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• 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 $B$ 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 $\int F \to B$ associated to a functor $F = B(-,X)$ by the standard Grothendieck construction:

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

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

$\array{ b_1 &&\stackrel{h}{\to}&& b_2 \\ & {}_{\mathllap{f_1}}\searrow && \swarrow_{\mathrlap{f_2}} \\ && X }$

commutes.