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1. • CommentRowNumber1.
   • CommentAuthorFinnLawler
   • CommentTimeSep 9th 2011
   • PermaLink
   Author: FinnLawler
   Format: MarkdownItexEveryone knows that you can define (co)ends as certain (co)equalizers, as described at [[end]]. A while ago I added the section [Connecting the two definitions](http://ncatlab.org/nlab/show/end#connecting_the_two_definitions_30), which derives the equalizer description from the weighted-limit description by observing that the weight $\hom_C$ for an end is presented as a certain split coequalizer in $[C^{op} \times C, V]$. Now I'm puzzled, though, because this would seem to imply that ends are _split_ equalizers, and dually, split coequalizers being absolute colimits. But it's easy to cook up an example where this seemingly can't happen, say with a pair of categories and functors $F,G$ such that $Nat(F,G) = \int_c [F c, G c]$ is empty but $\prod_c [F c, G c]$ isn't, so there can't be any morphism from the latter to the former to make the whole thing a split equalizer. I must have made a mistake somewhere, and knowing me it'll be a stupidly obvious one, but can anyone point it out?

   Everyone knows that you can define (co)ends as certain (co)equalizers, as described at end. A while ago I added the section Connecting the two definitions, which derives the equalizer description from the weighted-limit description by observing that the weight $\hom_C$ for an end is presented as a certain split coequalizer in $[C^{op} \times C, V]$. Now I’m puzzled, though, because this would seem to imply that ends are split equalizers, and dually, split coequalizers being absolute colimits. But it’s easy to cook up an example where this seemingly can’t happen, say with a pair of categories and functors $F,G$ such that $Nat(F,G) = \int_c [F c, G c]$ is empty but $\prod_c [F c, G c]$ isn’t, so there can’t be any morphism from the latter to the former to make the whole thing a split equalizer.

   I must have made a mistake somewhere, and knowing me it’ll be a stupidly obvious one, but can anyone point it out?

2. • CommentRowNumber2.
   • CommentAuthorMike Shulman
   • CommentTimeSep 9th 2011
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   Author: Mike Shulman
   Format: MarkdownItexAre you sure you mean "split" and not [[reflexive coequalizer|reflexive]]?

   Are you sure you mean “split” and not reflexive?

3. • CommentRowNumber3.
   • CommentAuthorFinnLawler
   • CommentTimeSep 9th 2011
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   Author: FinnLawler
   Format: MarkdownItexOK, so not one but two stupidly obvious mistakes: 1. The monad on $[ob C \times ob C, V]$ for which $\hom_C$ is an algebra is not the one I said it was. 2. The coequalizer presenting $\hom_C$ is indeed reflexive but not split (the splitting I had in mind is not natural). It's still a coequalizer, though, so the basic idea stands. I'll fix [[end]] accordingly. Thanks.

   OK, so not one but two stupidly obvious mistakes:

   1. The monad on $[ob C \times ob C, V]$ for which $\hom_C$ is an algebra is not the one I said it was.
   2. The coequalizer presenting $\hom_C$ is indeed reflexive but not split (the splitting I had in mind is not natural). It’s still a coequalizer, though, so the basic idea stands.

   I’ll fix end accordingly. Thanks.