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I asked this question on math.stackexchange, but no one answered.

Let $(X,\le)$ be a totally ordered set. Regarding it as a category, it has a classifying space $B(X,\le)=|N_\bullet(X,\le)|$. This should be the (possible infinite) simplex with vertices $X$, hence I expect it to be contractible.

However, I was not able to explicitly prove contractibility starting from the definition $$B(X,\le)=(\coprod_{i\in\mathbb{N}_0}N_i(X,\le)\times\Delta^i)/\tilde{} $$

Can anyone help me?

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  • $\begingroup$ A finite totally ordered set has a maximum so it has a final object as a category and its classifying space is contractible. On the other hand $X$ can be considered as a direct limit of its finite (ordered) subsets and so its classifying space is a direct limit of contractible spaces hence contractible. $\endgroup$
    – Mostafa - Free Palestine
    Commented May 17, 2015 at 10:52
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    $\begingroup$ Any filtered category has contractible classifying space. First source is, I guess, Quillen's "Higher algebraic K-theory" (in the Springer LNM 341), with very nice proof. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented May 17, 2015 at 10:54
  • $\begingroup$ Here is a proof sketch. First, a lemma: if a $T_1$-space $Z$ is a filtered colimit of subspaces $Z_i$ which are closed under intersection, and each $Z_i$ contains only finitely many other $Z_j$'s, then every compact subspace of $Z$ is contained in some $Z_i$. Corollary: Under these conditions, if the $Z_i$'s are contractible, then $Z$ must be weakly contractible, since every sphere in $Z$ factors through some $Z_i$. In conclusion, the space $BX$ is weakly contractible since it is a filtered colimit of contractible simplices. Hence, the CW-complex $BX$ is contractible. $\endgroup$
    – Ricardo Andrade
    Commented May 17, 2015 at 13:27
  • $\begingroup$ @Mostafa Do you have a reference for the claim that a directed colimit of contractible spaces is contractible? Am I missing something obvious? $\endgroup$
    – Todd Trimble
    Commented May 17, 2015 at 14:02
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    $\begingroup$ @Mostafa In an earlier comment I used the phrase "directed colimit", which I believe is pretty standard. (I thought you might have meant this. The nLab ncatlab.org/nlab/show/directed+colimit notes that "direct limit" is also sometimes used, so I guess my earlier comment was overly harsh, although I do find that "direct limit" is confusing because of the synonymous usage noted above). $\endgroup$
    – Todd Trimble
    Commented May 17, 2015 at 15:46

2 Answers 2

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Let $P$ be any partially ordered set. For any map $x\colon P\to [0,1]$, put $\sigma(x)=\{p\in P:x(p)>0\}$. Then $BP$ can be identified with the set of maps $x$ such that $\sigma(x)$ is finite and totally ordered, and $\sum_px(p)=1$. Now suppose that $a$ is an element of $P$ which is comparable with every other element, and define $e\colon P\to [0,1]$ by $e(a)=1$ and $e(p)=0$ for $p\neq a$. It is then easy to see that the map $h_t(x)=(1-t)x+te$ preserves $BP$ and gives a contraction. This construction is most often used when $a$ is largest or smallest in $P$, but you really only need it to be comparable with every element of $P$. In particular, if $P$ is nonempty and totally ordered then you can choose $a$ arbitrarily.

As another way to look at this, if $f,g\colon P\to Q$ are two poset maps, and $f(p)\leq g(p)$ for all $p$, then it is standard that $Bf$ is homotopic to $Bg$. If $P$ is totally ordered, we can define $f,g,h\colon P\to P$ by $f(p)=p$ and $g(p)=\max(a,p)$ and $h(p)=a$. Then $f\leq g\geq h$, so $Bf$, $Bg$ and $Bh$ are homotopic, but $Bf$ is the identity and $Bh$ is constant.

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  • $\begingroup$ Which topology should one use on the set of maps $x$, which are identified with $BP$? The one coming from the (possible infinite) product $[0,1]^P$? $\endgroup$
    – Tatjana Popow
    Commented May 17, 2015 at 12:11
  • $\begingroup$ The space $BP$ is topologised as the colimit of the spaces $BQ$, as $Q$ runs over finite subsets of $P$. So a subset $F\subset BP$ is closed iff $F\cap BQ$ is closed with respect to the obvious topology on $BQ$, for all $Q$. $\endgroup$
    – Neil Strickland
    Commented May 17, 2015 at 12:16
  • $\begingroup$ Thank you. Do you have a reference for the standard fact about homotopic maps on classifying spaces induced by order-preserving maps between posets? $\endgroup$
    – Tatjana Popow
    Commented May 17, 2015 at 12:34
  • $\begingroup$ The two maps combine to give a poset map $\{0,1\}\times P\to Q$, and $B(\{0,1\}\times P)=[0,1]\times BP$. $\endgroup$
    – Neil Strickland
    Commented May 17, 2015 at 13:01
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    $\begingroup$ So if (nonempty) $P$ merely has binary joins, the same argument as in the second paragraph shows $BP$ is contractible. This hadn't occurred to me before now. $\endgroup$
    – Todd Trimble
    Commented May 17, 2015 at 13:54
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I am not sure if you consider this explicit, but here you go:

Choose a point $x_0\in X$, and let $F\colon X\to X$ be the functor that sends $x$ to itself if $x\geq x_0$, and to $x_0$ otherwise. There is a zig-zag of natural transformations $$x_0\leq F(x)\geq x$$ between the constant functor $x_0$ and the identity functor, because $X$ is totally ordered. On classifying spaces this gives a homotopy between the constant map $x_0$ and the identity.

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    $\begingroup$ Emanuele, isn't this just the argument given in Neil's second paragraph? $\endgroup$
    – Todd Trimble
    Commented May 17, 2015 at 15:48
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    $\begingroup$ Yes, that's exactly the same... I should have read more carefully. $\endgroup$
    – Emanuele Dotto
    Commented May 17, 2015 at 15:50

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