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It seems obvious that, if the sphere $S^n$ is homeomorphic to a product $X\times Y$ of topological spaces, then either $X$ or $Y$ is a point. How can one prove that?

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Quelle joie de vous retrouver ici, Tryphon! –  Georges Elencwajg Oct 30 '11 at 13:56
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I can't quite justify a few steps, so I'm posting this as a comment. Note that $H_n(S^n)\cong \mathbb Z$ so $H_n(X) \times H_n(Y) \cong \mathbb Z$. Since $\mathbb Z$ can't be written as a non-trivial direct product, either $X$ or $Y$ is $n$-simply connected, assume $Y$ is. In particular $X$ and $Y$ must both be manifolds of dim $m,k\leq n$ where $m+k=n$. Suppose that $\dim X =m < n$, then we have that $H_n(X)$ is nontrivial where $n>m$, which isn't possible. It follows that $X$ must be an $n$-dimensional manifold. So $Y$ must be a $0$ dimensional connected manifold, I.E. a point. –  JSchlather Oct 30 '11 at 15:24
    
@Jacob: Thanks! I think your argument quite correctly shows that, assuming that $X$ and $Y$ are CW-complexes, $Y$ is indeed a point. –  Tournesol Oct 31 '11 at 16:47
    
For those who didn't quite get Georges's reference: you may know Professor Tryphon Tournesol better as Professor Cuthbert Calculus... @Georges: I noticed only now that you and the author of that cartoon are namesakes! :) –  J. M. Nov 1 '11 at 1:58
    
@JacobSchlather: That's not quite how homology interacts with products. See my response below. –  Aaron Mazel-Gee Nov 1 '11 at 20:42

2 Answers 2

up vote 7 down vote accepted

Suppose $S^n = X \times Y$, where $X$ and $Y$ are arbitrary topological spaces (that are homotopy-equivalent to CW complexes). With integer coefficients, there is the (unnaturally) split Kunneth sexseq

$0 \rightarrow \bigoplus_i (H_i(X) \otimes H_{m-i}(Y)) \rightarrow H_m(X \times Y) \rightarrow \bigoplus_i \mbox{Tor}(H_i(X),H_{m-i-1}(Y)) \rightarrow 0$.

Clearly $X$ and $Y$ must be path-connected since $\pi_0$ takes products to products (and obviously we should assume $n \geq 1$), so $H_0(X) = H_0(Y)=\mathbb{Z}$. Since $-\otimes \mathbb{Z}$ does nothing to an abelian group, this means that we're getting a copy of $H_*(X)$ in $H_*(X\times Y)$ from the inclusion above when it's tensored against $H_0(Y)$, and similarly for $H_*(Y)$. Moreover, all the homology of $S^n$ must come from the inclusions in the above sexseq, since $\mbox{Tor}$ always consists entirely of torsion. So without loss of generality, $H_*(X) \cong H_*(S^n)$ and $H_*(Y) \cong H_*(\mbox{pt})$. Since $\pi_1$ takes products to products, both $X$ and $Y$ are simply-connected. So the projection $S^n =X \times Y \rightarrow X$ is a homology isomorphism of simply-connected spaces and hence is a (weak) homotopy equivalence, while $Y$ is a simply-connected space with trivial integral homology so it must be (weakly) contractible. Thus the factorization $S^n = X \times Y$ is trivial.

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Neat! I guess you could do everything over, say $\mathbb{Z}/2$ - that would just kill the Tor term straight away –  Juan S Nov 1 '11 at 7:57
    
Over any field, the Kunneth sexseq becomes a Kunneth isomorphism. However, there's no immediate Relative Hurewicz sort of relationship between homotopy and Z/2 homology, so this wouldn't suffice. I was considering doing it at Z/p for all p and also at $\mathbb{Q}$, which would suffice, but this seemed like more of a pain because there didn't seem to be any immediate reason why it couldn't be that among the various torsion you get, some comes from $X$ and some comes from $Y$. –  Aaron Mazel-Gee Nov 1 '11 at 18:02
    
I should amend my second sentence: There is a framework called "Serre's C-theory" in which one ignores certain "classes of abelian groups", e.g. abelian groups consisting of odd torsion (or possibly just no even torsion, I'm forgetting which is the right thing). For example, a homomorphism is a "mod-C epimorphism" if the cokernel is in C. In this setting, there is a "mod C Hurewicz theorem", which says what you'd hope. (See the excellent book by Mosher & Tangora for details.)... –  Aaron Mazel-Gee Nov 1 '11 at 20:39
    
This led to the first really serious bite (by Serre) that anyone took out of the homotopy groups of spheres, and it also set off a long string of generalizations of the idea of "working one prime at a time" in homotopy theory. –  Aaron Mazel-Gee Nov 1 '11 at 20:41
    
Opps, you are right - I had kind of glossed over the fact that you are using the Hurewicz theorem to get that $S^n \to X$ is a weak homotopy equivalence. (I love Mosher & Tangora too. I guess you could ignore all odd torsion, since 2 is prime) –  Juan S Nov 1 '11 at 22:08

Assuming that everything is a connected manifold.

Let $X$, $Y$ such that $X \times Y = S^n$. Before we apply the Kunneth theorem lets note that the dimensions of $X$ and $Y$ must add up to to $n$. $$\bigoplus_{i+j=k} H^i(X)\otimes H^j(Y)=H^k(S^n) $$

Suppose $\dim X = i < n$. Then $H^i(X)=\mathbb{Z}=H^{n-i}(Y)$ as these are the only possible non-zero cohomology groups that add up to $n$ which is required by the Kunneth theorem. But therefore $H^i(S^n)=\bigoplus H^k(X)\oplus H^j(Y) \neq 0 $ as the summation contains $H^i(X) \oplus H^0(Y)=\mathbb{Z}$ which is a contradiction. Therefore, wlog, $i=n$ and $Y$ is a point.

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You are assuming the manifolds are connected and oriented. –  Mariano Suárez-Alvarez Nov 1 '11 at 0:08
    
Well, a product of manifolds is orientable iff both factors are. My complaint is that we're assuming $X$ and $Y$ are manifolds! –  Aaron Mazel-Gee Nov 1 '11 at 2:34
    
Is it possible for $X \times Y \simeq M$ (in the category of topological spaces) and for $M$ to be a manifold, and $X,Y$ not to be? –  Juan S Nov 1 '11 at 7:59
    
Certainly! Take $X$ to be homotopy equivalent to a manifold but not homeomorphic to one, and take $Y$ to be a point. Being a manifold is a very strong condition; among other things, it means that you have Poincare duality (possibly twisted). Thus, it is a much more restrictive answer to show that $S^n$ is not the product of two manifolds. –  Aaron Mazel-Gee Nov 1 '11 at 17:59
    
@Sven: Integral (co)homology doesn't admit Kunneth isomorphisms; see my closely-related answer. –  Aaron Mazel-Gee Nov 1 '11 at 18:04

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