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I'm reading the paper "Configuration Spaces and Braid Groups on Graphs in Robotics" by Robert Ghrist, in which he states and proves the following theorem:

Theorem: Given a tree $T$, the configuration space $C^N(T)$ (of $N$ distint ordered points on $T$), and a connected subset $K\subset C^N(T)$, if the homomorphism $\pi_1(K)\rightarrow \pi_1(C^N(T))$ induced by inclusion is trivial, then $K$ is nullhomotopic in $C^N(T)$.

and then as a corollary states the following without proof or citation:

Corollary: The configuration space $C^N(T)$ is an Eilenberg-MacLane space of type $K(\pi_1,1)$: i.e., $\pi_k(C^N(T))=0$ for all $k>1$.

To get from the theorem to the corollary, I'm guessing he is using some sort of general sufficient condition for a space to be Eilenberg-Maclane, namely: If every inclusion of a subspace $K$ into a space $X$ is nullhomotopic whenever it induces a trivial map on $\pi_1$, then $X$ is an Eilenberg-Maclane space. I don't remember having seen anything like this, nor have I been able to find anything. I hope I'm not missing something obvious. Anyone have any ideas?

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    $\begingroup$ Without some additional niceness hypothesis on $K$, I don't believe Ghrist's theorem is true. For instance, for an annulus $X$, there is a simply connected subset $K$ that is not nullhomotopic in $X$ (take a Warsaw circle around the hole). It should be easy to do something similar inside $C^N(T)$. $\endgroup$
    – Eric Wofsey
    Commented Aug 16, 2013 at 21:24
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    $\begingroup$ @Eric: In fact, it appears that you can topologically embed by a non-null map $W\to C^2(T)$ a Warsaw circle $W$ in the space of pairs of distinct points in the tree $T$ given by the cone on three points. Note that $C^2(T)$ has the homotopy type of a circle. $\endgroup$
    – Ricardo Andrade
    Commented Aug 16, 2013 at 23:36
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    $\begingroup$ Having looked at Ghrist's argument, it seems that the flaw is in the final sentence: having homotoped the inclusion $K\to C^N(T)$ onto a graph, he claims that graph must be nullhomotopic in $C^N(T)$. This need not be true since that graph might have loops that don't lift to $K$ (which will be the case if $K$ is a Warsaw circle). But I believe the argument does work if you assume $K$ is locally path-connected. $\endgroup$
    – Eric Wofsey
    Commented Aug 17, 2013 at 6:23

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As I said in my comment, Ghrist's theorem is stated in too much generality to possibly be true. But if, say, you restrict to requiring $K$ to be a finite CW complex, here's a counterexample to your question. Let $X$ be $S^2$ with two points identified. Then it is not hard to see that any finite complex $K\subset X$ for which $\pi_1(K)\to \pi_1(X)$ is trivial must be nullhomotopic in $X$. But $X\simeq S^2\vee S^1$ is not a $K(\pi,1)$.

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  • $\begingroup$ Where do you use the finiteness of $K$ in the part that "is not hard to see"? $\endgroup$
    – Vidit Nanda
    Commented Aug 16, 2013 at 23:00
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    $\begingroup$ The way I imagine proving the "not hard to see" part is to show (eg, by induction on the number of cells in $K$) that if $K$ contains the singular point of $X$, $K$ is contained in a contractible subset whose boundary is a union of two loops contained in the 1-skeleton of $X$. It is quite possible that you can do away with the finiteness assumption with a little more work. $\endgroup$
    – Eric Wofsey
    Commented Aug 16, 2013 at 23:07
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Let $f:S^n\to C$ and let $K$ be its image. If $n>1$ then the Theorem implies $f\simeq *$.

EDIT: Clearly this is over-simple and wrong, as Omar Antolín-Camarena has pointed out. In fact, it looks like $C^N(T)$ would be a compact CW complex, and so there will be surjective continuous functions $S^n \to C^N(T)$ for any $n\geq 1$; then $K = C^N(T)$.

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    $\begingroup$ Why is $i_\ast:\pi_1(K) \to \pi_1(C)$ trivial? If for example, $F$ were a homeomorphism $S^n \to K$, sure. But in general, $i_\ast$ need not be trivial, for example, $f$ could squash $S^n$ to an interval and wrap it a along an essential $1$-dimensional loop in $C$. I guess if $C$ were a regular CW-complex (which it almost surely is), applying your observation to the attaching maps does show that $C$ is a $K(\pi,1)$. $\endgroup$
    – Omar Antolín-Camarena
    Commented Aug 16, 2013 at 19:16
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    $\begingroup$ A minor nitpick: $C^n(T)$ is not a compact space. $\endgroup$
    – Ricardo Andrade
    Commented Aug 16, 2013 at 23:48

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