is there a continuous surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H1(T,x) \rightarrow H1(T,x)$is the zero-map. Let $T = \mathbb{S}^1 \times \mathbb{S}^1$ be a torus and $x \in T$. Prove or disprove: There exists a continuous
surjective map $f : T \rightarrow T$ such that the induced homomorphism $f^* : H_1(T,x) \rightarrow
H_1(T,x)$is the zero-map.  
I have no idea how to solve this kind of problems. All I know is that since the fundamental group of the torus is abelian we may think of $f^*$ as a map of fundamental groups instead. We can also say by the lifting lemma that our maps lifts to its universal cover $\mathbb{R}^2$ which is contractible, but I don't know if  this is relevant.  
Any help will be much appreciated
 A: Here is an outline on constructing such a map.
I'm thinking of $S^1 = \{z\in \mathbb{C}: |z|=1\}$.  Then define $g:S^1\rightarrow S^1$ by $g(z) = \begin{cases} z^2 & \operatorname{Im}(z)\geq 0\\ \overline{z}^2 & \operatorname{Im}(z)\leq 0\end{cases}$.
Note that if $\operatorname{Im}(z) = 0$, then $z = \pm 1$, so $z^2 = \overline{z}^2  =1$ in this case.  Thus, $g$ really is continuous.
Intuitively, $g$ wraps the top half of $S^1$ fully around $S^1$ one way, and wraps the bottom half of $S^1$ fully around $S^1$ the other way.
Can you prove $g$ is surjective?  Can you prove that $g$ is the $0$ map on $H_1$?  Can you use $g$ to construct $f:T\rightarrow T$ with the properties you're looking for?
A: You can try also the following solution: Take any surjective continuous map from $T$ to $I$ - unit interval, for example a height function. Then using Peano curve you can find surjective continuous map from $I$ to $I^2$, and then the quotient map from $I^2$ to $T$. This map will be continuous and surjective, and induces $0$ in $H_1(X)$, since it factors through contractible space. 
Actually, this map is null-homotopic.
To be totally fair, my friend gave me this solution, so I am not claiming I had this idea myself. Thanks Andrzej :)
