# Coend of full subcategory

$$\require{AMScd}$$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?

• 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 $C$. – Tim Campion Feb 12 at 15:42
• 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 – Bipolar Minds Feb 12 at 16:09