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Given a map between two manifolds that induces an isomorphism on integral cohomology in the top dimension, it follows from naturality of the cup product and Poincaré duality and universal coefficient theorem that all maps on cohomology in every dimension are injective with torsion free cokernel.

Is there an example where one of these maps is not surjective?

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For any orientable manifold M of dimension n, map M to $S^n$ by sending some suitable neighborhood of a point homeomorphically to $\mathbb{R}^n$ and the rest to $\infty$.

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    $\begingroup$ E.g. let $M = {\mathbb CP}^2$, and collapse the line at infinity to a point, mapping to $S^4$. $\endgroup$ – Allen Knutson Jun 29 '10 at 15:02
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I think I have one: consider a torus in $\mathbb{R}^{3}$ embedded as a surface of rotation, e.g. rotate a circle of radius $1$ in the zy plane with center $(0,2,0)$ around the $z$ axis. Now put a small sphere with center $(0,2,0)$ and radius $\epsilon$. Then the map $T \rightarrow S^{2}$ given by projection towards the center of the sphere should give an isomorphism in $H^{2}$ (since the degree is 1) but it is obviously not surjective in $H^{1}$

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