Functors preserve homotopy pushouts and homotopy sums

category-theoryhomotopy-theorymodel-categories

Let $\mathcal {M}$ and $\mathcal {N}$ be two cofibrantly generated model categories. Let $I$ be a small category. Then $\mathcal M^I$ has the projective model structure. The colimit functor is hence a left Quillen functor and we can define the homotopy colimit functor to be the left derived functor of the ordinary colimit functor.

Now let $F:\mathcal M\rightarrow \mathcal N$ be a functor preserving homotopy pushouts and homotopy infinite sums. I want to show that $F$ also preserves homotopy colimits.

So let $X:I\rightarrow M$ be a diagram in $\mathcal M$. Then we take the cofibrant replacement $X'=Q(X)$. Let $P=\oplus_{i\in I}{X'_i}$ and $Q=\bigoplus\limits_{m:i\rightarrow j\\\in\ I}P_m$ where $P_m$ is a copy of $P$ for each $m$. Define the codiagonal map $\nabla:Q\rightarrow P$ to be induced by the identity map of $P$ for each $m$ and $\mu:Q\rightarrow P$ defined by $\mu\circ q_m\circ p_i=p_j\circ X'(m) $ where $p_i:X'_i\rightarrow P$ and $q_m:P_m\rightarrow Q$ are the canonical morphisms. Then $\varinjlim X'$ is the pushout of the diagram defined by $\nabla$ and $\mu$. In order to show the pushout of this diagram is the homotopy pushout, I need to show that at least one of $\mu$ and $\nabla$ is cofibration. I will be appreciated for any hint and comment.

Best Answer

Neither map is a cofibration in general. First, for $Q$ you must want $\oplus_m X'_i$. Now consider a cofibrant diagram $x\to y \to z$. In general the induced maps $x\oplus y\to z$ and $x\oplus x \to x$ are not cofibrations.

The proof of this result in a general model category is a bit complex. One proves it for simplicial homotopy colimits via a skeletal induction, then for categories of simplices by Kan extending to push a diagram down to $\Delta^{op}$, then for all categories by proving that the projection of the category of simplices of $I$ to $I$ is homotopy final. One can also combine the first two steps by working skeletally in an abstract Reedy category. I'm not sure whether there's an easier proof for the cofibrantly generated case.

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