# Factorization of morphisms in a diagram category

Let us suppose that $I$ is a small category and $\mathcal{E}$ a combinatorial model category. Then there exists two Quillen equivalent combinatorial model category structures on the diagram category $\mathcal{E}^I$, the projective one (fibrations and weak equivalences are pointwise) and the injective one (cofibrations and weak equivalences are pointwise). One can look for instance in Appendix A of Higher Topos Theory, Jacob Lurie, for a proof.

I would like to know if it is possible, maybe under some hypotheses on $I$, to obtain the following factorization property that mixes both model structures: any morphism factors through an acyclic projective cofibration followed by an injective fibration.

It is not possible in the case $\mathcal{E} = \mathbf{sSets}$, $\mathcal{I} = (\cdot \to \cdot)$.

To see this, note that since $\mathcal{I}$ is both an inverse and a direct category, the projective cofibrations (resp. injective fibrations) are exactly the Reedy cofibrations (resp. fibrations).


$$\begin{array}{c} A & \ra{h} & C \\ \da{f} & & \da{g} \\ B & \ra{k} & D \end{array}$$

factors as $$\begin{array}{c} A & \ra{h_1} & A' & \ra{h_2} & C \\ \da{f} & & \da{f'} & & \da{g} \\ B & \ra{k_1} & B' & \ra{k_2} & D \end{array}$$ with the left square a trivial Reedy cofibration, and the right square a Reedy fibration.

In particular, if $f$ is a cofibration and $g$ a fibration, then $f'$ must be both a cofibration and a fibration; so (in $\mathbf{sSets}$) it must just be an inclusion of connected components, and in particular must induce isomorphisms on $\pi_k$ for $k > 0$.

But $f'$ is weakly equivalent to $f$, so if $f$ acts non-trivially on some $\pi_k$, $k > 0$, this is impossible. For instance, take $A = C = D = 1$, and $B = S^1$; then such a factorisation is impossible.

• We can exclude many more categories by observing that such a factorization property implies that the injective and the projective model structures coincide. (Factor an acyclic injective cofibration and use the retract argument to see that it is projective.) I suspect that the only $I$s that work are (equivalent to) the discrete ones, but I don't really see how to prove it. – Karol Szumiło Nov 25 '13 at 17:08