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• CommentRowNumber1.
• CommentAuthordanlior2
• CommentTimeMay 17th 2010

Hello,

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

Is $colimit(nerve(- \downarrow C)) = nerve(C)$ ?

• 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 $(c \downarrow C)$ 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:

$C^{op} \stackrel{(- \downarrow C)}{\to} P C \hookrightarrow Cat \stackrel{nerve}{\to} Set^{\Delta^{op}},$

assuming that the first arrow makes sense?

• CommentRowNumber3.
• CommentAuthordanlior2
• CommentTimeMay 18th 2010

Hi Todd,

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

I think of $(c \downarrow C)$ as a subcategory of $C$ by sending the object $c \rightarrow c_0$ to $c_0$ and the morphism $c \rightarrow c_0 \rightarrow c_1$ to the morphism $c_0 \rightarrow c_1$.

• 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 $P C$ and go straight to $(- \downarrow C): C^{op} \to Cat$. 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. $(c \downarrow C)$ 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 $C$, $(c \downarrow C)$ 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 $Cat$.

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 $Cat$ of the $(c \downarrow C)$ 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 $colim_{c: C^{op}} (c \downarrow C)$ are equivalence classes of arrows $c \to d$ where the equivalence $\sim$ is generated by

$(c \stackrel{f}{\to} d) \sim (c' \stackrel{g}{\to} c \stackrel{f}{\to} d)$

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 $nerve(C)$ to the colimit of

$C^{op} \stackrel{(c \downarrow C)}{\to} Cat \stackrel{nerve}{\to} Set^{\Delta^{op}}$

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

$C^{op} \stackrel{(c \downarrow C)}{\to} Cat \stackrel{nerve}{\to} Set^{\Delta^{op}} \stackrel{ev_n}{\to} Set,$

where $ev_n$ is evaluation at an object $n$, agrees with $nerve(C)_n$. This is

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

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

$c \to (d_0 \to \ldots \to d_n)$

is equivalent to

$d_0 \stackrel{1_{d_0}}{\to} (d_0 \to \ldots \to d_n)$

but the collection of such $d_0 \to \ldots \to d_n$ is the same as $nerve(C)_n$. 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 $B(*,C,*)$ (where $*$ denotes the functor constant at a terminal object), while the nerve of $c\downarrow C$ is the bar construction $B(C(c,-),C,*)$. Since colimits of a functor $F:C^{op}\to D$ 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

$\colim^c N(c\downarrow C) = * \otimes_{c\in C} B(C(c,-),C,*) = B(* \otimes_{c\in C} C(c,-), C, *) = B(*,C,*) = N C$

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

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeMay 20th 2010

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