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edited Sep 5, 2014 at 4:23

Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_3(\tilde N)=0$$H_n(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)

Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_3(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)

Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_n(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)

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created Sep 4, 2014 at 14:43

Allen Hatcher

Some further comments on homology spheres: First, for any closed connected orientable $n$-manifold $M$ there is always a degree $1$ map $M\to S^n$ obtained by collapsing the complement of an open ball in $M$ to a point. In the reverse direction, any degree $1$ map $f:M\to N$ of closed connected orientable $n$-manifolds must induce a surjection on $\pi_1$, for otherwise $f$ could be lifted to the covering space $\tilde N \to N$ corresponding to the proper subgroup $f_*(\pi_1M) \subset \pi_1N$. If this covering is finite-sheeted, then the degree of the projection $\tilde N\to N$ is equal to the number of sheets (which is the index of $f_*(\pi_1M)$ in $\pi_1N$) since an orientation of $N$ lifts to an orientation of $\tilde N$, making the local degrees at all points in a fiber have the same sign. Thus the degree of $f$ is divisible by the number of sheets, so it can't be $1$. If the covering is infinite-sheeted then $\tilde N$ is noncompact so $H_3(\tilde N)=0$, forcing $f$ to have degree $0$. Applying this fact to a degree $1$ map $S^n\to N$ we see that $\pi_1N=0$ so it can't be a nonsimply-connected homology sphere. (These are classical arguments, incidentally.)