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$\require{AMScd}$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?

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    $\begingroup$ 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
    Commented Feb 12, 2019 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
    Commented Feb 12, 2019 at 16:09

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