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• CommentRowNumber1.
• CommentAuthordanlior2
• CommentTimeMay 17th 2010

Hello,

Let $C$ be a category and let $\mathrm{PC}$ be the category of subcategories of $C$ . Then $\left(-↓C\right):{C}^{\mathrm{op}}\to \mathrm{PC}$ is a functor and it’s colimit is $C\in \mathrm{PC}$.

Is $\mathrm{colimit}\left(\mathrm{nerve}\left(-↓C\right)\right)=\mathrm{nerve}\left(C\right)$ ?

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeMay 17th 2010

Sorry, I’m having some trouble parsing this. If I have an object $c$ of $C$, which subcategory is $\left(c↓C\right)$ supposed to be? Ordinarily I would interpret the notation as denoting the comma category whose objects are morphisms $c\to d$ where $d$ is an object of $C$, and whose morphisms are commutative triangles with vertex $c$, but that’s not a subcategory of $C$.

As for the next part, am I to interpret the nerve we’re taking the colimit of as this composite:

${C}^{\mathrm{op}}\stackrel{\left(-↓C\right)}{\to }PC↪\mathrm{Cat}\stackrel{\mathrm{nerve}}{\to }{\mathrm{Set}}^{{\Delta }^{\mathrm{op}}},$

assuming that the first arrow makes sense?

• CommentRowNumber3.
• CommentAuthordanlior2
• CommentTimeMay 18th 2010

Hi Todd,

Your interpretation of $\left(c↓C\right)$ is the same as mine. So is your interpretation of the the the nerve we’re taking colimits of.

I think of $\left(c↓C\right)$ as a subcategory of $C$ by sending the object $c\to {c}_{0}$ to ${c}_{0}$ and the morphism $c\to {c}_{0}\to {c}_{1}$ to the morphism ${c}_{0}\to {c}_{1}$.

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeMay 18th 2010

Well, it’s not a subcategory by any definition I know of (certainly not injective on objects; it’s faithful but not full), but it seems not to matter since we can just bypass $PC$ and go straight to $\left(-↓C\right):{C}^{\mathrm{op}}\to \mathrm{Cat}$. Someone around here may know right away, but I’ll try to give it a think when I get a chance.

• CommentRowNumber5.
• CommentAuthordanlior2
• CommentTimeMay 18th 2010

Yes of course, you’re right. $\left(c↓C\right)$ is not a subcategory or $C$ for general categories $C$. However, I neglected to mention that in the particular situation that I’m considering, $C$ is a partially ordered set. In particular, it’s hom-sets have cardinality at most 1. I think that for such categories $C$, $\left(c↓C\right)$ is a subcategory of $C$.

I also agree with you that this point doesn’t really matter anyway since we can bypass PC and go straight to $\mathrm{Cat}$.

I would have saved confusion if I simply stated things that way to start with. Thanks for helping me clarify my question. I just hope that someone can give me an answer.

dan

• CommentRowNumber6.
• CommentAuthorDavidRoberts
• CommentTimeMay 18th 2010

This is related to questions about test categories or the like. I recall seeing some result like this in the book on the homotopy theory of Grothendieck by Maltsiniotis, but my memory may be faulty.

• CommentRowNumber7.
• CommentAuthorTodd_Trimble
• CommentTimeMay 19th 2010
• (edited May 19th 2010)

Dan, I’ve thought a little about your question, and I think the answer is ’yes’, and the answer is not hard to see. Let’s see if I have this right:

Your statement about the colimit in $\mathrm{Cat}$ of the $\left(c↓C\right)$ being isomorphic to $C$ intrigued me – I had never seen that before – but on reflection it was something fairly obvious, in fact basically the Yoneda lemma in disguise. The objects of ${\mathrm{colim}}_{c:{C}^{\mathrm{op}}}\left(c↓C\right)$ are equivalence classes of arrows $c\to d$ where the equivalence $\sim$ is generated by

$\left(c\stackrel{f}{\to }d\right)\sim \left(c\prime \stackrel{g}{\to }c\stackrel{f}{\to }d\right)$

and it is immediate that every $g:c\to d$ is equivalent to ${1}_{d}:d\to d$; this of course is just a form of the Yoneda lemma.

Now let’s look at your problem, which compares the $\mathrm{nerve}\left(C\right)$ to the colimit of

${C}^{\mathrm{op}}\stackrel{\left(c↓C\right)}{\to }\mathrm{Cat}\stackrel{\mathrm{nerve}}{\to }{\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}$

Since colimits in ${\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}$ are computed pointwise, we just have to show the colimit of

${C}^{\mathrm{op}}\stackrel{\left(c↓C\right)}{\to }\mathrm{Cat}\stackrel{\mathrm{nerve}}{\to }{\mathrm{Set}}^{{\Delta }^{\mathrm{op}}}\stackrel{{\mathrm{ev}}_{n}}{\to }\mathrm{Set},$

where ${\mathrm{ev}}_{n}$ is evaluation at an object $n$, agrees with $\mathrm{nerve}\left(C{\right)}_{n}$. This is

${C}^{\mathrm{op}}\stackrel{\left(c↓C\right)}{\to }\mathrm{Cat}\stackrel{\mathrm{hom}\left(\left[n\right],-\right)}{\to }\mathrm{Set}$

Now an $n$-simplex in the comma category $\left(c↓C\right)$, which is an element of this composite, is the same as an $\left(n+1\right)$-simplex beginning with the vertex $c$, and the colimit (in $\mathrm{Set}$) consists of equivalence classes of $\left(n+1\right)$-simplices where a simplex beginning with $c$ is deemed equivalent to a simplex beginning with $c\prime$ obtained by pulling back along any $g:c\prime \to c$. And again, it is a triviality that each $\left(n+1\right)$-simplex

$c\to \left({d}_{0}\to \dots \to {d}_{n}\right)$

is equivalent to

${d}_{0}\stackrel{{1}_{{d}_{0}}}{\to }\left({d}_{0}\to \dots \to {d}_{n}\right)$

but the collection of such ${d}_{0}\to \dots \to {d}_{n}$ is the same as $\mathrm{nerve}\left(C{\right)}_{n}$. This proves your conjecture.

Edit: By the way, this reminds me of the tangent category stuff that originated at the n-Café in a discussion that included Urs, David Roberts, and me, and which was developed further by Schreiber-Roberts. I think I recall now remarking on the Yoneda lemma in this connection.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeMay 20th 2010
• (edited May 20th 2010)

Now that I’ve had a chance to think about this properly, Todd is exactly right. It is also a special case of a general fact about (2-sided) bar constructions. Specifically, the nerve of C is the bar construction $B\left(*,C,*\right)$ (where $*$ denotes the functor constant at a terminal object), while the nerve of $c↓C$ is the bar construction $B\left(C\left(c,-\right),C,*\right)$. Since colimits of a functor $F:{C}^{\mathrm{op}}\to D$ are given by tensor products of functors $*{\otimes }_{C}F$, and such tensor products come inside a bar construction (since colimits commute with colimits), we have

${colim}^{c}N\left(c↓C\right)=*{\otimes }_{c\in C}B\left(C\left(c,-\right),C,*\right)=B\left(*{\otimes }_{c\in C}C\left(c,-\right),C,*\right)=B\left(*,C,*\right)=NC$

where $*{\otimes }_{c\in C}C\left(c,-\right)=*$ by the co-Yoneda lemma.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeMay 20th 2010

Ah, ah, ah – excellent point, Mike. Thanks.