= Solution
Let the integer <degree of a map between oriented manifolds> be $d\ne0$, so $f_*[X]=d[Y]$. For any nonzero $\alpha\in H^k(Y;\mathbb Q)$, nondegeneracy of the <Poincare duality pairing> supplies $\gamma\in H^{n-k}(Y;\mathbb Q)$ with $\langle\alpha\smile\gamma,[Y]\rangle\ne0$. Naturality of the <cup product> and its evaluation gives
$$
\langle f^*\alpha\smile f^*\gamma,[X]\rangle=\langle\alpha\smile\gamma,f_*[X]\rangle=d\langle\alpha\smile\gamma,[Y]\rangle\ne0.
$$
Thus $f^*\alpha\ne0$, proving injectivity of pullback on rational <cohomology>. By the <universal coefficient theorem for cohomology>, these finite-dimensional rational <cohomology> groups are the duals of the corresponding <homology (mathematics)> groups, and the dual of $f_*$ is $f^*$. Injectivity of the dual map therefore proves
$$
\boxed{f_*:H_k(X;\mathbb Q)\twoheadrightarrow H_k(Y;\mathbb Q)\quad\text{for every }k.}
$$
One can see the surjection constructively. Given $u\in H_k(Y;\mathbb Q)$, choose its <Poincare dual> $\eta\in H^{n-k}(Y;\mathbb Q)$ with $\eta\cap[Y]=u$. Then $v=d^{-1}(f^*\eta\cap[X])$ satisfies $f_*v=d^{-1}(\eta\cap f_*[X])=u$ by <naturality of the cap product>. Division by $d$ is the reason rational coefficients suffice.
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