How are simplicial sets with Quillen model structure a simplicial model category?

Question

I got very lost in checking that simplicial sets with Quillen model structure are indeed a simplicial model category.

Recall that a model category $\mathcal{M}$ is simplicial if it is enriched in $\mathbf{SSet}$, powered by $\mathbf{SSet}$ and tensored by $\mathbf{SSet}$ so that the natural adjunctions exist and and also the enrichment is compatible with the model structure -- namely, for $i: A \to B$ a cofibration in $\mathcal{M}$ and $p: X \to Y$ a fibration in $M$, the map of simplicial sets $i^*\times p_*: \operatorname{Map}(B,X) \to \operatorname{Map}(A,X)\underset{\operatorname{Map}(A,Y)}{\times}\operatorname{Map}(B,Y)$ is required to be a fibration in $\mathbf{SSet}_{Quillen}$, and it is required to be trivial if either $i$ or $p$ is.

Recall that for simplicial sets, the enrichment is given by $\operatorname{Map}(K,M)_n = \mathbf{SSet}(\Delta^n \times K, M).$

$\mathbf{SSet}_{Quillen}$ is a cofibrantly generated model category, so to check the compatibility one must fill the horns: $\require{AMScd}$ \begin{CD} \Lambda^n_i @>h>> \operatorname{Map}(B,X) \\ @V \iota V V @VV i^* \times p_* V\\ \Delta^n @>(\alpha,\beta)>> \operatorname{Map}(A,X)\underset{\operatorname{Map}(A,Y)}{\times}\operatorname{Map}(B,Y) \end{CD}

The filling would be the lift in the diagram

$\require{AMScd}$ \begin{CD} \Delta^n \times A @>\alpha>> X \\ @V 1\times i V V @VV p V\\ \Delta^n \times B@>\beta>> Y \end{CD}

The natural idea is to consider the diagram

$\require{AMScd}$ \begin{CD} \Lambda^n_i \times B @>h>> X \\ @V \iota\times 1 V V @VV p V\\ \Delta^n \times B@>\beta>> Y \end{CD}

where the left map is a trivial cofibration, the right map is a fibration and thus from the lifting properties I obtain $\sigma: \Delta^n \times B \to X$ satisfying that $p\sigma = \beta$. However, if I want this $\sigma$ to be the answer, I must check $\sigma(1 \times i) = \alpha$... Would that be so?

I see that I have not made use of $i$ being an inclusion, but I don't see how it helps.

Answer 1

The trick is to check that the corner map $$\lambda^n_k\bar{\times}\delta^m:\Lambda^n_k \times \Delta^m \coprod_{\Lambda^n_k\times \partial \Delta^m} \Delta^n \times \partial \Delta^m \hookrightarrow \Delta^n\times \Delta^m$$ is anodyne for all $k, m, n$ appropriate. This is proven in Higher Topos Theory chapter 2, for your reference.

The fact that $i$ was an inclusion allows you to build it up as a relative cell complex of boundary inclusions, so it is enough to prove the case where $i$ is a boundary inclusion.