There is hint: Prove $H^{n-1}(N) \to H^{n-1}(M)$ is trivial. Just don't know how to prove this.
$M$ is a compact manifold with boundary $N$,then $M$ can't retract onto $N$.
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algebraic-topology
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1Further hint: Think about homology instead of cohomology. What *are* homology classes, anyways? – 2011-06-24
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1Related: [Is there a retraction of a non-orientable manifold to its boundary?](http://math.stackexchange.com/q/880016/10014), [Retraction to the Boundary on Compact Manifold](http://math.stackexchange.com/q/1257814/10014), [A manifold such that its boundary is a deformation retract of the manifold itself.](http://math.stackexchange.com/q/1359563/10014) – 2016-05-25