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Mar 4, 2011 at 21:41 comment added Eric Wofsey For purposes of dualizing, any homology class in $H_n(X)$ is pushed forward from a homology class of a space $Y$ such that $H_n(Y)$ is finitely generated and free (proof: glue together the simplices that make up the chain in $X$ to obtain an $n$-dimensional complex $Y$). The dualizing argument should then work on $Y$, and by naturality you then get the result for $X$.
Mar 4, 2011 at 19:24 comment added Johannes Ebert I had this argument in mind. However, I was unable to make the step "follow by duality" precise.
Mar 4, 2011 at 17:11 history answered Eric Wofsey CC BY-SA 2.5