pullback square in Goerss-Jardine

Question

In proving the the existence of the Reedy model structure on the category of simplicial objects in a model category $\mathcal{C}$, Goerss-Jardine prove there is a pullback square induced by a map of simplicial objects $f:X\to Y$, namely $$\begin{array}{ccc} Y_n\times_{M_{n,k+1}}M_{n,k+1}X& \to & X_{n-1}\\ \downarrow & & \downarrow\\ Y_n\times_{M_{n,k}Y}M_{n,k}X& \to & Y_n\times_{M_{n-1,k}Y}M_{n-1,k}X \end{array}$$ and I am having some trouble seeing exactly how they derive this pullback. For reference, this is lemma VII.2.2 in Goerss-Jardine.

Here is what I have attempted so far. Here, Goerss-Jardine define $$ M_{n,k}X = M_{\Delta^{n,k}}X $$ where $\Delta^{n,k}$ is the simplicial set $d^0\Delta^{n-1}\cup \cdots \cup d^k\Delta^{n-1}$ and for a simplicial set $K$ and a simplicial object $X$ in $\mathcal{C}$, you can take $$ M_KX:= \hom^{\Delta^\mathrm{op}}(K,X) $$ From a standard pushout $$\begin{array}{ccc} \Delta^{n-1,k}& \to & \Delta^{n-1}\\ \downarrow & & \downarrow\\ \Delta^{n,k}& \to & \Delta^{n,k+1} \end{array}$$ Since $M_{(-)}X$ takes pushouts to pullbacks, we get a pullback $$\begin{array}{ccc} M_{n,k+1}X& \to & X_{n-1}\\ \downarrow & & \downarrow\\ M_{n,k}X& \to & M_{n-1,k}X \end{array}$$

and we get a similar pullback for the partial matching objects on $Y$. My attempt at getting the desired pullback was to enlarge the diagram to the following 3-by-3 diagram $$ \begin{array}{ccccc} Y_n & \to & Y_{n-1} & \leftarrow & X_{n-1} \\ \downarrow & & \downarrow & & \downarrow \\ M_{n,k}Y & \to & M_{n-1,k}Y & \leftarrow & X_{n-1}\\ \uparrow & & \uparrow & & \uparrow \\ M_{n,k}X & \to & M_{n-1, k}X & \leftarrow & X_{n-1} \end{array} $$ The desired pullback diagram is then supposed to be obtained from this 3-by-3 diagram by taking the horizontal limits and then taking the corresponding pullback. I tried commuting these two limits, so I thought of taking the limits of the row and then taking the corresponding pullback, but this didn't really seem to simplify the problem.

I am not sure if my attempt can be made to work and I am just not seeing something, or perhaps I am unaware of some category theory trick that Goerss-Jardine are using. Anyway, any hints would be greatly appreciated.

Answer 1

I would understand this proof as describing the limit of a "deleted 3-cube" in two different ways.

We have a square involving maps $X_n\to X_{n-1}$, $X_n\to M_{n,k}X$, $X_{n-1}\to M_{n-1,k}X$, and $M_{n,k}X\to M_{n-1,k}X$; this square maps to a similar square with $X$ replaced with $Y$. Altogether it is a cube. I can't draw this cube here.

Now let's completely forget about the "maximal vertex" $X_n$. The pullback you want to compute is actually the inverse limit of the deleted cube. (It is also equal to the limit of your 3-by-3 diagram.)

When you compute the limit of a deleted 3-cube, you can slice it up in several different ways, so that it is equal to a pullback of three objects, one of which is a vertex, and the other two are pullbacks. Goerss-Jardine have set it up one way. But we can also slice it up as the limit of $$ Y_n \to Y_{n-1}\times_{M_{n-1,k}Y}M_{n,k}Y \leftarrow X_{n-1}\times_{M_{n-1,k}X} M_{n,k}X. $$ GJ identify the rightmost term as $M_{n,k+1}X$. The same argument shows that the middle term is equal to $M_{n,k+1}Y$.