I'm reading a paper and here the authors say that a connected 4-manifold with zero rational top homology has a homotopy type of 3-dimensional CW-structure. I can't figure out how it can be done.
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4$\begingroup$ For this to be true, you need to assume that your 4-manifold is orientable. Assuming this, your conditions imply that your 4-manifold is not compact (otherwise the 4th homology group would be Q). It is a general fact that smooth noncompact $n$-manifolds are homotopy equivalent to $(n-1)$-dimensional CW complexes. For details, see Mohan Ramachandran's answer to my question here : mathoverflow.net/questions/18454 $\endgroup$– Andy PutmanOct 11, 2010 at 19:39
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3$\begingroup$ Consider $M=RP^4$ (or any other non-orientable closed 4-manifold). It has $H_4(M;\mathbb Q)=0$, but $H_4(M,\mathbb Z/2)\ne 0$, hence it is not homotopy equivalent to a 3-dimensional CW-complex. $\endgroup$– Sergei IvanovOct 11, 2010 at 19:44
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$\begingroup$ @Andy @Sergei Could someone post an answer so that the question gets closed? $\endgroup$– Greg KuperbergOct 12, 2010 at 19:59
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$\begingroup$ Sure Greg, I'll make it into an answer. $\endgroup$– Andy PutmanOct 13, 2010 at 2:39
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For this to be true, you need to assume that your $4$-manifold $M$ is not a compact nonorientable manifold. Otherwise, you would have $H_4(M;\mathbb{Q}) = 0$ but $H_4(M;\mathbb{Z}/2) \neq 0$, so there is no hope that your manifold is homotopy equivalent to a $3$-dimensional CW-complex.
Assuming this, your conditions imply that your $4$-manifold is not compact (otherwise the 4th homology group would be $\mathbb{Q}$). It is a general fact that smooth noncompact $n$-manifolds are homotopy equivalent to $(n−1)$-dimensional CW complexes. For details, see Mohan Ramachandran's answer to my question here.
answered Oct 13, 2010 at 2:44