TITLE: A sufficient condition for a space to be an Eilenberg-Maclane space
QUESTION [8 upvotes]: 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?

REPLY [5 votes]: 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 $K(\pi,1)$.