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Let $X$ be a strict $\infty$-category (not $(\infty,1)$, I am talking about true $\infty$-categories (Grothendieck modules (exact presheaves (finite-limit preserving functors $\Theta^{op}\to \mathrm{Set}$) over Joyal's category $\Theta$ (see Dimitri Ara's thesis)). Is there a notion of a Grothendieck fibration between two such $\infty$-categories or a notion of a 1-cell in inside such a strict $\infty$-category being a Grothendieck fibration? If so, how is it formulated?

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up vote 3 down vote accepted

It depends on whether you want to mimic the 1-categorical case strictly, or take into account the homotopy nature of the $\infty$-categories. Clearly (or I would hope so!) there is an underlying 1-category of an $\infty$-category, and so a trivial way of arriving at a fibration. I guess this is not what you are after.

My guess would be to imitate the definition along the lines of what I think Street gave in 'Fibrations in bicategories', i.e.

  • For every morphism $f: X \to B$, the canonical map $i: f/_{\cong} p \to f/p$ has a right adjoint in the slice 2-category $K / X$

where $f/_{\cong} p$ is the 2-pullback, and $f/p$ is the comma object (from the nLab). So we should just replace 2-categorical concepts here by $\infty$-categorical concepts.

This presumes the definition of adjoint $\infty$-functors :)

As to your second question, if we know that $\infty$Cat is enriched over itself, we use the $\infty$-categorical version of the first definition at fibration in a 2-category (nLab again)

EDIT: Since you are after Grothendieck fibrations, not Street fibrations, you would want the $\infty$-versions of strict 2-pullbacks and strict comma objects. This should be a significant simplification. I"m not sure, but you may have strict $\infty$-pullbacks = strict $\infty$-iso-comma object, and if both the comma- and iso-comma-objects can be constructed using cotensors, then to me this would be the best approach. But it of course depends on what you want to do next (aside: yes, I saw your other post in that other place).

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I think i's actually not so clear how a weak ∞-category has an "underlying" 1-category. The 1-arrows don't necessarily compose strictly associatively, of course. You can identify equivalent 1-arrows to get a 1-category, but then it's a bit more of a stretch to call the result an "underlying" anything, and it probably isn't going to give you a useful notion of fibration. – Mike Shulman Jul 26 '11 at 8:34
Note that Harry said strict $\infty$-categories. If a strict $\infty$-category doesn't have associative composition, then it doesn't deserve being called strict. – David Roberts Jul 26 '11 at 9:34
Oh. People actually think about strict $\infty$-categories? (-: – Mike Shulman Jul 29 '11 at 14:01

I don't know of anyone having written down such a definition for the particular model of ∞-categories that you appear to be interested in, but the general form of the notion of fibration for higher categories is known at an imprecise level. See for instance the nLab page on n-fibrations.

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There is some ambiguity on the lab page: In the definition of a fibration, are we requiring all three properties, or just one of them? If we require all three, how can the recursive definition work, since part $2$ of the definition explicitly relies on a notion of an $n-1$-fibration. – Harry Gindi Jul 26 '11 at 8:56
(That is, how can it work for $n=\infty$?) – Harry Gindi Jul 26 '11 at 8:57
Yes, we require all three properties. It's remarked afterwards that "if we unravel the definition, it makes perfect sense for $n=\omega$" and I think this is true. That is, saying that f is a fibration requires some things about cartesian 1-cells, and also that its action on hom-categories be a fibration, which requires some things about cartesian 2-cells, and also that its action on hom-categories be a fibration, which requires some things about cartesian 3-cells, and so on. After $\omega$ steps of unraveling, we have a list of conditions on cartesian n-cells for every n. – Mike Shulman Jul 29 '11 at 13:59
Equivalently, in that case we can simply regard it as a coinductive definition. – Mike Shulman Jul 29 '11 at 14:00

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