retract. Since $I$ is compactly generated, $(-\times I) \dashv (I\Rightarrow -)$, so $(-\times I)$ preserves colimits, hence $D\times I$ is the pushout of $B\times I$ and $C\times I$ over $A\times I$. Let $K:B\times I\rightarrow B$ be a homotopy that makes $i$ into the inclusion of a deformation retract. Then $gK$ together with the map $C\times I\rightarrow D$ defined by $(c,t)\mapsto j(c)$ define, by the universal property of the pushout, an arrow $H:D\times I\rightarrow D$. By construction , $H(c,t)=j(c)$ for all $c\in C$, and $H(d,0)=d$ for all $d\in D$. Since $K(b,1)\in iA$ for all $b\in B$, it follows that $H(d,1)\in jC$ for all $d\in D$. Since $j$ is an inclusion map, $H$ is a deformation retraction, as required. $\square$

My problem: From the commutativity of the triangles from the universal diagram I did obtain $H(c,t)=c$. The second equation I got however is $H(g(b),t)=g(b)$ instead of $H(d,0)=d$. Why does the latter hold and how can I prove it? Also, how does $H(d,1)\in jC$ for all $d\in D$ follow?

Update: I managed to prove the theorem using the fact retracts are closed under pushouts along with the uniqueness of the induced arrow. I'm not sure whether this is what the author meant, so my question remains.

  • $\begingroup$ I think you confused the letters $B$ and $C$ since you compose $K$ with $g$ and you apply $j$ to an element of $C$. $\endgroup$ Mar 7 '15 at 2:21
  • $\begingroup$ @StefanHamcke thanks, I corrected the question. Also, could you take a look at my comments here? $\endgroup$
    – user153312
    Mar 7 '15 at 8:36

The fact $H(d,0) = d$ follows from the fact that $H(g(b),0) = gK(b,0) = g(b)$ and $H(j(c), 0) = j(c)$. Thus $H(-, 0)$ is the identity on both parts of $D = g(B) \cup j(C)$, so it is the identity (you can reformulate that in terms of universal property of the pushout: $H(-,0) : D \to D$ is the map induced by $g : B \to D$ and $j : C \to D$, but by definition of the pushout that map is the identity).

Similarly, $H(g(b), 1) = gK(b,1)$ belongs to $g(K(B,1)) \subset g(i(A)) = j(f(A)) \subset j(C)$, and again by construction $H(j(c),1) = j(c) \in j(C)$. Since $H(-,1) : D \to D$ is the map induced on $D$ by $gK(-,1)$ and $j$, and since both maps factors through $\operatorname{im}(j)$, it follows that $H(-,1)$ factors through it too.


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