Given the map on the torus, $f : T=\Bbb R^2 / \Bbb Z^2 \to T$ given by $f[x,y]=[x+y,y],$ I want to show that the map induced by $f$ on the homotopy group, $f_*:\pi_1(T)\to \pi_1(T),$ is not the identity.
(Here, $[x,y]$ denotes $q(x,y)$, where $q : \Bbb R^2 \to T $ is the quotient map.)
Intuitively, identifying $\pi_1(T)$ with $\Bbb Z^2$, I think that $f_* : \Bbb Z^2 \to \Bbb Z^2$ is the map $(x,y) \mapsto (x+y,y)$. But how do I have to show this?
Thomas Andrews provided a sketch of proof in the comments, with an explicit homotopy in mind (probably).
Here's another sketch of a solution, using the Eckmann-Hilton argument : $f$ can be described as $T \to T\times T \to T$ where the first map is $g:(x,y) \mapsto ((x,y), (y,0))$ and the second map is the "multiplication" $m:T\times T\to T$ (which is actually addition here)
It follows that $f_* = m_*\circ g_*$. But now the Eckmann-Hilton argument allows us to identify $m_*$ as simply the addition map $\pi_1(T)\times\pi_1(T)\to\pi_1(T)$, and we can compute $g_*$ easily because it decomposes as $(id_T, p)$ where $p$ is well-known and $p_*$ is easily computable. This yields an easy computation for $f_*$.
If you go through the Eckmann-Hilton argument and write down the homotopies every time, you probably end up with the same thing as what Thomas Andrews offered.