It is well-known that for any presheaf $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Set}$, the category of elements (obtained by the so-called Grothendieck construction) of $F$ is a comma category $y/\ulcorner F \urcorner$ in the category $\mathrm{CAT}$ of

$$\mathcal{C} \xrightarrow{y} \widehat{\mathcal{C}} \xleftarrow{\ulcorner F \urcorner} 1.$$

Question: does this generalise to presheaves $F \colon \mathcal{C}^{\mathrm{op}} \to \mathrm{Gpd}$ of groupoids (or even categories)? Mere comma categories yield discrete fibrations, hence won't give the expected answer. So the question is whether some other construction with a similar flavour could do.

Probably, if the construction does generalise, then it will also work for pseudo-functors to $\mathrm{Gpd}$ or $\mathrm{Cat}$, but I'm really interested in strict functors.

Note: I'm half-aware of another universal property of the Grothendieck construction as an oplax colimit. Is it related?

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    $\begingroup$ It's not really a "lax pullback" but rather a comma category. $\endgroup$ – Zhen Lin Jan 22 '14 at 10:22
  • $\begingroup$ I thought this was the same in $\mathrm{CAT}$, what's the subtlety i'm missing? $\endgroup$ – Tom Hirschowitz Jan 22 '14 at 11:44
  • $\begingroup$ The lax pullback is symmetric with respect to the given two functors, the comma category is not. $\endgroup$ – Zhen Lin Jan 22 '14 at 12:04
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    $\begingroup$ There is a general definition of "lax limit", and lax pullbacks are the evident special case. $\endgroup$ – Zhen Lin Jan 22 '14 at 13:26
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    $\begingroup$ I see. For the record, a definition of lax limit may be found in Kelly's "Elementary Observations on 2-Categorical Limits". Edited the question to avoid the ambiguity, thanks for pointing this out. $\endgroup$ – Tom Hirschowitz Jan 22 '14 at 16:31

Possibly the comma 2-category gives what you want.

If you think of $C^{op}\rightarrow Gpd$ as a functor of 2-categories (e.g. pass to the nerve for some model of $(\infty,2)$ categories) and then take the comma object $$ \matrix{(r\downarrow F)&\rightarrow& C \\ \downarrow & & \downarrow _{r} \\ 1 & \rightarrow^F & Gpd^{C^{op}}} $$

(where $r$ is the Yoneda imbedding.)

then an object is then a pair $(c\in C,\varphi:rc\rightarrow F)$, and a morphism is a pair $(\Psi:c\rightarrow c',M:\varphi\rightarrow \varphi'\circ r\Psi)$.

There is a canonical map from the Grothendiek construction into this comma category, and I suspect it is a homotopy equivalence, so that when you take the corresponding usual category you get the same thing.

  • $\begingroup$ In which general sense is this a comma object? I only know what a comma object is in a 2-category, but you seem to be using a more general notion here, right? $\endgroup$ – Tom Hirschowitz May 27 '14 at 7:29
  • $\begingroup$ it's pretty much the same idea. The square is final in the category of squares with the same bottom right corner. In this case it's the comma object in the 3-category of 2-categories. $M$ here is a non trivial 3-morphism. $\endgroup$ – Adam Gal May 27 '14 at 11:09

The answer seems to be "no": Consider a strong-/pseudofuntor $F:C^\mathrm{op}\to \mathrm{Cat}$.

The object "sets" are isomorphic: they are given by pairs $(x,a_x)$ with $x\in C$ and $a_x\in F(x)$. But arrows in the Grothendieck construction / category of elements are given by pairs

$$(f:x\to y,\varphi: a_x\to f^*a_y)$$

where $f^*:=F(f)$.

On the other hand: Arrows in the respective comma-category are given by triples

$$(f:x\to y, \Phi: F\to F, [\varphi]: \Phi_x(a_x) \to f^*a_y).$$

Composition is similar but the endo-transformation $\Phi$ does not need to be the identity.

These categories are in general not equivalent: Consider the case where $C=1$ for suitable $F$.

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    $\begingroup$ This is precisely what i meant by: "Mere comma categories yield discrete fibrations, hence won't give the expected answer." My question was thus about what should replace comma categories to yield the correct construction. $\endgroup$ – Tom Hirschowitz Jan 24 '14 at 10:17
  • $\begingroup$ Ah sry; misread the question then. $\endgroup$ – Gerrit Begher Jan 24 '14 at 13:32

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