# 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.

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What are your proposed classes of weak equivalences, fibrations, and cofibrations? – Kevin Carlson Oct 17 '12 at 21:59
W.e. is diagram that comutes $h_1=h_2f$ and f is w.e. , $f:A->B, h_1:A->X, h_2:B->X$.Analog for fibration and cofibration – dandan Oct 17 '12 at 22:06
You're making potential answerers work hard to understand your problem. That's suboptimal for getting responses. Anyway, do you mean $f$ is a weak equivalence between $A$ and $B$ in the base category $C$ (apparently a model category) we're taking slices over $X$ in? – Kevin Carlson Oct 17 '12 at 22:09
I'm sorry because I'm not very clear in explaining,english is not my mother language, I'll try to do so now on. Yes,f is weak equivalence between A and B in the base C and slice category is $C/X$. – dandan Oct 17 '12 at 22:16
No worries. Next time, maybe it would help to write your question out in your native language, and decide whether it includes all the necessary information before translating into English. I've answered below. – Kevin Carlson Oct 18 '12 at 1:17

## 1 Answer

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 $\mathcal{C}/X$.

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