Circle to circle homotopic to the constant map? How to prove that a continuous function, homotopic to the constant map $f:S^1\to S^1$  (a) has a fixed point and that (b) has a point $x$, such that $f$ maps $x$ to its antipodal point $-x$?

For the first part, let $i\colon S^1\to D^2$ be the inclusion of the circle into the unit disk and, since $f$ is null-homotopic, let $\tilde{f}\colon D^2\to S^1$ be an extension of $f$ to the whole disk (which exists). Since $f$ has no fixed points, and the image of $\tilde{f}$ lies within $S^1$, what can we say about $i\circ \tilde{f}\colon D^2\to D^2$ and what theorem about maps on disks does this contradict?
For the second part, just prove that the composition of a nullhomotopic map with the map which rotates the circle by $\pi$ is also nullhomotopic (hint: rotation is homotopic to the identity and if $f\simeq f'$ and $g\simeq g'$ then $f\circ g\simeq f'\circ g'$), and then use part a.
