Suppose that $f:S^n \to M$ is a map from the $n$-sphere to a simply-connected $n$-dimensional manifold that induces an isomorphism on top homology. I wonder if it's true that $f$ is already a homotopy equivalence.

I will first present the arguments of a fellow user which show that $M$ can have no non-torsion elements in reduced homology below dimension $n$. Assuming the contrary, there exists a primitive non-torsion element $\alpha \in H^k(M,\mathbb Z)$ for $0 < k < n$. By Poincaré-duality, there exists some $\beta \in H^{n-k}(M,\mathbb Z)$ such that $\alpha \cup \beta$ is generator of $H^n(M,\mathbb Z)$. But $f^*$ is natural with respect to cup products and an isomorphism in top cohomology. We arrive at a contradiction, since $H^k(S^n) = 0$ for $0 < k < n$.

Taking $\mathbb Q$ coefficients (thereby killing torsion), we have that $f$ induces an isomorphism on all homology groups with $\mathbb Q$ coefficients. Since $M'$ was assumed to be simply connected, it follows by the rational Hurewicz theorem that $f$ induces an isomorphism on all homotopy groups tensored with $\mathbb Q$. The Whitehead theorem, however, cannot be imediately applied now. Is there any other way to conclude that $f$ is a homotopy equivalence ?

  • $\begingroup$ It would be nice if you referenced the answer this comes from, where I suggested in the answer that you deal with torsion by taking $\Bbb Z/p$ coefficients and doing the same argument. $\endgroup$ – user98602 Oct 30 '15 at 19:48
  • $\begingroup$ Sorry, I am still quite new on this site and wasn't sure how to reference correctly. Also, I explained why you cannot do the same argument on $\mathbb Z/p$ coefficients by giving a concrete example and pointing out that that particular part of your argument just uses the fact that $f^*$ is injective. $\endgroup$ – H1ghfiv3 Oct 30 '15 at 20:02
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    $\begingroup$ The point is that $f^*: H^n(M;\Bbb Z/p) \to H^n(S^n;\Bbb Z/p)$ is an isomorphism by the assumption that $f$ is degree 1 (if you like, use the universal coefficient theorem). Now we are in the same situation. The reason it did not work in your example is that a map of degree 2 induces zero in mod 2 coefficients. $\endgroup$ – user98602 Oct 30 '15 at 20:07
  • $\begingroup$ Thanks, this was my mistake, after all. $\endgroup$ – H1ghfiv3 Oct 30 '15 at 20:08
  • $\begingroup$ BTW, I apologize for the hostile tone of my previous couple comments. I was grumpy for independent reasons and that shone through. $\endgroup$ – user98602 Oct 30 '15 at 20:11

While the question as asked has already been resolved, let me remark that the hypothesis that $M$ is simply connected is unnecessary. Indeed, suppose $M$ is possibly not simply connected, and let $p:\tilde{M}\to M$ be its universal cover. We may assume $n>1$, so that our degree $1$ map $f:S^n\to M$ lifts to a map $\tilde{f}:S^n\to\tilde{M}$. It follows that the isomorphism $f_*:H_n(S^n,\mathbb{Z})\to H_n(M,\mathbb{Z})$ factors through $H_n(\tilde{M},\mathbb{Z})$. In particular, this means that $H_n(\tilde{M},\mathbb{Z})\neq 0$ so $\tilde{M}$ must be compact; it follows that $p$ is a finite cover. The degree of $p$ (i.e., the map induced on $H_n$) is then equal to the degree of $p$ as a covering map. Since $1=\deg f=\deg \tilde{f}\deg p$, we must have $\deg p=1$. Thus $p$ is a degree $1$ cover and hence a homeomorphism, so $M\cong \tilde{M}$ is simply connected.


Mike Miller has already given a perfect answer to this question in the context on another similiar question, which can be found here.

I will just present his argument in a bit more detail, the first part of which can be found in the second paragraph of my original question. This proofs that for $0 < k < n$, the homology groups $H_k(M,\mathbb Z)$ are all torsion groups. In order to proof that they are indeed trivial, the crucial observation lies in seeing that since $f_* : H_n(S^n,\mathbb Z) \to H_n(M,\mathbb Z)$ is an isomorphism, so is $f^*: H^n(M, \mathbb Z/p) \to H^n(M,\mathbb Z/p)$ for all $p$ prime. Now assuming that $H_k(M, \mathbb Z)$ had an element of finite order $p$, it would give rise to primitive element in $\alpha \in H^k(M,\mathbb Z/p)$. By the very same argument as before, using Poincaré-duality, naturality of cup product and the fact that $f^*$ is an isomorphism in top $\mathbb Z/p$-cohomology, one would obtain a non-trivial element $\alpha' \in H^k(S^n,\mathbb Z/p)$, a contradiction, since we assumed $0 < k < n$.

It follows that $f$ induces an isomorphism on all Homology groups with $\mathbb Z$ coefficients, and, since $M$ is assumed to be simply connected, the standard Hurewicz/Whitehead chain proofs that $f$ is a homotopy equivalence

Edit: You don't actually need to distinguish between torsion and non-torsion elements. For suppose there exists some non-trivial $\alpha \in H_k (M)$ Poincaré dual to some $\gamma \in H^{n-k}(M) $. Then we would have $0 = f_*([S^n] \cap f^*(\gamma)) = [M] \cap \gamma = \alpha $.


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