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I am stuck on the following qualifying Exam question.  I would appreciate any help. 

Let $RP^2$ and $T$ denote the eal projective plane and the torus.  Prove that any map $f: RP^2 \rightarrow T$ is homotopic to a constant map.

Ethan Hawver
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Any map $f:RP^2 \longrightarrow T$ induces a map on the fundamental groups, and hence we get a map $f_*:\mathbb{Z}/2\mathbb{Z} \longrightarrow \mathbb{Z} \oplus \mathbb{Z}$. Any such map is trivial.

Now consider the universal cover of the torus, $p:\mathbb{R^2} \longrightarrow T$. Since the image of our map is trivial, it certainly lies inside $p_*(\pi_1(\mathbb{R}^2)) = \{{1}\}$. So by the lifting criterion, there exists a lift $\tilde f:RP^2 \longrightarrow \mathbb{R}^2$ such that $f = p \circ \tilde f$. Noting that $\mathbb{R^2}$ is contractible, this might help you find your nullhomotopy.

squiggles
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