Very good cylinder and strong deformation retract

Question

Let $\mathcal{M}$ be a model category and let $C:\mathcal{M}\to\mathcal{M}$ be a very good cylinder object. The natural transformations coming with $C$ are denoted by $\gamma^\epsilon_X:X\to CX$ with $\epsilon=0,1$ and $\sigma_X:CX\to X$. In this setting, a deformation retract is a map $i:Z\to Y$ such that there exists a retract $r:Y\to Z$ ($ri=1_Y$) such that the maps $ir$ and $1_Y$ are left homotopic, i.e., there exists a map $h:CY\to Y$ such that $h\gamma^0_Y=ir$ and $h\gamma^1_Y=1_Y$. The deformation retract is strong if moreover $h$ can be chosen so that $\sigma_Y C(i) = h C(i)$.

I can prove that $\gamma^0_X:X\to CX$ is a deformation retract, I would like to prove that it is a strong deformation retract.

Proof that it is a deformation retract. Consider the commutative diagram $\require{AMScd}$ \begin{CD} CX \sqcup CX @>\gamma^0_X\sigma_X \sqcup 1_{CX} >> CX\\ @V \gamma_{CX} V V= @VV \sigma_X V\\ CCX @>>\sigma_X \sigma_{CX}> X. \end{CD} The left vertical map is a cofibration and the right vertical map is a trivial fibration since the cylinder is very good. Thus, the existence of a lift $h:CCX\to CX$.

Answer 1

This follows from Corollary XIII.5 of Williamson's thesis Cylindrical model structures, where he proves that a morphism is a trivial cofibration (in his model structure determined by the cylinder object) if and only if it admits a strong deformation retraction. Since you already know $\gamma_X^0$ is a trivial cofibration, you're done.

You might also be interested in Section XII of Williamson's paper might also be of interest, as it provides conditions on the cylinder object in order that various associated maps are strong deformation retractions.