2
$\begingroup$

$\require{AMScd}$Let $F:\mathcal{C}^{op}\times \mathcal{C} \to \mathcal{D}$ be a functor and $\mathcal{C}' \subseteq \mathcal{C}$ a full subcategory. Assume that the coends $C$ over $F$ and $C'$ over $F|_{\mathcal{C}'}$ exist. By the universal property of $C'$, we obtain a map $i:C' \to C$.

I am looking for a necessary and sufficient condition for this to be a monomorphism.

As it is pointed out in the comments, the following part doesn't really make sense:

I think we can consider $C$ as a pushout of the diagram $$ \begin{CD} \coprod_{c\in\mathcal C'} F(c,c) @>>> C' \\ @VVV \\ \coprod_{c\in\mathcal C} F(c,c) \end{CD} $$ in the category of cocones $Cocone(F)$ of $F$. Hence, if coprojections are monos in $\mathcal{D}$ and pushouts preserve monos in $Cocone(F)$, $i$ is a monomorphism. However, the last condition seems to be a bit strong, so is there another argument I'm not seeing? In particular, is it true for abelian categories?

$\endgroup$
  • 1
    $\begingroup$ I'm not sure how to regard the above pushout diagram as "in $Cocone(F)"$. Anway, $C$ is not the pushout of that diagram regarded as a diagram in $\mathcal D$. For the map $\amalg_{c \in C} F(c,c) \to C$ is a coequalizer of relations coming from each morphism in $C$, and even if $C'$ is a full subcategory of $C$, it will not "see" all of these relations. Eg, consider the case where $C'$ is the empty category -- then $\amalg_{c \in C'} F(c,c) \to C'$ is the identity map on the initial object, so the pushout is likewise isomorphic to $\amalg_{c \in C} F(c,c)$, and thus is not the same as $C$. $\endgroup$ – Tim Campion Feb 12 at 15:42
  • $\begingroup$ Sorry, what I wrote in the last paragraph is complete nonsense, I thought I could make sense of this as a diagram of cocones (or sometimes called cowedges since we are dealing with dinatural transformations).. however, the coproducts are no cowedges, so it doesn't work $\endgroup$ – Bipolar Minds Feb 12 at 16:09

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.