How to prove that the slice category over a Quillen model category is also a Quillen model category? I have to prove that slice category $\mathcal{C}/X$ for $\mathcal{C}$ a Quillen model category is also a Quillen model category. I proved 2 out of 3 axiom, but I'm stuck with the retract axiom. For me, it's obvious that diagram commutes (morphism in slice category), because of definition of retract.
 A: It looks like you've got the idea. Working in the slice category $\mathcal{C}/X$, where $\mathcal{C}=\{\mathcal{C},F,\text{co}F,W\}$ is a model category with fibrations $F$, cofibrations $\text{co}F$, and weak equivalences $W$, we want to show $\mathcal{C}/X$ satisfies the axiom that retracts of morphisms in $F,\text{co}F,$ or $W$ are also in $F,\text{co}F,W$ respectively. 
So suppose we have 
$$\begin{matrix} A&&\stackrel{f}{\to}&&A'\\&\searrow& &\swarrow&\\&&X&&\end{matrix}$$
as a retract of
$$\begin{matrix} B&&\stackrel{g}{\to}&&B'\\&\searrow& &\swarrow&\\&&X&&\end{matrix}$$
That means we have maps $i:A\to B, r:B\to A, j:A'\to B',s: B'\to A'$ such that $ri=\text{id}_A, sj=\text{id}_A'$, everything commutes with the maps from $A,A',B,B'$ to $X$, and the squares commute: $gi=jf, fr=sg$.
But we can send this diagram to $\mathcal{C}$ just by forgetting the slice structure: ignore the bit about commuting with maps to $X$ in the previous sentence and it's expressing exactly that $f$ is a retract of $g$ in $\mathcal{C}$. Then $f$ is a weak equivalence (fibration, cofibration) between $A$ and $A'$ in $\mathcal{C}$, so that it'll be a weak equivalence (fibration, cofibration) between $A\to X$ and $A'\to X$ in $\mathcal{C}/X$ whenever this same diagram commutes:
$$\begin{matrix} A&&\stackrel{f}{\to}&&A'\\&\searrow& &\swarrow&\\&&X&&\end{matrix}$$
And in the given case we required that it does commute, so we have a weak equivalence, fibration, or cofibration in $\mathcal{C}/X$.
