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1. • CommentRowNumber1.
   • CommentAuthordanlior2
   • CommentTimeMay 17th 2010
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   Author: danlior2
   Format: MarkdownItexHello, 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 $colimit(nerve(- \downarrow C)) = nerve(C)$ ?

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

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 17th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexSorry, 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?

   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?

3. • CommentRowNumber3.
   • CommentAuthordanlior2
   • CommentTimeMay 18th 2010
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   Author: danlior2
   Format: MarkdownItexHi Todd, Thanks for you quick reply. 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 $c_0 \rightarrow c_1$.

   Hi Todd,

   Thanks for you quick reply.

   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 $c_0 \rightarrow c_1$.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 18th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexWell, 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.

   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.

5. • CommentRowNumber5.
   • CommentAuthordanlior2
   • CommentTimeMay 18th 2010
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   Author: danlior2
   Format: MarkdownItexYes 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

   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

6. • CommentRowNumber6.
   • CommentAuthorDavidRoberts
   • CommentTimeMay 18th 2010
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   Author: DavidRoberts
   Format: MarkdownItexThis 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.

   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.

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 19th 2010
   • (edited May 19th 2010)
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   Author: Todd_Trimble
   Format: MarkdownItexDan, 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.

   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.

8. • CommentRowNumber8.
   • CommentAuthorMike Shulman
   • CommentTimeMay 20th 2010
   • (edited May 20th 2010)
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   Author: Mike Shulman
   Format: MarkdownItexNow 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]].

   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.

9. • CommentRowNumber9.
   • CommentAuthorTodd_Trimble
   • CommentTimeMay 20th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexAh, ah, ah -- excellent point, Mike. Thanks.

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