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.



Factorization of morphisms in a diagram category
**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).


So your condition would imply that any commutative square
$\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}$


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


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.


