2
$\begingroup$

$\require{AMScd}$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 $\mathcal A \to \mathcal X$.

Is this rewriting correct?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ ${\cal A}^I$ 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 Apr 18 '17 at 16:05
  • 1
    $\begingroup$ 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 Apr 19 '17 at 15:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.