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Let $X$ be a fibrant pointed cosimplicial space.
Following Bousfield-Kan, let $\text{lim}^{\partial \Delta_{n+1}} X = M^{n}X$ be the nth matching object of $X$. Onecan then show that there is a pullback square: $$\require{AMScd} \begin{CD} M^{n-1,k+1}X @>>> X^{n-1} \\ @VVV @VVV \\ M^{n-1,k}X @>>> M^{n-2,k}X \end{CD}$$ where $M^{n,k}X = lim^{\partial \Delta_{{n+1},k}} X$ where $\partial \Delta_{n+1,k} $ is the smallest cosimplicial set containing the images of $\Delta^{n-1}$ under $s_i$ for $i=0, \ldots , k$. I hope I didn't make any errors here, but the idea is that we can build the matching object inductively.
Consider then the following two statements, for $t \geq 2$:
- H(n,k): The natural map $\pi_t M^{n-1,k}X \rightarrow M^{n-1,k} \pi_t X$ is an isomorphism.
- I(n,k): Let $N^{n,k}X$ be the fibre of $s:X^n \rightarrow M^{n-1,k}X$. Then
$$ 0 \rightarrow \pi_t N^{n,k}X \rightarrow \pi_t X^n \rightarrow \pi_t M^{n-1,k}X \rightarrow 0$$ is short exact.

We want to prove these two statements for all n and k by double induction. One easily verifies that $I(0,0)$ , $H(n,0)$ and $I(n,0)$ are true. So suppose that we know that $H(n,k)$,$H(n-1,k)$ and $I(n-1,k)$ yields $H(n,k+1)$. Goerss-Jardine, on pg. 392 writes that this follows since $\pi_t$ applied to our above diagram gives a pullback diagram

$$\require{AMScd} \begin{CD} \pi_tM^{n-1,k+1}X @>>> \pi_t X^{n-1} \\ @VVV @VVV \\ M^{n-1,k}\pi_t X @>>> M^{n-2,k}\pi_t X \end{CD}.$$

So, my question is: why is this a pullback?

I can see that if we apply $\pi_t$ and use $H(n,k)$ and $H(n-1,k)$ we have that we can identify $\pi_t M^{n-1,k} X $ with $M^{n-1,k} \pi_t X$. Then, I agree that we have a pullback diagram as follows, since weighted limits are cocontinuous:

$$\require{AMScd} \begin{CD} M^{n-1,k+1}\pi_t X @>>> \pi_t X^{n-1} \\ @VVV @VVV \\ M^{n-1,k}\pi_t X @>>> M^{n-2,k}\pi_t X \end{CD}$$ but I don't see why this implies that the previous diagram is a pullback. Maybe I am missing something obvious, but any help would be welcome!

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