8
$\begingroup$

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?

$\endgroup$
3
  • 2
    $\begingroup$ 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 $C^N(T)$. $\endgroup$ Aug 16, 2013 at 21:24
  • 1
    $\begingroup$ @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. $\endgroup$ Aug 16, 2013 at 23:36
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 17, 2013 at 6:23

2 Answers 2

5
$\begingroup$

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 $K(\pi,1)$.

$\endgroup$
2
  • $\begingroup$ Where do you use the finiteness of $K$ in the part that "is not hard to see"? $\endgroup$ Aug 16, 2013 at 23:00
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 16, 2013 at 23:07
1
$\begingroup$

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 $K = C^N(T)$.

$\endgroup$
2
  • 5
    $\begingroup$ 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 $K(\pi,1)$. $\endgroup$ Aug 16, 2013 at 19:16
  • 2
    $\begingroup$ A minor nitpick: $C^n(T)$ is not a compact space. $\endgroup$ Aug 16, 2013 at 23:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.