Continuous map from Projective Plane to Torus 
Is there a continuous map from the $\mathbb{R}P^2$ to the torus which
  is not homotopic to a constant map?

My working: I am pretty sure the answer is no. But I am not sure how to present the answer "properly".
Is it correct to say that any map $f:\mathbb{R}P^2\to T$ induces $f^*:\mathbb{Z}/2\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$.
$f^*(1+1)=f^*(1)+f^*(1)$
But $f^*(1+1)=f^*(0)=0$
This means $f^*(1)=0$, and $f^*(0)=0$, and so $f$ must be the constant map.
Thanks for any help.
 A: It is not true in general that a map that induces 0 on the level of fundamental groups is null-homotopic. (Take the identity map of $S^2$!) You need further argument, such as:
Because $f_*$ is zero, you can factor $f$ through the universal cover $\Bbb R^2 \to T^2$. Because $\Bbb R^2$ is contractible, the map is null-homotopic. 
E: The same argument works in more generality. Suppose $X,Y$ are decent path-connected spaces such that $Y$ has contractible universal cover. Then $f: X \to Y$ is null-homotopic iff it induces the zero map on the level of fundamental groups. Actually, much more is true: if $X$ and $Y$ are path-connected CW complexes with $Y$ having contractible universal cover, then continuous maps $X \to Y$ are homotopic iff they induce the same map on the level of fundamental groups, and every map is induced. This takes a bit more care and requires one to think about the actual cell decomposition, but it's true, and it's fantastic.
A: observe that $\mathbb{R^2}$ is the universal cover of torus, and since $f^*$ is the zero map, so by map lifting lemma, you can lift $f$ in $\mathbb{R^2}$, and since $\mathbb{R^2}$ is contractible, so image is contractibe , i.e image is homotopic to zero. Now compose the homotopy with covering map will give a null homotopic map in $\mathbb{T^2}$.
