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This question might have an easy answer, but my research is far from the region of topology that makes use of classifying spaces of categories, so I can't find it.

For (possibly infinite) integers $0 \leq k_1 \leq k_2 \leq \infty$, define a category $X(k_1,k_2)$ as follows. There are objects $S_n$ for any finite integer $n$ satisfying $k_1 \leq n \leq k_2$. The morphisms from $S_n$ to $S_m$ are maps $f:\{0,\ldots,n\} \rightarrow \{0,\ldots,m\}$ such that $f(i) < f(j)$ whenever $i < j$.

I remark that for $k_1=0$ and $k_2 = \infty$, this is almost the usual category of simplices. The only difference is that we require strict inequality $f(i) < f(j)$ in our morphisms rather than merely inequality $f(i) \leq f(j)$.

The classifying space of the category $X(k_1,k_2)$ is clearly $(k_2-k_1)$-dimensional. My question is whether or not this classifying space is $(k_2-k_1-1)$-connected? I can verify by hand that it is connected if $k_2-k_1 \geq 1$ and simply-connected if $k_2-k_1 \geq 2$, but my techniques are too ad-hoc to deal with the general case.


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up vote 4 down vote accepted


Claim 1: $X(k,\infty)$ is contractible.

Proof: There is a functor $X(k,\infty)\times X(k,\infty)\to X(k,\infty)$ given by laying two ordered sets end to end. It admits a natural map from each of the two projection functors, and this implies that the resulting map of spaces (realizations of nerves) is homotopic to each of the two projections. But when a nonempty space $X$ is such that the two projections $X\times X\to X$ are homotopic then $X$ is contractible.

Claim 2: The relative homology $H_j(X(k_1,k_2+1),X(k_1,k_2))$ is trivial for $j\ne k_2-k_1+1$.

Proof: Let $P(k_1,k_2+1)$ be the poset of subsets of $\lbrace 0,\dots ,k_2+1\rbrace $ of cardinality at least $k_1+1$, and let $bP(k_1,k_2+1)$ be the subposet of proper subsets. There is a functor $P(k_1,k_2+1)\to X(k_1,k_2+1)$ inducing an isomorphism of quotients of nerves $P(k_1,k_2+1)/bP(k_1,k_2+1)\to X(k_1,k_2+1)/X(k_1,k_2)$. $P(k_1,k_2+1)$ is contractible, while $bP(k_1,k_2+1)$, being isomorphic to the $(k_2-k_1)$-skeleton of a $(k_2+1)$-simplex, has its reduced homology all in dimension $k_2-k_1$.

Claim 2 implies by induction on $m$ that the relative homology $H_j(X(k_1,k_2+m),X(k_1,k_2))$ is trivial for $j\le k_2-k_1$. Thus $H_j(X(k_1,\infty),X(k_1,k_2))$ is trivial for $j\le k_2-k_1$. But this is isomorphic to the (reduced) homology $\tilde H_{j-1}X(k_1,k_2)$, by Claim 1.

It follows that $X(k_1,k_2)$ has no homology below dimension $k_2-k_1$. Since you say you have checked that it is also $1$-connected (if $k_2-k_1\ge 2$), this makes it $(k_2-k_1-1)$-connected.

EDIT More simply: The inclusion map $X(k_1,k_2-1)\to X(k_1,k_2)$ is homotopic to a constant because it admits a natural map from a constant map. Therefore the cofiber $X(k_1,k_2)/X(k_1,k_2-1)$ is homotopy equivalent to $\Sigma X(k_1,k_2-1)\vee X(k_1,k_2)$. And this cofiber is isomorphic to a wedge of $(k_2-k_1)$-dimensional spheres, by the proof of Claim 2. So up to homotopy $X(k_1,k_2)$ is a retract of a wedge of such spheres. I suppose you could also figure out the rank of $\tilde H_{k_2-k_1}X(k_1,k_2)$ by this method, by induction on $k_2$.

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Thank you very much! – T- the confused Dec 30 '10 at 6:05

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