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Let $M$ and $N$, with $M$ compact, a couple of topological (differential) manifolds.
What is the easiest/fastest/most elementary way to show that $M\times N$ cannot be homeomorphic (diffeomorphic) to the standard ${\mathbb R}^n$ ?

flag	asked Feb 22 at 14:33 Carlo Mantegazza 578●10

I don't think he's suggesting that $M\times N$ is never $\mathbb{R}^n$, Zev. – Richard Kent Feb 22 at 14:44

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Compare cohomology with coefficients in $\mathbb Z/2\mathbb Z$ (if $M$ is not a point, of course:) – Serge Lvovski Feb 22 at 14:45

Ah, I see I misinterpreted the question. Thanks for mentioning it Professor Kent, and apologies to Professor Mantegazza. – Zev Chonoles Feb 22 at 14:53

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In the topological case, surely the most elementary way is simply to note that the product of two cofibrant spaces (e.g. manifolds) is only contractible when each of the two spaces is contractible. – Sam Nolen Feb 22 at 14:56

no problem Zev – Carlo Mantegazza Feb 22 at 15:20

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Lee Mosher · Answer 1 · 2013-02-22 15:40:00Z UTC

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You can avoid the Kunneth formula, "all" you need is the existence of a single nontrivial homology group of $M$ in positive dimensions, namely $H_{\text{dim} \, M}(M;\mathbb{Z}/2\mathbb{Z}) = \mathbb{Z}/2\mathbb{Z}$. Once you know that, it follows that the identity map on $M$ is not homotopic to a constant, and so the inclusion map $M \to M \times N$ is not homotopic to a constant.

answered Feb 22 at 15:40

Lee Mosher
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	If one of $M$ or $N$ isn't simply connected, the same argument will work with $\pi_1$; this result isn't as general but might be more accessible for undergraduates, as it's probably easier for them to quickly grasp the idea of fundamental groups than of homology (if they haven't seen either before). Then you can explain to them that there are "similar but a little more complicated" things one can use if both spaces are simply connected. – Greg Friedman Feb 22 at 21:22
	@Greg: if $M$ or $N$ is not simply connected, it suffices to observe that $\pi_1(M\times N)$ is non-trivial:) – Serge Lvovski Feb 23 at 5:25

Serge Lvovski · Answer 2 · 2013-02-22 14:49:54Z UTC

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$H^{\dim M}((M\times N),\mathbb Z/2\mathbb Z)\ne 0$ (by virtue of Kunneth's formula), which is not the case for $\mathbb R^n$ if $\dim M>0$.

answered Feb 22 at 14:49

Serge Lvovski
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I had in mind something in this spirit. I was just wondering if it was possible something easier, the question came from one of my undergraduate students. – Carlo Mantegazza Feb 22 at 15:16

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@Carlo: but how are they supposed to prove that two manifolds are not homeomorphic? They are to use some topological invariant anyway. If it is not singular (co)homology, what else could it be if we want it to be as elementary as possible? – Serge Lvovski Feb 22 at 15:36

Topology of ${\mathbb R}^n$

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