Apparently it's 'well known' that if $P$ is a presheaf on $C$ then there is an equivalence $\widehat{C}/P \simeq \widehat{\int P}$, where $\int P$ is the usual category of elements and $\widehat{C} = [C^{\rm op},{\rm Set}]$. (I've seen a reference to Johnstone's Topos Theory for this, but I don't have easy access to the book. It's also Exercise III.8(a) in Mac Lane--Moerdijk)

Now, I've come across comma categories $\widehat{C}/H$ for functors $H \colon D \to \widehat{C}$ (mainly when $H = D(F-,-)$ for $F \colon C \to D$), and I'd like to have a similarly useful/interesting result for them. The equivalence doesn't generalize to $\widehat{C}/H \simeq \widehat{\int H}$, or at least I can't see any way to make that work. So I want to understand the first equivalence from a more abstract-nonsensical point of view, to figure out what's really going on.

Is there a way to see the above equivalence as living in a fibrational cosmos or something similar? If not, is there any other kind of machinery that might help me understand categories of the form $\widehat{C}/H$?

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    $\begingroup$ "I don't have easy access to the book." - Finn Lawler. Yes you do! Perhaps you don't know where to find it, but if you knew where, you could find it in fewer than two minutes. I guess you could shoot me an e-mail and I could tell you where to go and/or send you the book. $\endgroup$ – Harry Gindi May 7 '10 at 19:36
  • $\begingroup$ what is $\widehat{C}$? $\endgroup$ – Martin Brandenburg May 7 '10 at 20:46
  • $\begingroup$ $\widehat{C} = [C^{\rm op},{\rm Set}]$ $\endgroup$ – Finn Lawler May 7 '10 at 21:08
  • $\begingroup$ By the way, Finn, that should be literally "C^". This is Grothendieck's (bad) notation from SGA. Psh(C) is much better notation. I've even seen it used in French, despite the fact that they use Fais(C) to denote the category of sheaves. $\endgroup$ – Harry Gindi May 7 '10 at 21:36
  • $\begingroup$ Oh, by the way, are you asking why the category of discrete opfibrations over a category is equivalent to the category of presheaves? I can't tell because you're using nonstandard names for things, but if this is equivalent to what you're asking, it is in chapter 1 of Tom Leinster's book. $\endgroup$ – Harry Gindi May 7 '10 at 21:42

Yes, you can see this as happening in a "fibrational cosmos." I'll describe how it goes, but then we'll see that the description of $\widehat{C}/H$ that comes out could also be deduced pretty naively.

The universal property of $\widehat{C}$ is that the category of functors $A\to \widehat{C}$ is equivalent to the category of discrete fibrations from $A$ to $C$, via pullback of the universal such. (A discrete fibration from $A$ to $C$ is a span $A\leftarrow E \to C$ such that $E\to C$ is a fibration, $E\to A$ is an opfibration, the two structures are compatible, and $E$ is discrete in the slice 2-category over $A\times C$.)

Now the slice category $\widehat{C}/P$, for $P\in\widehat{C}$, also has a universal property: it is the comma object of $\mathrm{Id}_{\widehat{C}}$ over $P:1\to \widehat{C}$. Thus a functor $A\to \widehat{C}/P$ is equivalent to a functor $A\to \widehat{C}$ and a natural transformation from it to the composite $A \to 1 \overset{P}{\to} \widehat{C}$. By the universal property of $\widehat{C}$, this is the same as giving a discrete fibration $A\leftarrow E \rightarrow C$ together with a map to the discrete fibration classified by $P:1\to \widehat{C}$, which is just $1\leftarrow \int P \rightarrow C$.

Next, (discrete) fibrations have the special property that not only is the composite of two (discrete) fibrations again a (discrete) fibration, but if $g$ is a discrete fibration and $g\circ f$ is a fibration, then $f$ is a fibration. Therefore, given a discrete fibration from $A$ to $C$ together with a map $E\to \int P$ over $C$, the map $E\to \int P$ is itself a fibration, and we can actually show that $A\leftarrow E \to \int P$ is itself a discrete fibration from $A$ to $\int P$, and that this is an equivalence. Hence, the slice category $\widehat{C}/P$ has the same universal property as $\widehat{\int P}$, so they are equivalent.

Now we can ask about replacing $P:1\to \widehat{C}$ with a more general functor $H:D\to \widehat{C}$. Here the fibrational-cosmos argument fails, because in the discrete fibration $D \leftarrow \int H \to C$ it is no longer true that the solitary map $\int H \to C$ is itself a discrete fibration, so cancellability no longer holds and $E\to \int H$ is no longer necessarily a fibration. However, at least in the 2-category $Cat$, we can still use it to figure out what $\widehat{C}/H$ should look like, by looking at the case $A=1$. In this case, to give a map $1 \to \widehat{C}/H$ means to give a map $d:1\to D$ (i.e. an object of $D$) along with a discrete fibration $E\to C$ and a map from $E$ to $D \leftarrow \int H \to C$ over $d$ and $\mathrm{Id}_C$. This is the same as a map from $E$ to $d^*(\int H) = \int H(d)$ over $C$, and we can then apply the previous argument here since both are discrete fibrations over $C$.

Thus, an object of $\widehat{C}/H$ consists of an object $d\in D$ together with a presheaf on $\int H(d)$. This is not a surprise, though, since we're just applying the original fact $\widehat{C}/P \simeq \widehat{\int P}$ objectwise. Since $d$ can vary between objects, what we're saying is really that $\widehat{C}/H$ is the category of elements of the functor $D\to Cat$ defined by $d\mapsto \widehat{\int H(d)}$, or more evocatively $$ \widehat{C}/H = \int_d \widehat{\int H(d)} $$ as a "double integral".

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  • $\begingroup$ Lucid and elegant as ever, Mike. Thanks! I had a feeling that the equivalence could be applied objectwise, but it was the 'indexing' by D that was tripping me up. As usual, if I'd stared at the problem for longer before reaching for the heavy artillery I might have had a better chance of figuring it out! $\endgroup$ – Finn Lawler May 8 '10 at 18:23

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