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

Question

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.

Answer 1

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 $\mathbb Z \to \mathbb Z/m$.

Answer 2

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 $H^{n+1}$