A map which is trivial on homology but not on cohomology?

Is there a map $f:X\to Y$ of connected CW-complexes which induces the trivial map $f_*=0:H_i(X,\Bbb Z) \to H_i(Y,\Bbb Z)$ for all $i\ge 1$, but with the property that the induced map on cohomology $f^*:H^i(Y,\Bbb Z)\to H^i(X, \Bbb Z)$ is nonzero for at least one $i \ge 1$?

Note that the universal coefficient theorem does not a priori imply that the induced map is trivial because the splitting $H^i(X, \mathbb{Z}) = H_i(X, \mathbb{Z})^* \oplus \operatorname{Ext}^{i-1}(X, \mathbb{Z})$ is not natural. Since I don't see any other reason against it, I believe such a counterexample should exist.

• Tiny remark: the identity map of the real projective plane gives a trivial answer to the question you didn't ask: it is trivial in cohomology (because the cohomology is itself trivial) but not in homology. Feb 6 '15 at 19:09
• @Pseudo: that's not true. $H^2(\mathbb{RP}^2, \mathbb{Z}) \cong \mathbb{Z}_2$ by universal coefficients, and the identity induces the identity on this. Feb 6 '15 at 19:47
• @PseudoNeo Trivial cohomology $\implies$ trivial homology (see eg. this), if that helps you not repeat that mistake :) Feb 6 '15 at 19:53
• Oh, right. Thank you very much... Feb 6 '15 at 20:59

Take the $n$-sphere and attach a $n+1$-cell with a degree $m$ map. Consider $X \to X/S^{n+1}$ where $X$ is the described space.
$$\begin{array}{c} \cdots & \to& 0 &\to& \mathbb Z&\stackrel {* m} \to& \mathbb Z & \to &0& \cdots \\ &&\downarrow&&\downarrow&&\downarrow \\ \cdots & \to &0&\to&\mathbb Z& \to&0& \to & 0 &\cdots \end{array}$$
Apply $hom(-,\mathbb Z)$ and then homology (ie cohomology), you will get on degree $n+1$ the quotient map $\mathbb Z \to \mathbb Z/m$.
you can see exercise 11 of chapter 3.1 of hatcher.let X obtained from$S^n$ by attaching a cell of degree m.you can easily see (by cellular homology and cohomology) the map $X\to X/S^n$ is trivial on $H_i$ but not on $H^{n+1}$