The 2 skeleton of a 3 manifold is the 2 skeleton of a $K(\pi, 1)$

Question

Recall that a topological space $M$ is called aspherical if is path connected and the homotopy groups $\pi_n(M) $ vanish for $n\geq 2$.

A (smooth) 3-manifold $M$ is an homology sphere if $H_*(M,\mathbb{Z})\simeq H_*(\mathbb{S}^3,\mathbb{Z})$.

Is true that if a closed 3-manifold is an homology sphere then its 2-skeleton is aspherical? How can we see this?

I was reading a paper that at a certain point says assumes that the 2-skeleton is a $K(\pi,1)$ but I can't see how to use the assumption on the homology groups of $M$ to prove asphericity.

Answer 1

I think you misunderstood what was being claimed.

Since the $2$-skeleton on $M$ is the $2$-skeleton of a $K(\pi_1(M), 1)$ ...

It is not claimed that the $2$-skeleton of $M$ is a $K(\pi_1(M), 1)$. Rather, there is a $K(\pi_1(M), 1)$ which has the same $2$-skeleton as $M$.

This has nothing to do with homology spheres or even manifolds. If $X$ is a two-dimensional CW complex, then we can add cells in dimensions at least three to obtain a $K(\pi_1(X), 1)$. The two-skeleton of this $K(\pi_1(X), 1)$ is $X$ itself (because we only added cells of higher dimension).