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I have the following question, which I dont really know if its true: Let $g : X \rightarrow Y$ be a continous map between two closed, oriented $n-$dimensional manifolds such that $g^{*} : H^{n}(Y, \mathbb{Q}) \rightarrow H^{n}(X, \mathbb{Q})$ is non-zero (here in this case we have $H^{n}(X, \mathbb{Q}) = \mathbb{Q}$ and $H^{n}(Y, \mathbb{Q}) = \mathbb{Q}$). How can one show that the map $g^{*} : H^{k}(Y, \mathbb{Q}) \rightarrow H^{k}(X, \mathbb{Q})$ is injective for $k

ronald

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    Aahhh! I've made a huge mistake. I posted a "counterexample" using the Hopf manifold with its map to $\mathbb P^n$, but forgot that the dimensions of the manifolds don't match. Sorry about that.2012-05-29
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    does someone have any ideas ?2012-05-29
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    What have you tried? If you see closed, oriented manifold then you should probably think about Poincare duality. Then to get back cohomology information, you have the universal coefficient theorem. So follow your nose...it might all work!2012-05-29
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    well ok. this actually is equivalent to showing that $g_{*}: H_{k}(X, \mathbb{Q}) \rightarrow H_{k}(Y, \mathbb{Q})$ is surjective. But how to do that? I dont see any reason why this should be surjective. Do you ?2012-05-29
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    @ronald: Dear ronald, Your reformulation still hasn't taken into account Poincare duality, which is why it doesn't get you any further. Regards,2012-05-30

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We will use Corollary 3.39 of Hatcher's Algebraic Topology book, paraphrased to fit this problem better. (It is a corollary of Poincar$\acute{\text{e}}$ duality.)

Suppose $M$ is a closed orientable $n$-manifold. Choose once and for all a nonzero element $z\in H^n(M;\mathbb{Q})$. Then, for any $0\neq x\in H^k(M;\mathbb{Q})$ (with $k\leq n$), there is a cohomology class $y\in H^{n-k}(M;\mathbb{Q})$ such that $x\cup y = z$.

How does this help with your specific problem?

Well, choose $0\neq z\in H^n(Y;\mathbb{Q})$ once and for all. The hypotheses of the problem guarantee $f^\ast(z) \neq 0$. Now, let $0\neq x\in H^k(Y;\mathbb{Q})$. We want to show $f^\ast(x)\neq 0$. By the above corollary, we know $x\cup y = z$ for some appropriate choice of $y$. Then $0\neq f^\ast(z) = f^\ast(x\cup y) = f^\ast(x)\cup f^\ast(y)$, so $f^\ast(x)\cup f^\ast(y) \neq 0$. This shows $f^\ast(x)\neq 0$.

Edit I misinterpeted ronald's last comment to be about cohomology rather than homology. I believe his comment is correct and should follow from naturality of Poincar$\acute{\text{e}}$ duality. Nonetheless, I'm leaving the example up because there may be some pedagogical value in doing so.

Finally, as to your last comment, the problem is not equivalent to showing $f^\ast$ is surjective, since we don't know $H^k(Y;\mathbb{Q})$ and $H^k(X;\mathbb{Q})$ have the same dimension (as $\mathbb{Q}$-vector spaces). In fact, the map $f:S^1\times S^1\rightarrow S^2$ collapsing the one skeleton of $S^1\times S^1$ to a point is injective on $H^2$ (and thus, by the problem we just did it's injective on $H^0$ and $H^1$), but it is not surjective on $H^1$.

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    +1: very nice. Lucky are the students in your cohomology class...2012-05-29
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    Nice! ${}{}{}{}{}{}$2012-05-29
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    His claim was rather $f_*$ is surjective (at homology level, rather than cohomology), which is probably true?2012-05-29
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    @Sanchez: I think you're right. I'll edit.2012-05-30