Currently I am studying for a topology exam next week and came across an exercise where I could need some hints (cf. here):
Let $X$ be a path-connected space with $\pi := \pi_1(X,*)$ abelian and contractible universal cover $\widetilde{X}$. Suppose $f \colon Y \rightarrow X$ is a continuous map so that the induced map $f^* \colon H^1(X, \pi) \rightarrow H^1(Y, \pi)$ is zero. Prove $f$ is null-homotopic.
Some thoughts I had so far: From Hurewicz and the fact that the fundamental group is already abelian we get that $\pi \cong \pi^{ab} \cong H_1(X;\mathbb{Z})$. Also from the universal coefficient theorem we get (since $H_0$ is always free and so the $Ext$-term vanishes, $G$ abelian group) an isomorphism $H^1(X; G) \rightarrow Hom_{\mathbb{Z}}(H_1(X;\mathbb{Z}), G)$. Similarly for $Y$, and we could take $G = \pi$ in the isomorphism above.
Since we somehow have to know something about the homotopy type of $f$ and we have the assumption that the universal covering is contractible, maybe we are looking for some lift
\begin{array} & & & \widetilde{X} \\ & \nearrow & \downarrow{p} \\ Y & \xrightarrow{f} & X \end{array}
of $f$ to the universal cover (then the map is nullhomotopic because it factors through a contractible space).
We can probably get that the map induced by $f$ on the fundamental groups is trivial and with more assumtions on $Y$ (path-connected and locally path-connected) that would give us the lift. But in the exercise Y seems to be a general topological space.
What approach would you take to this?
They mean to have the locally path-connected assumption, presumably, but didn't feel the need to state it (often "space" can be taken to mean "non-terrible space").
As a counterexample in general, take the Warsaw circle $X$. This has trivial homotopy groups and homology groups (exercise), but collapsing the 'bad part' gives us a non-null map $X \to S^1 =: Y$. This is Hatcher's exercise 1.3.7.