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- CommentRowNumber1.
- CommentAuthorUrs
- CommentTimeDec 21st 2009
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Author: Urs
Format: MarkdownExpanded the examples section at [[end]] (prodded by discussion in the blog), stating Kan extension and geometric realization. Also added a toc.

Expanded the examples section at end (prodded by discussion in the blog), stating Kan extension and geometric realization.

Also added a toc.
- CommentRowNumber2.
- CommentAuthorFinnLawler
- CommentTimeAug 18th 2011
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Author: FinnLawler
Format: MarkdownItexI've (tidied up a bit and) added a new subsection at [[end]] explaining why the hom-weighted (co)limit definition implies the (co)equalizer one. I wanted to add labels, à la AMSLatex's `equation` environment, to a couple of displayed diagrams, but I can't remember/never knew if iTeX can do that. Anyone know? Incidentally, the hom-weighted limit definition of ends follows, at least when V=Set, from the universal-extranatural one once you know that there is a Yoneda-type lemma for extranaturals giving a bijection between Exnat(pt, F) and Nat(hom_C, F) (maths rendering doesn't seem to be working here, at least in the preview). Has anyone seen this extranatural-Yoneda written down elsewhere? I came up with it myself a while ago, but it must have been noticed before.

I’ve (tidied up a bit and) added a new subsection at end explaining why the hom-weighted (co)limit definition implies the (co)equalizer one.

I wanted to add labels, à la AMSLatex’s equation environment, to a couple of displayed diagrams, but I can’t remember/never knew if iTeX can do that. Anyone know?

Incidentally, the hom-weighted limit definition of ends follows, at least when V=Set, from the universal-extranatural one once you know that there is a Yoneda-type lemma for extranaturals giving a bijection between Exnat(pt, F) and Nat(hom_C, F) (maths rendering doesn’t seem to be working here, at least in the preview). Has anyone seen this extranatural-Yoneda written down elsewhere? I came up with it myself a while ago, but it must have been noticed before.
- CommentRowNumber3.
- CommentAuthorTobyBartels
- CommentTimeAug 18th 2011
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Author: TobyBartels
Format: MarkdownItexTo add numbers to a displayed equation, use `\[ ... \label{foo} \]`. Then `(eq:foo)` will produce ‘(1)’ (or whatever number it is) and an internal link.

To add numbers to a displayed equation, use \[ ... \label{foo} \]. Then (eq:foo) will produce ‘(1)’ (or whatever number it is) and an internal link.
- CommentRowNumber4.
- CommentAuthorFinnLawler
- CommentTimeAug 18th 2011
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Author: FinnLawler
Format: MarkdownItexThat works perfectly, thanks!

That works perfectly, thanks!
- CommentRowNumber5.
- CommentAuthorFinnLawler
- CommentTimeAug 19th 2011
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Author: FinnLawler
Format: MarkdownItexAnother couple of minor edits/additions.

Another couple of minor edits/additions.
- CommentRowNumber6.
- CommentAuthorzskoda
- CommentTimeSep 29th 2012
- (edited Sep 29th 2012)
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Author: zskoda
Format: MarkdownItexEntry [[end]] requires that the definition of an end of $V$-enriched functor $C^{op}\otimes C\to V$ requires that $V$ is _symmetric_. Is this really necessary? (I think I know some examples at least where it is not, but maybe I misunderstand) On the other hand, Kelly's book starts requiring that $V$ is symmetric precisely before defining the [[tensor product of enriched categories]] (entry not existing yet; the notation of tensor product of enriched categories is freely used at [[end]] without warning). What is the problem (if any) with defining the tensor product of $V$-enriched categories if $V$ is not symmetric ? I see that [[Matsuhiro Takeuchi]] is in his paper introducing what is now [[Takeuchi product]] used the notation for end and coend interchanged, what is unfortunately followed later by modern writers in Hopf algebras, for that context. This comes in fact from earlier paper by Sweedler who assigns the notation to MacLane. See page 87 in * M. E. Sweedler, _Groups of simple algebras_, Publ. IHES __44__, 79--189, [MR51:587](http://www.ams.org/mathscinet-getitem?mr=51:587), [numdam](http://www.numdam.org/item?id=PMIHES_1974__44__79_0) Worse, e.g. the label $\int_{r\in R}\, M_r\otimes {}_r N$ would suggest in category community that $r$ is an _object_ in some category $R$ and $M:R\to Ab$ a functor. For Sweedler and Takeuchi $r$ will be a morphism in category theoretic interpretation. In fact, as MacLane explains in his section on coends, the usual coend $\int^R M\otimes N$ reproduces the tensor product $M\otimes_R N$ over rings, where ring $R$ is understood as a category with one object and $R$ worth of morphisms, and modules are understood as functors $R\to Ab$; and $M\otimes N$ is understood as a bifunctor $R^{op}\times R\to Ab$.
Entry end requires that the definition of an end of $V$ -enriched functor $C^{op}\otimes C\to V$ requires that $V$ is symmetric. Is this really necessary? (I think I know some examples at least where it is not, but maybe I misunderstand) On the other hand, Kelly’s book starts requiring that $V$ is symmetric precisely before defining the tensor product of enriched categories (entry not existing yet; the notation of tensor product of enriched categories is freely used at end without warning). What is the problem (if any) with defining the tensor product of $V$ -enriched categories if $V$ is not symmetric ?

I see that Matsuhiro Takeuchi is in his paper introducing what is now Takeuchi product used the notation for end and coend interchanged, what is unfortunately followed later by modern writers in Hopf algebras, for that context. This comes in fact from earlier paper by Sweedler who assigns the notation to MacLane. See page 87 in
- M. E. Sweedler, Groups of simple algebras, Publ. IHES 44, 79–189, MR51:587, numdam
Worse, e.g. the label $\int_{r\in R}\, M_r\otimes {}_r N$ would suggest in category community that $r$ is an object in some category $R$ and $M:R\to Ab$ a functor. For Sweedler and Takeuchi $r$ will be a morphism in category theoretic interpretation. In fact, as MacLane explains in his section on coends, the usual coend $\int^R M\otimes N$ reproduces the tensor product $M\otimes_R N$ over rings, where ring $R$ is understood as a category with one object and $R$ worth of morphisms, and modules are understood as functors $R\to Ab$ ; and $M\otimes N$ is understood as a bifunctor $R^{op}\times R\to Ab$ .
- CommentRowNumber7.
- CommentAuthorTodd_Trimble
- CommentTimeSep 29th 2012
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Author: Todd_Trimble
Format: MarkdownItexI believe you can get away with a braiding, but I'm not sure how you'd get away with less. I suppose that if for example the homs of $C$ live in the **center** of the monoidal category $V$, you can get away with strictly less and speak of a $V$-enriched functor $C^{op} \otimes C \to V$. Maybe the best way of answering the question is to ask another. If $C$ and $D$ are $V$-enriched, how do you define $C \otimes D$ as a $V$-category, and in particular how are you supposed to define the composition law? If $(C \otimes D)((c, d), (c', d')) \coloneqq C(c, c') \otimes D(d, d')$, then _some_ sort of interchange seems to be necessary to get the composition.

I believe you can get away with a braiding, but I’m not sure how you’d get away with less. I suppose that if for example the homs of $C$ live in the center of the monoidal category $V$ , you can get away with strictly less and speak of a $V$ -enriched functor $C^{op} \otimes C \to V$ .

Maybe the best way of answering the question is to ask another. If $C$ and $D$ are $V$ -enriched, how do you define $C \otimes D$ as a $V$ -category, and in particular how are you supposed to define the composition law? If $(C \otimes D)((c, d), (c', d')) \coloneqq C(c, c') \otimes D(d, d')$ , then some sort of interchange seems to be necessary to get the composition.
- CommentRowNumber8.
- CommentAuthorMike Shulman
- CommentTimeSep 29th 2012
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Author: Mike Shulman
Format: MarkdownItexAs Todd says, you do need some sort of symmetry or braiding in order to define the tensor product of $V$-categories in general. However, you *can* define the notion of $V$-profunctor without any symmetry or braiding, by using separate left and right actions. I wonder whether you could define the end or coend of a $V$-profunctor directly without needing to regard it as a $V$-functor $C^{op}\otimes C \to V$?

As Todd says, you do need some sort of symmetry or braiding in order to define the tensor product of $V$ -categories in general. However, you can define the notion of $V$ -profunctor without any symmetry or braiding, by using separate left and right actions. I wonder whether you could define the end or coend of a $V$ -profunctor directly without needing to regard it as a $V$ -functor $C^{op}\otimes C \to V$ ?
- CommentRowNumber9.
- CommentAuthorzskoda
- CommentTimeSep 30th 2012
- (edited Sep 30th 2012)
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Author: zskoda
Format: MarkdownItexThank you guys, I will have to extensively think of you suggestions. Let me mention just the trivial case of a functor of the form $V(F,G):C^{op}\otimes C\to V$ where $F,G:C\to V$ are just $V$-enriched. If $F= G$ this end of $V(F,G)$ gives the usual "inner end" of a functor $F$ which is in that case a monoid. Instead of $\pi_X: E = \int_C V(F,G)\to V(F X,G X)$ one usually by adjunction considers map $\tilde\pi_X: \int_C V(F,G)\otimes F X \to G X$, which become actions in the special case $F=G$. The universal property is that for every object $A$ in $V$ and every $V$-natural transformation $g: A\otimes F\to G$ there is a unique morphism $a:A\to E$ in $V$ such that $g = \tilde\pi\circ (a\otimes F)$. This special case, in this adjoint form, makes sense for biclosed but not symmetric $V$, it seems to me (and one does not care how the domain category of $V(F,G)$ be called, as at least enriched hom makes sense always by definition). Though I am not sure, if I translated fully the conditions for universal wedges of morphisms $E\otimes C(X,Y)\to V(F X,G X)\otimes V(G X,G Y)\to V(F X,G Y)$ and along another line $E\otimes C(X,Y)\to V(F Y,G Y)\otimes V(F X,F Y)\to V(F X,G Y)$.

Thank you guys, I will have to extensively think of you suggestions.

Let me mention just the trivial case of a functor of the form $V(F,G):C^{op}\otimes C\to V$ where $F,G:C\to V$ are just $V$ -enriched. If $F= G$ this end of $V(F,G)$ gives the usual “inner end” of a functor $F$ which is in that case a monoid. Instead of $\pi_X: E = \int_C V(F,G)\to V(F X,G X)$ one usually by adjunction considers map $\tilde\pi_X: \int_C V(F,G)\otimes F X \to G X$ , which become actions in the special case $F=G$ . The universal property is that for every object $A$ in $V$ and every $V$ -natural transformation $g: A\otimes F\to G$ there is a unique morphism $a:A\to E$ in $V$ such that $g = \tilde\pi\circ (a\otimes F)$ . This special case, in this adjoint form, makes sense for biclosed but not symmetric $V$ , it seems to me (and one does not care how the domain category of $V(F,G)$ be called, as at least enriched hom makes sense always by definition). Though I am not sure, if I translated fully the conditions for universal wedges of morphisms $E\otimes C(X,Y)\to V(F X,G X)\otimes V(G X,G Y)\to V(F X,G Y)$ and along another line $E\otimes C(X,Y)\to V(F Y,G Y)\otimes V(F X,F Y)\to V(F X,G Y)$ .
- CommentRowNumber10.
- CommentAuthorTodd_Trimble
- CommentTimeSep 30th 2012
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Author: Todd_Trimble
Format: MarkdownItexIn the case of a tensor product of a functor $F: C^{op} \to V$ and $G: C \to V$, where $V$ is merely monoidal biclosed and cocomplete, we have a coequalizer diagram $$\sum_{c, d} G(c) \otimes C(c, d) \otimes F(d) \stackrel{\to}{\to} \sum_c G(c) \otimes F(c) \to G \otimes_C F = \int^c G(c) \otimes F(c)$$ with not a trace of symmetry in sight. We can do something similar for the hom of two functors $F, G: C \to V$, phrased in terms of actions $F(c) \otimes C(c, d) \to F(d)$. If we are careful and say $- \otimes v$ is left adjoint to $v \multimap -$ (so we are just assuming closed monoidal here), then one can carefully check there is a canonical isomorphism $$(u \otimes v) \multimap w \cong u \multimap (v \multimap w)$$ and we can also curry the action on $G$ as an extranatural family of maps $G(c) \to C(c, d) \multimap G(d)$. In that case we have a family $$F(d) \multimap G(d) \to (F(c) \otimes C(c, d)) \multimap G(d) \cong F(c) \multimap (C(c, d) \multimap G(d))$$ (using the action on $F$) and a family $$F(c) \multimap G(c) \to F(c) \multimap (C(c, d) \multimap G(d))$$ (using the curryed action on $G$), and take the equalizer $$\hom_C(F, G) = \int_c F(c) \multimap G(c) \to \prod_c F(c) \multimap G(c) \stackrel{\to}{\to} \prod_{c, d} F(c) \multimap (C(c, d) \multimap G(d))$$ using these two families to define the parallel arrows. I haven't investigated the situation for more general ends and coends but without symmetry.

In the case of a tensor product of a functor $F: C^{op} \to V$ and $G: C \to V$ , where $V$ is merely monoidal biclosed and cocomplete, we have a coequalizer diagram
$\sum_{c, d} G(c) \otimes C(c, d) \otimes F(d) \stackrel{\to}{\to} \sum_c G(c) \otimes F(c) \to G \otimes_C F = \int^c G(c) \otimes F(c)$
with not a trace of symmetry in sight.

We can do something similar for the hom of two functors $F, G: C \to V$ , phrased in terms of actions $F(c) \otimes C(c, d) \to F(d)$ . If we are careful and say $- \otimes v$ is left adjoint to $v \multimap -$ (so we are just assuming closed monoidal here), then one can carefully check there is a canonical isomorphism
$(u \otimes v) \multimap w \cong u \multimap (v \multimap w)$
and we can also curry the action on $G$ as an extranatural family of maps $G(c) \to C(c, d) \multimap G(d)$ . In that case we have a family
$F(d) \multimap G(d) \to (F(c) \otimes C(c, d)) \multimap G(d) \cong F(c) \multimap (C(c, d) \multimap G(d))$
(using the action on $F$ ) and a family
$F(c) \multimap G(c) \to F(c) \multimap (C(c, d) \multimap G(d))$
(using the curryed action on $G$ ), and take the equalizer
$\hom_C(F, G) = \int_c F(c) \multimap G(c) \to \prod_c F(c) \multimap G(c) \stackrel{\to}{\to} \prod_{c, d} F(c) \multimap (C(c, d) \multimap G(d))$
using these two families to define the parallel arrows.

I haven’t investigated the situation for more general ends and coends but without symmetry.
- CommentRowNumber11.
- CommentAuthorMike Shulman
- CommentTimeSep 30th 2012
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Author: Mike Shulman
Format: MarkdownItexAn even more general statement is that if $W$ is a locally cocomplete bicategory whose composition has right adjoints in each variable (i.e. all right extensions and right liftings exist), then there is another such bicategory $Mod(W)$ whose objects are $W$-enriched categories and whose morphisms are bimodules. An advantage of phrasing and proving it like that is that it doesn't even make sense to talk about a bicategory being "symmetric", so there's less danger that you're using symmetry without realizing it.

An even more general statement is that if $W$ is a locally cocomplete bicategory whose composition has right adjoints in each variable (i.e. all right extensions and right liftings exist), then there is another such bicategory $Mod(W)$ whose objects are $W$ -enriched categories and whose morphisms are bimodules. An advantage of phrasing and proving it like that is that it doesn’t even make sense to talk about a bicategory being “symmetric”, so there’s less danger that you’re using symmetry without realizing it.
- CommentRowNumber12.
- CommentAuthorzskoda
- CommentTimeSep 30th 2012
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Author: zskoda
Format: MarkdownItexAmazing answers :)

Amazing answers :)
- CommentRowNumber13.
- CommentAuthorIngoBlechschmidt
- CommentTimeMay 8th 2015
- (edited May 8th 2015)
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Author: IngoBlechschmidt
Format: MarkdownItexAdded to _[[end]]_ a pointer to Fosco's [beautiful introduction](http://arxiv.org/abs/1501.02503) which was also discussed [here](http://nforum.mathforge.org/discussion/6376/coend-calculus/).

Added to end a pointer to Fosco’s beautiful introduction which was also discussed here.
- CommentRowNumber14.
- CommentAuthorUrs
- CommentTimeMay 12th 2015
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Author: Urs
Format: MarkdownItexIngo, thanks for adding the pointer! I could have sworn that I had added this earlier, but clearly I didn't.

Ingo, thanks for adding the pointer! I could have sworn that I had added this earlier, but clearly I didn’t.
- CommentRowNumber15.
- CommentAuthorTim_Porter
- CommentTimeJun 5th 2015
- (edited Jun 5th 2015)
- PermaLink
Author: Tim_Porter
Format: MarkdownItexThere is some strange wording at [[end]]: > [where is customary notation for the enriched hom of in ] What was intended? It sort of looks like a half completed edit.
There is some strange wording at end:
```
       [where is customary notation for the enriched hom of in ]
```
What was intended? It sort of looks like a half completed edit.
- CommentRowNumber16.
- CommentAuthorDexter Chua
- CommentTimeAug 20th 2016
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Author: Dexter Chua
Format: MarkdownItexAdded an explicit definition of ends in terms of wedges; natural transformations as ends, and an example of (co)end calculus.

Added an explicit definition of ends in terms of wedges; natural transformations as ends, and an example of (co)end calculus.

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