There's something basic about the torus that I don't understand.
Given an integral-linear map $\varphi : \mathbb{Z}^2 \rightarrow \mathbb{Z}^2$, we get a real-linear map $\psi : \mathbb{R}^2 \rightarrow \mathbb{R}^2,$ which descends to a continuous function $\tilde{\psi} : \mathbb{R}^2/\mathbb{Z}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2$ from the torus $T^2$ to itself. This yields a group homomorphism $\tilde{\psi}_* : \pi_1(T_2) \rightarrow \pi_1(T_2).$
Now given $p \in \mathbb{Z}^2$, we get a path $\omega_p$ in $\mathbb{R}^2$ by moving from the origin to $p$ in a straight line. We can push this path across the quotient map $\mathbb{R}^2 \rightarrow T^2$ and since it's start and end points are elements of $\mathbb{Z}^2$ the result will be a loop based at $[0,0] \in T^2$. This means that $p \mapsto [\omega_p]$ is a map $\mathbb{Z}^2 \rightarrow \pi_1(T_2).$
So far so good, but then something amazing happens. It seems to be the case that: $$\tilde{\psi}_*([\omega_p]) = [\omega_{\varphi(p)}].$$
In more detail, we seem to have $$\tilde{\psi}_*([\omega_p]) = [\tilde{\psi} \circ \omega_{p}] = [\omega_{\varphi(p)}],$$ except that I don't know how to justify the second equality.
Question. Is this true? If so, why is it true?
As I wrote in the comments, $\psi\circ \omega_p$ is a path starting at $0$ (because $\omega_p$ does and $\psi(0)=0$) and ending at $\psi(p)=\varphi(p)$.
Moreover $\psi$ is linear so it sends segments to segments, hence $\psi\circ\omega_p$ is a segment. By definition, $\psi\circ\omega_p = \omega_{\varphi(p)}$.
Passing to the quotient, this gives the desired equality.
Note that you could equally well get this equality by writing down the explicit formulas for $\omega_q$, you would perhaps see more clearly how the linearity of $\psi$ gets involved