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$\require{AMScd}$If $\mathcal{X}$ is a category and $I$ a small category, the category of functors $\mathcal{X}^I$ inherits a (orthogonal) factorization system for each (orthogonal) factorization system on $\mathcal{X}$, defining the two classes objectwise.

It seems to me that I can define this factorization system "formally" in this way. Call $(\cal A,B)$ the factorization system on $\mathcal X$. Then there is a pullback diagram $$ \begin{CD} \mathcal{A}^I @>>> \mathcal{A}^{|I|} \\ @VVV @VVp_*V\\\ \mathcal{X}^I @>>j^*> \mathcal{X}^{|I|} \end{CD} $$ where $j : |I|\hookrightarrow I$ is the inclusion of the discrete subcategory of $I$ into $I$ itself, $j^*$ is the induced functor, and $\mathcal{A}$ is regarded a the nonfull subcategory of $\mathcal{X}$ on the arrows of $\cal A$ and $p_*$ comes from the obvious functor $\mathcal A \to \mathcal X$.

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If by $\mathcal{A}^I$ you mean the non-full subcategory of $\mathcal{X}^I$ corresponding to the left class of the induced factorization system (which is not the functor category of $I$ into $\mathcal{A}$), then yes, it does fit into such a pullback square. This doesn't construct the whole factorization system however.

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  • $\begingroup$ ${\cal A}^I$ is precisely what you said (the notation was ambiguous, I must admit). Why it doesn't construct the FS? Once you have a class you get the other by orthogonality. $\endgroup$
    – fosco
    Commented Apr 18, 2017 at 16:05
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    $\begingroup$ And then you have to prove that it actually is a factorization system. I'm not saying that it's hard, just that there's more to it than just writing down a pullback square. $\endgroup$
    – Mike Shulman
    Commented Apr 19, 2017 at 15:39

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