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    • CommentRowNumber1.
    • CommentAuthorUrs
    • CommentTimeDec 21st 2009

    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

    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

    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

    That works perfectly, thanks!

    • CommentRowNumber5.
    • CommentAuthorFinnLawler
    • CommentTimeAug 19th 2011

    Another couple of minor edits/additions.

    • CommentRowNumber6.
    • CommentAuthorzskoda
    • CommentTimeSep 29th 2012
    • (edited Sep 29th 2012)

    Entry end requires that the definition of an end of VV-enriched functor C opCVC^{op}\otimes C\to V requires that VV 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 VV 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 VV-enriched categories if VV 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 rRM r rN\int_{r\in R}\, M_r\otimes {}_r N would suggest in category community that rr is an object in some category RR and M:RAbM:R\to Ab a functor. For Sweedler and Takeuchi rr will be a morphism in category theoretic interpretation. In fact, as MacLane explains in his section on coends, the usual coend RMN\int^R M\otimes N reproduces the tensor product M RNM\otimes_R N over rings, where ring RR is understood as a category with one object and RR worth of morphisms, and modules are understood as functors RAbR\to Ab; and MNM\otimes N is understood as a bifunctor R op×RAbR^{op}\times R\to Ab.

    • CommentRowNumber7.
    • CommentAuthorTodd_Trimble
    • CommentTimeSep 29th 2012

    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 CC live in the center of the monoidal category VV, you can get away with strictly less and speak of a VV-enriched functor C opCVC^{op} \otimes C \to V.

    Maybe the best way of answering the question is to ask another. If CC and DD are VV-enriched, how do you define CDC \otimes D as a VV-category, and in particular how are you supposed to define the composition law? If (CD)((c,d),(c,d))C(c,c)D(d,d)(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

    As Todd says, you do need some sort of symmetry or braiding in order to define the tensor product of VV-categories in general. However, you can define the notion of VV-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 VV-profunctor directly without needing to regard it as a VV-functor C opCVC^{op}\otimes C \to V?

    • CommentRowNumber9.
    • CommentAuthorzskoda
    • CommentTimeSep 30th 2012
    • (edited Sep 30th 2012)

    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 opCVV(F,G):C^{op}\otimes C\to V where F,G:CVF,G:C\to V are just VV-enriched. If F=GF= G this end of V(F,G)V(F,G) gives the usual “inner end” of a functor FF which is in that case a monoid. Instead of π X:E= CV(F,G)V(FX,GX)\pi_X: E = \int_C V(F,G)\to V(F X,G X) one usually by adjunction considers map π˜ X: CV(F,G)FXGX\tilde\pi_X: \int_C V(F,G)\otimes F X \to G X, which become actions in the special case F=GF=G. The universal property is that for every object AA in VV and every VV-natural transformation g:AFGg: A\otimes F\to G there is a unique morphism a:AEa:A\to E in VV such that g=π˜(aF)g = \tilde\pi\circ (a\otimes F). This special case, in this adjoint form, makes sense for biclosed but not symmetric VV, it seems to me (and one does not care how the domain category of V(F,G)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 EC(X,Y)V(FX,GX)V(GX,GY)V(FX,GY)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 EC(X,Y)V(FY,GY)V(FX,FY)V(FX,GY)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

    In the case of a tensor product of a functor F:C opVF: C^{op} \to V and G:CVG: C \to V, where VV is merely monoidal biclosed and cocomplete, we have a coequalizer diagram

    c,dG(c)C(c,d)F(d) cG(c)F(c)G CF= cG(c)F(c)\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:CVF, G: C \to V, phrased in terms of actions F(c)C(c,d)F(d)F(c) \otimes C(c, d) \to F(d). If we are careful and say v- \otimes v is left adjoint to vv \multimap - (so we are just assuming closed monoidal here), then one can carefully check there is a canonical isomorphism

    (uv)wu(vw)(u \otimes v) \multimap w \cong u \multimap (v \multimap w)

    and we can also curry the action on GG as an extranatural family of maps G(c)C(c,d)G(d)G(c) \to C(c, d) \multimap G(d). In that case we have a family

    F(d)G(d)(F(c)C(c,d))G(d)F(c)(C(c,d)G(d))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 FF) and a family

    F(c)G(c)F(c)(C(c,d)G(d))F(c) \multimap G(c) \to F(c) \multimap (C(c, d) \multimap G(d))

    (using the curryed action on GG), and take the equalizer

    hom C(F,G)= cF(c)G(c) cF(c)G(c) c,dF(c)(C(c,d)G(d))\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

    An even more general statement is that if WW 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)Mod(W) whose objects are WW-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

    Amazing answers :)

    • CommentRowNumber13.
    • CommentAuthorIngoBlechschmidt
    • CommentTimeMay 8th 2015
    • (edited May 8th 2015)

    Added to end a pointer to Fosco’s beautiful introduction which was also discussed here.

    • CommentRowNumber14.
    • CommentAuthorUrs
    • CommentTimeMay 12th 2015

    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)

    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

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