Let $M,N$ be a closed (connected, without boundary, say smooth) manifolds which are homotopy equivalent. Does it follows that they are of the same dimension? One should be aware of examples of contractible manifolds (say, open balls), which can be of arbitrary dimension but are homotopy equivalent to a point: however those manifolds are not closed.
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$\begingroup$ Poincare duality $\endgroup$– Alex DegtyarevJul 18, 2015 at 20:23
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For a closed, connected topological $n$-manifold, H_n(X)≠0 (in fact, it's either Z or Z/2Z, depending on orientability) $H_n(X;\mathbb{Z}/2\mathbb{Z})= \mathbb{Z}/2\mathbb{Z}$ and $H_m(X)=0$ if $m>n$, hence dimension can be characterised homologically.
Since homology is a homotopy equivalence invariant, so is dimension (for closed, connected topological manifolds).
answered Jul 18, 2015 at 13:06
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5$\begingroup$ Isn't $H_2(\mathbb{RP}^2, \mathbb{Z}) = 0$? $\endgroup$ Jul 18, 2015 at 13:35
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4$\begingroup$ If $M$ is not $R$-orientable, the top homology with coefficients in $R$ is isomorphic to the 2-torsion of $R$ hence $0$ in the case $R=\mathbb{Z}$. You can test the dimension e.g. with homology with coefficients in $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$ Jul 18, 2015 at 14:00
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$\begingroup$ Ooops. You are absolutely right, I had things mixed up, and I have now edited accordingly. $\endgroup$ Jul 19, 2015 at 9:14