Homotopy 4-spheres and degree 1 maps

Question

In 3-dimensions a compact connected orientable 3-manifold $M$ is a homotopy 3-sphere (and hence by the Poincaré conjecture $S^3$) if and only if there is a degree 1 map $f: S^3 \to M$. I want to know some examples that show me how this result fails in dimension 4. Namely, I would like a 4-manifold $X$ that is not a homotopy 4-sphere and a degree 1 map $f: S^4 \to X$.

Answer 1

As is shown in this question, if $M$ is a closed, connected, orientable $n$-manifold and $f : S^n \to M$ is a degree $d$ map, then for $0 < i < n$ and $x \in H^i(M; \mathbb{Z})$, $dx = 0$. So if there is a degree one map $f : S^n \to M$, $M$ must be an integral homology sphere.

Suppose $f : M \to N$ is a degree one map between closed, connected, orientable manifolds of the same dimension and consider the subgroup $f_*(\pi_1(M)) \subseteq \pi_1(N)$. Associated to this subgroup, there is a covering $\pi : \widetilde{N} \to N$ and a lift $\tilde{f} : M \to \widetilde{N}$ such that $\pi\circ\tilde{f} = f$. As $f$ has degree one, $f_*[M] = [N]$, but on the other hand, $f_*[M] = (\pi\circ\tilde{f})_*[M] = \pi_*(\tilde{f}_*[M])$. If $\pi$ is an infinitely sheeted covering (i.e. $f_*(\pi_1(M))$ has infinite index in $\pi_1(N)$), then $\widetilde{N}$ would be non-compact so $\tilde{f}_*[M] = 0$ as $H_n(\widetilde{N}; \mathbb{Z}) = 0$, but this is impossible as $\pi_*(\tilde{f}_*[M]) = [N] \neq 0$. Therefore $\pi$ is a finite covering and the number of sheets is equal to the index of $f_*(\pi_1(M))$ in $\pi_1(N)$, moreover, it is equal to the degree of $\pi$. As $1 = \deg f = \deg(\pi\circ\tilde{f}) = \deg\pi\deg\tilde{f}$, we see that $\deg\pi = 1$ and therefore $f_*(\pi_1(M)) = \pi_1(N)$. That is, a degree one map induces a surjection on fundamental groups.

Therefore, if $f : S^n \to M$ is a degree one map, $M$ must be simply connected. So $f$ is a map between simply connected CW complexes which induces an isomorphism on all homology groups (note, $f_* : H_n(S^n; \mathbb{Z}) \to H_n(M; \mathbb{Z})$ is an isomorphism because $f$ has degree one). By Whitehead's Theorem, $f$ is a homotopy equivalence and hence $M$ is homotopy equivalent to a sphere. By the solution of the topological Poincaré conjecture, $M$ must be homeomorphic to a sphere.

In conclusion, if $f : S^n \to M$ is a degree one map, $M$ is homeomorphic to $S^n$.

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You might be interested in Mike Miller's answer to a recent question of mine: If $M$ is a rational homology sphere, is there a map $S^n \to M$ of non-zero degree?