8
$\begingroup$

Let $C$ be a category with an object $X$ such that there are no non-trivial endomorphisms $X\rightarrow X$. Consider a simplex $\sigma$ of the nerve $NC$ of $C$. It is just a string of composable arrows in $C$. Define a $X$-component of $\sigma$ to be a maximal substring of the form $X\rightarrow X\rightarrow...\rightarrow X$. Let $N^r(C)$ be the simplicial subset of $N(C)$ consisting of the simplices with at most one $X$-component. Roughly speaking, you only consider those simplices where you travel over $X$ only once. For example, $A\rightarrow X\rightarrow B$ is allowed, but not $X\rightarrow A\rightarrow X$.

Question: Is the inclusion $N^r(C)\rightarrow N(C)$ a weak equivalence?

Improve this question
$\endgroup$
8
  • $\begingroup$ Is $N^r(C)$ even a simplicial set? Let $C$ be the groupoid with 2 objects $x$, $y$, and two nontrivial morphisms $f\colon x\to y$, $f^{-1}\colon y\to x$. Let $\sigma = (\operatorname{id}_x, f, f^{-1}, f, f^{-1})\in N(C)$. Then $\sigma\in N^r(C)$, but $d_4(\sigma) = (\operatorname{id}_x,f,f^{-1},\operatorname{id}_x)\not\in N^r(C)$ $\endgroup$
    – Roman Bruckner
    Commented Mar 19, 2014 at 8:21
  • 2
    $\begingroup$ A single $X$ is also a $X$-component, so your example is $X\rightarrow X\rightarrow Y\rightarrow X\rightarrow Y\rightarrow X$ and has thus three $X$-components. $\endgroup$
    – Werner Thumann
    Commented Mar 19, 2014 at 9:00
  • 2
    $\begingroup$ I guess a proof would go like this: define $N^r_k(C)$ to be the simplicial subset of $N^r(C)$ with at most $k$ $X$-component. Then the inclusion $N^r_k(C)\subset N^r_{k+1}(C) $ would be a homotopy equivalence, because collapsing all cells between, let's say, the first and second occurence of X would be homotopic to the identity. Since $N^r(C)$ is the colimit of $N^r_k(C)$'s, one would get desired weak equivalence by passing to the limit. $\endgroup$
    – user43326
    Commented Mar 20, 2014 at 14:39
  • 4
    $\begingroup$ User43326, I don't see how that helps, as instead of collapsing all cells between the first and second occurence of X, one could always collapse all cells between the first and last occurence of X, yielding a direct proof. The content we require is that such a 'collapse' is a homotopy equivalence, but I can't see that the 'collapse' is even a map of simplicial sets, assuming the version of collapse I'm using is the same as yours. $\endgroup$
    – James Griffin
    Commented Mar 22, 2014 at 18:23
  • 1
    $\begingroup$ Sorry, I was not aware that the definition could cause so much confusion, but the idea is really simple: Take a simplex of the form $A_1\rightarrow A_2\rightarrow A_3\rightarrow ...\rightarrow A_k$ and look at it as a linear graph. Then delete all objects $A_i$ which are not equal to $X$. Then the number of connected components you get is the number of $X$-components. $\endgroup$
    – Werner Thumann
    Commented Mar 23, 2014 at 18:00

1 Answer 1

Reset to default
1
$\begingroup$

I found a proof for this statement. But the reason is another one that one might think. Let $C^-$ be the full subcategory generated by the objects of $C$ minus the $X$. I claim: If there is a simplex with more than one $X$-component at all, then both the inclusions $NC^-\rightarrow NC$ and $NC^-\rightarrow N^rC$ are homotopy equivalences. I give a very short explanation. First observe that the existence of a simplex with more than one $X$-component implies the existence of an object $A$ and arrows $A\rightarrow X$ and $X\rightarrow A$. From this one can show that the comma category $C^-\downarrow X$ is filtered and therefore contractible. Quillen's A implies that the inlusion $NC^-\rightarrow NC$ is a homotopy equivalence. On the other hand $N^rC$ is the pushout of $$NC^-\leftarrow N((C^-\downarrow X) * (X\downarrow C^-))\rightarrow \operatorname{Cone}N((C^-\downarrow X) * (X\downarrow C^-))$$ (That's in fact the reason why I defined $N^rC$). So also $NC^-\rightarrow N^rC$ is a homotopy equivalence.

Probably the following is the upshot of this question: Whenever you see a category with an object $X$ with no non-trivial endomorphisms but an object $A$ with $A\rightarrow X$ and $X\rightarrow A$, then you can kick the $X$ and the homotopy type of the category doesn't change. This follows easily, as pointed out above, from Quillen's Theorem A which we all love so much.

Improve this answer
$\endgroup$
10
  • $\begingroup$ Depending on what "kick the $X$" means, the final upshot in your answer does not appear to be true. Consider a category with $6$ objects $a \leftrightarrow b \rightarrow c \leftarrow \cdots \rightarrow f \leftarrow a$ where the maps between $a$ and $b$ are invertible (hence bidirectional) but the other maps are not. The classifying space with $a$ present is homotopy-equivalent to $S^1$ but if we remove $a$ altogether we get a contractible category. $\endgroup$
    – Vidit Nanda
    Commented Mar 24, 2014 at 18:35
  • 1
    $\begingroup$ @ViditNanda These cannot be the only morphisms - e. g. there must be composite $b\to a\to f$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 24, 2014 at 18:58
  • $\begingroup$ @მამუკაჯიბლაძე Agreed, but that doesn't change the homotopy type of the classifying space. Essentially, you can contract $b$ to $a$ via the obvious natural transformation and update all the morphisms accordingly -- this is not quite the same as deleting $a$ from the category altogether! $\endgroup$
    – Vidit Nanda
    Commented Mar 24, 2014 at 19:19
  • 1
    $\begingroup$ @ViditNanda Both identifying isomorphic objects and leaving less (but at least one) in each isomorphism class gives categories equivalent to the original one $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 24, 2014 at 19:25
  • 1
    $\begingroup$ As pointed out by მამუკაჯიბლაძე, the arrow $b\rightarrow f$ closes the circle in the full subcategory where $a$ is not present, it is also a $S^1$. $\endgroup$
    – Werner Thumann
    Commented Mar 24, 2014 at 22:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .