Coend of $\mathscr{D}(F(\bullet), G(\bullet))$

Asked Sep 24 '16 at 23:33

Active Sep 25 '16 at 01:14

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(Crosspost from Overflow)

Given categories $\mathscr{C}$ and $\mathscr{D}$ and functors $F,G: \mathscr{C} \to \mathscr{D}$, we can form a bifunctor $$\mathscr{D}(F(\bullet), G(\bullet)): \mathscr{C}^\text{op} \times \mathscr{C} \to \mathsf{Set}$$ and the end of this functor is the set of natural transformations from $F \Rightarrow G$. (I guess we need $\mathscr{C}$ to be small in order to guarantee there is a set of such natural transformations, in general.)

Can I say anything about the coend? Is it some familiar thing?

edited Apr 13 '17 at 12:58

Community

asked Sep 24 '16 at 23:33

Eric Auld

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Just a comment: I have seen something looking a bit like this recently in a talk, in the context of derived functors/homology of functors, where the functors went to an abelian category and they considered the coend of the (Deligne) box tensor product of the two functors. I don't know much about the subject, though. You might try to crosspost to MO. – Daniel Robert-Nicoud Sep 24 '16 at 23:59
@DanielRobert-Nicoud Thanks, I crossposted it – Eric Auld Sep 25 '16 at 00:28
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You're welcome :) Put a link in each question redirecting to the other one, please. – Daniel Robert-Nicoud Sep 25 '16 at 00:29
I'm voting to close this question as off-topic because it has been cross-posted and answered at mathoverflow. – Apr 03 '18 at 10:18
Possible duplicate of Question on the proof of the existence of rank function – Eric Auld Apr 04 '18 at 01:51

Coend of $\mathscr{D}(F(\bullet), G(\bullet))$

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