I am trying to find a map between the torus and the circle which is not nullhomotopic.
There is a theorem that says
Let $p:E\rightarrow B$ be a covering map. Let $f:Y\rightarrow B$ be a continuous map. The map can be lifted to a map $\tilde{f}:Y\rightarrow E$ if and only if, $f_*(\pi_1(Y))\subseteq p_*(\pi_1(E))$.
My idea/guess:
We can see the torus as $S^1\times S^1$. We can use the projection map $f:S^1\times S^1\rightarrow S^1$ from the first coordinate onto the circle. Since $f$ is surjective we have $$f_*(\pi_1(Y))\subsetneq p_*(\pi_1(E)).$$ So I guess that the map $f$ should be non-nullhomotopic. But how to prove this?
You have $f$ is projection from $S^1\times S^1\to S^1$. Define $g$ in the reverse direction by $g(z)=(z,*)$ where $*$ is your basepoint. Then $f\circ g$ is the identity on $S^1$. So it induces the identity on $\pi_1(S^1,*)$. So some element of $\pi_1(S^1\times S^1)$ is mapped by $f_*$ to the generator of $\pi_1(S^1,*)$, an infinite cyclic group.