Is There an Aspherical 3-Manifold with $H_i(M^3) \cong \mathbb{Z}$ for i = 0 to 3? The title says it all: Is there an aspherical 3-manifold with $H_i(M^3) \cong \mathbb{Z}$ for i = 0 to 3?
 A: Motivated by Moishe's claim in the comments, let's consider the mapping torus $X$ of the map $f : T^2 \to T^2$ given by the matrix $M = \left[ \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right]$, which is listed on the Wikipedia article on geometrization as an example of a Sol manifold. I believe but am not sure how to prove that any mapping torus of an endomorphism of $T^2$ has universal cover $\mathbb{R}^3$, so this is a closed aspherical $3$-manifold; I also believe but am not sure how to prove that $X$ is orientable because $f$ preserves the orientation of $T^2$ (since $\det M = 1$). Hopefully someone else can fill in these details.
Given that we get $H_0(X) \cong H_3(X) \cong \mathbb{Z}$; it remains to show that $H_1(X) \cong \mathbb{Z}$ and then we'll get $H_2(X) \cong H^1(X) \cong \text{Hom}(H_1(X), \mathbb{Z}) \cong \mathbb{Z}$ by Poincare duality. $H_1(X)$ is the abelianization of $\pi_1(X)$ which is the semidirect product $\mathbb{Z}^2 \rtimes \mathbb{Z}$ with $\mathbb{Z}$ acting on $\mathbb{Z}^2$ by the matrix $M$. The abelianization of this semidirect product is the product of the coinvariants of this action on $\mathbb{Z}^2$ and $\mathbb{Z}$. Taking coinvariants means quotienting by $M - I = \left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]$, which is invertible, so the quotient is trivial; hence the abelianization is $\mathbb{Z}$ as desired.
A: Here's a supplement to Qiaochu's answer, proving that the mapping torus is indeed aspherical and orientable.
To see this, recall that if $T_f = M\times [0,1]/(m,0)\sim (f(m),1)$ is the mapping torus of a homeomorphism $f:M\rightarrow M$, we have a fiber bundle $M\rightarrow T_f\rightarrow S^1$, where the projectoin $\pi:T_f\rightarrow S^1$ is $\pi(m,t) = t \text{ mod } 1$.
This gives a long exact sequence in homotopy groups $$...\rightarrow \pi_k(M)\rightarrow \pi_k(T_f)\rightarrow \pi_k(S^1) \rightarrow \pi_{k-1}(M)\rightarrow ...$$
Since $S^1$ is aspherical, for $k\geq 2$, we thus obtain isomorphisms $\pi_k(M)\rightarrow \pi_k(T_f)$.  Thus, if $M$ is also aspherical, then so is $T_f$.
As far as orientability is concerned, we will use Mayer-Vietoris where $U$ denotes the primage in $T_f$ of the northern hemisphere of $S^1$, and $V$ is the preimage of the southern hemisphere.  (Technically, we use slightly more than the northern and southern hemispheres, so we actually cover all of $T_f$.)
Because both hemispheres are contractible, both $U$ and $V$ are diffeomorphic to $X\times (-1,1)$, and $U\cap V$ has the homotopy type of two disjoint copies of $X$.  I'm thinking of $U$ as having preimage in $X\times [0,1/2]$ in $X\times [0,1]$, and I'm thinking of $V$ has having preimage $X\times [1/2,1]$ in $X\times [0,1]$.
Thus, the inclusion map of $U\cap V\cong X\coprod X\rightarrow U$ is homotopic to the identity on each copy of $X$, while the inclusion map $U\cap V\rightarrow V$ is homotopic to the identity on one copy of $X$, and homotopic to $f$ on the other.
Now, the Mayer-Vietoris homology sequence for this is $$...\rightarrow H_k(U\cap V)\rightarrow H_k(U)\oplus H_k(V)\rightarrow H_k(T_f)\rightarrow H_{k-1}(U\cap V)\rightarrow ...$$
The map $H_k(U\cap V)\rightarrow H_k(U)\oplus H_k(V)$ is a map $H_k(X)\oplus H_k(X)\rightarrow H_k(X)\oplus H_k(X)$ incuded by the inclusions.  Based on the above description of the inclusions, this takes an element $(a,b)$ to $(a+b,a+f_\ast(b))$.
Now, focus on the case where $k=\dim T_f = \dim M + 1$ and $M$ is orientable.  Mayer-Vietoris tells us that $H_k(T_f)$ is the kernel of hte map $H_{k-1}(U\cap V)\rightarrow H_{k-1}(U)\oplus H_{k-1}(V)$.  But note that $H_{k-1}(U)\cong H_{k-1}(V)\cong H_{k-1}(M)\cong \mathbb{Z}$.  The map $f_\ast$ is then an isomorphism of $\mathbb{Z}$, so it multiplication by $\pm 1$, and is $-1$ iff $f$ reverses orientation.
Thus, assuming $f$ preserves orientation, the map $(a,b)\mapsto (a+b,a+f_\ast(b))$ is simply $(a,b)\mapsto (a+b,a+b)$, which clearly has kernel $\mathbb{Z}$.  Thus, $H_k(T_f)\cong \mathbb{Z}$, so $T_f$ is orientable.
