Exhibit a map $X \to K(\pi_1(X),1)$?

Question

Is there a natural map $X \stackrel{f}{\to} K(\pi_1(X),1)$?

Can you tell me how it is constructed and what properties has the map $\pi_1(f)$?

Answer 1

Yes. One way to describe it is as follows: prove the more general statement that if $G$ is a discrete group, then maps $X \to BG$, which classify $G$-covers, correspond to maps $\Pi_1(X) \to BG$ from the fundamental groupoid of $X$ to $BG$. (Here $BG$ happens to be $K(G, 1)$, but this is special to the discrete case; if $G$ isn't discrete then $BG$ is a more complicated space.) If $X$ is pointed and path-connected, they furthermore correspond to maps $\pi_1(X) \to G$ (up to conjugation). This is a slight elaboration on covering space theory.

Now if $G = \pi_1(X)$ there is a distinguished such map, namely the identity. This gives a distinguished (homotopy class of) map $X \to B \pi_1(X)$. It has the property that its homotopy fiber is, up to homotopy, the universal cover of $X$.

This construction generalizes: if $X$ is $(n-1)$-connected, there is a distinguished map $X \to B^n \pi_n(X)$ inducing an isomorphism on $\pi_n$ whose homotopy fiber is the $n$-connected cover of $X$, a generalization of the universal cover. (Here $B^n \pi_n(X)$ is $K(\pi_n(X), n)$.) Iterating this construction for all $n$ gives the Whitehead tower of $X$; Serre famously used a version of this construction ("killing homotopy groups") to compute homotopy groups of spheres.