EDIT: After a little prompting by Mark Grant, I answered the first question in the comments. The second question remains open.
Let $M$ be a compact $3$-manifold with $\pi_2(M) \neq 0$. The sphere theorem says that there is an embedded 2-sphere in $M$ that is nontrivial in $\pi_2$.
In his book "Group theory and 3-dimensional topology", Stallings has a proof of this using his theorem about groups with infinitely many ends. He first has an easy reduction to the case where none of the boundary components of $M$ are $2$-spheres and all of them are incompressible. The next step is to prove that the universal cover $\tilde{M}$ has at least 2 ends, and I'm having trouble understanding his proof of this. I have two questions. The first one is probably pretty easy (I bet there is something silly I'm missing), but the second one seems fundamental.
What he says is as follows. Using Poincare duality and the fact that all the boundary components of the universal cover of $\tilde{M}$ are contractible, we have
$$\pi_2(\tilde{M}) = H_2(\tilde{M}) \cong H_2(\tilde{M},\partial \tilde{M}) \cong H^1_c(\tilde{M}),$$
so
$$H^1_c(\tilde{M}) \neq 0.$$
Let $H^{k}_e(\tilde{M})$ denote the homology of the cochain complex obtained by quotienting the using simplicial cochain complex of $\tilde{M}$ by the complex of cochains with compact (or, equivalently, finite) support. We then have a long exact sequence containing the segment
$$H^0_c(\tilde{M}) \rightarrow H^0(\tilde{M}) \rightarrow H^0_e(\tilde{M}) \rightarrow H^1_c(\tilde{M}) \rightarrow H^1(\tilde{M}).$$
We have $H^1(\tilde{M})=0$ since $\tilde{M}$ is simply-connected. He then says that $H^0_c(\tilde{M}) = 0$ since $\tilde{M}$ is noncompact. This is the first thing I don't understand:
- Question 1: Why can't $\tilde{M}$ be compact? This is equivalent to asking why the hypotheses imply that $\pi_1(M)$ is infinite. I know that geometrization implies that $\pi_1(M)$ being finite would imply that $M$ has universal cover $S^3$ and hence $\pi_2(M) = 0$, but obviously this should not depend on that!
Let's grant that $\tilde{M}$ is noncompact. Stallings then says that $H^1(\tilde{M})=0$ since $\tilde{M}$ is $1$-connected. We thus have a short exact sequence
$$0 \rightarrow \mathbb{Z} \rightarrow H^0_e(\tilde{M}) \rightarrow H^1_c(\tilde{M}) \rightarrow 0.$$
Since the cokernel of this is nontrivial $H^0_e(\tilde{M})$ has rank at least $2$. Stallings then asserts that this means that $\tilde{M}$ has at least $2$ ends.
- Question 2: Why is that? The number of ends of $\tilde{M}$ should equal the dimension of $H^0_e(\tilde{M};\mathbb{F}_2)$, but all we know is that the integral cohomology group $H^0_e(\tilde{M})$ has rank at least $2$. If $\pi_2(M) = H^1_c(\tilde{M})$ consisted entirely of $p$-torsion with $p \neq 2$, then its contribution would disappear when we work mod-$2$.
Best Answer
What you need for your second question is that $H^1_c(\tilde{M})$ is a free abelian group. As you noted, this is isomorphic to $H_2(\tilde{M})$. Now, $\tilde{M}$ is a connected non-compact 3-manifold. A basic theorem of Whitehead says that a connected noncompact PL $n$-manifold is homotopy equivalent to an $(n-1)$-dimensional simplicial complex. See my note here for a proof and references. Applying this to $\tilde{M}$, we find that $\tilde{M}$ is homotopy equivalent to a $2$-dimensional simplicial complex $X$, so $H_2(\tilde{M}) \cong H_2(X)$. Letting $C_{\bullet}(X)$ be the simplicial chain complex, we have $C_3(X) = 0$, so $$H_2(X) = \text{ker}(C_2(X) \stackrel{\partial}{\rightarrow} C_1(X)).$$ In particular, $H_2(X)$ is a subgroup of the free abelian group $C_2(X)$, and thus is itself free abelian.