# Bousfield-Kan: Cosimplicial Replacement of a fibration of diagrams is a fibration?

I've been working through Bousfield and Kan's book on Homotopy Limits for background work on my dissertation, and there is a statement that I've just not been able to solve (and my adviser doesn't really know how to fix the issue that I'm running into). I'll try and be brief about my question, since it is a bit involved. The statement I'm attempting to prove is this:

Chapter XI, Sect. 5, Proposition 5.3: Let $f: X \longrightarrow X'$ in $\mathcal{C}^{\mathcal{I}}$ be a morphism of diagrams such that $f_i : X_{i} \longrightarrow X'_{i}$ if a fibration for every $i \in \mathcal{I}$. Then if we apply cosimplicial replacement $\Pi^{\*}$, we have that $\Pi^{\*}(f): \Pi^{\*}X \longrightarrow \Pi^{\*}X'$ is a fibration of cosimplicial objects over $\mathcal{C}$.

So my question is simply this: How do we prove this statement?

Here are some of my thoughts:

Recall that the fibrations in the category of cosimplicial objects over $\mathcal{C}$ are morphisms $f: X \longrightarrow Y$ such that $(f^{n+1},s_{n}^{X}) : X^{n+1} \longrightarrow Y^{n+1} \times _{M^{n}Y} M^{n}X$ are all fibrations in $\mathcal{M}$ for all $n \ge -1$. Here $M^{n}X$ is the set of all tuples $(x^{0}, x^{1}, ..., x^{n}) \in X^{n} \times \cdots \times X^{n}$ such that $s^{i}x^{j}=s^{j-1}x^{i}$ for $0 \leq i < j \leq n$. (I've included this since this definition of matching space seems non-standard; it is not quite the same as the one given in Hirschhorn).

My idea has been to try and realize each morphism as a retract of a known fibration. Using the shorthand $X^{n+1}=(\Pi^{\*}(X))^{n+1}$, $Y^{n+1}=(\Pi^{\*}(Y))^{n+1}$, $M^{n}X=M^{n}(\Pi^{\*}(X))$ to denote the matching space (and similarly for $M^{n}Y$), and $Z = X^{n+1} \times_{Y^{n+1}} (Y^{n+1} \times_{M^{n}Y} M^{n}X)$, I can get the diagram with top row $X^{n+1} \longrightarrow Z \longrightarrow X^{n+1}$ where the first of these maps is a universal map for pullbacks and the second of these is just projection onto the first coordinate; and bottom row $Y^{n+1} \times_{M^{n}Y} M^{n}X \longrightarrow Y^{n+1} \times_{M^{n}Y} M^{n}X \longrightarrow Y^{n+1} \times_{M^{n}Y} M^{n}X$ where both maps are the identity. The vertical maps on the left and right are $f: X^{n+1} \longrightarrow Y^{n+1} \times_{M^{n}Y} M^{n}X$ where $f = (\alpha^{n+1},s_{n}^{X})$ (here $\alpha^{n+1} : X^{n+1} \longrightarrow Y^{n+1}$ and $s_{n}^{X}: X^{n+1} \longrightarrow M^{n}X$ is the natural map for matching spaces). This map $f$ is the map I want to show is a fibraion. The middle vertical map is just projection onto the second coordinate.

This gives me a possible retract diagram (I'm sorry it's so poorly described; I don't know how to get diagrams to work on this site). The left square commutes by construction (although I have brushed over a lot of the construction), but the right square does not, or at least not necessarily. It's here that I'm a bit lost...

In any event, does anyone know the full proof of the statement I gave above? Am I on the right track? And if not, what approach should I be taking?

Thank you so much!

-
Usually the trick is to realize that the cosimplicial remplacement of a diagram satisfies $X^{n+1}\cong F^{n+1}\times M^n X$ where $F^{n+1}$ is the 'cofree' part of $X$ in degree $n+1$. This part should be explicitly computable in terms of the original diagram. Once you know this, the fibered products simplify and the matching maps should be coordinatewise fibrations, hence a fibration. –  Justin Noel Dec 20 '11 at 8:58

The cosimplicial replacement of a diagram $F:J\rightarrow \mathcal C$ produces a codegeneracy free cosimplicial object $X^\bullet$. In particular, $X^{n}=Z^nX\times M^{n-1}X$ where $$Z^nX(j)=\prod_{f\in J_n} F(\mathrm{target}f)$$ where $J_n$ is the set of all $n+1$ composable arrows in $J$, with the first arrow originating at $j$ and none of which are the identity. The matching object can be identified with those arrows that have at least one identity morphism.
Now switching to your notation we are trying to see if the map: $$X^{n+1}\cong Z^{n+1}X \times M^n X\rightarrow Y^{n+1}\times_{M^n Y} M^n X\cong Z^{n+1}Y\times M^n X$$ is a fibration. The map is the identity on $M^n X$ and the map $Z^{n+1} X\rightarrow Z^{n+1} Y$ is just the product, indexed as above, of levelwise fibrations, hence a fibration. Note it is also easy to see that this construction takes levelwise weak equivalences to weak equivalences.