Not signed in (Sign In)

Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

  • Sign in using OpenID

Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

Welcome to nForum
If you want to take part in these discussions either sign in now (if you have an account), apply for one now (if you don't).
    • CommentRowNumber1.
    • CommentAuthorzskoda
    • CommentTimeApr 27th 2011
    • CommentRowNumber2.
    • CommentAuthorUrs
    • CommentTimeApr 27th 2011
    • (edited Apr 27th 2011)

    I have added the remark that under the Grothendieck construction this does correspond to a representable functor.

    • CommentRowNumber3.
    • CommentAuthorzskoda
    • CommentTimeApr 27th 2011

    What about representable pseudofunctors into Cat ?

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeApr 27th 2011
    • (edited Apr 27th 2011)

    What do you mean by “What about”?

    I could say: as long as BB is a 1-category, every representable functor is necessarily (equivalent to) an ordinary functor.

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeApr 27th 2011
    • (edited Apr 27th 2011)

    The notion of Grothendieck fibration is not stable under equivalence. Thus the notion of representable Grothendieck fibration is taken only up to isomorphism in the definition which I used (taken from Kock), not up to equivalence.

    The Yoneda lemma in strict 2-categories is also more precise (gives: iso or categories) than for bicategories (gives: equiv of categories). We can also work with internal categories or alike where not all fibrations will be split. Still the representability in both senses may make sense.

    In any case there seem to be some cases in between, so I was wondering what was the precise statement without various simplifications from axiom of choice and of changing the original definition of Grothendieck. So by what about I meant not wondering about which version is actually claimed, what is answered to large extent in 4.

    Also there are similarities with the notion of small fibration, where there is sort of categorification of the representability…

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeApr 27th 2011
    • (edited Apr 27th 2011)

    which version is actually claimed,

    I am not meaning to make a deep claim here. Just take the standard definition of the fibered category FB\int F \to B associated to a functor F=B(,X)F = B(-,X) by the standard Grothendieck construction:

    • an object over bBb \in B is an element of F(b)=B(b,X)F(b) = B(b,X), so is a morphism f:bXf : b \to X;

    • a morphism over h:b 1b 2h : b_1 \to b_2 is two elements f 1F(b 1)f_1 \in F(b_1) and f 2F(b 2)f_2 \in F(b_2) and a morphism h *b 2b 1h^* b_2 \to b_1. For the case F=B(,X)F = B(-,X) the morphism has to be an identity and the conditon says that

      b 1 h b 2 f 1 f 2 X \array{ b_1 &&\stackrel{h}{\to}&& b_2 \\ & {}_{\mathllap{f_1}}\searrow && \swarrow_{\mathrlap{f_2}} \\ && X }

      commutes.

Add your comments
  • Please log in or leave your comment as a "guest post". If commenting as a "guest", please include your name in the message as a courtesy. Note: only certain categories allow guest posts.
  • To produce a hyperlink to an nLab entry, simply put double square brackets around its name, e.g. [[category]]. To use (La)TeX mathematics in your post, make sure Markdown+Itex is selected below and put your mathematics between dollar signs as usual. Only a subset of the usual TeX math commands are accepted: see here for a list.

  • (Help)