TITLE: A map which is trivial on homology but not on cohomology?
QUESTION [7 upvotes]: 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.

REPLY [7 votes]: 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. 
You can write down the cellular chain complexes:
$$
\begin{array}{c}
\cdots & \to& 0 &\to& \mathbb Z&\stackrel {* m} \to& \mathbb Z &  \to  &0& \cdots \\
&&\downarrow&&\downarrow&&\downarrow \\
\cdots & \to &0&\to&\mathbb Z& \to&0&  \to  & 0 &\cdots 
\end{array}
$$
Apply homology, you will kill every possible map (the degrees where the non-trivial groups lie are disjoint, except for the trivial degree).
Apply $hom(-,\mathbb Z)$ and then homology (ie cohomology), you will get on degree $n+1$ the quotient map $\mathbb Z \to \mathbb Z/m$.