injective map in cohomology theory 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 $k<n$ ? Is this result true? If yes ow can one show it? thanks in advance!
ronald
 A: 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 $H^1$.
