Mapping homotopic to the identity map has a fixed point Suppose $\phi:\mathbb{S}^2\to\mathbb{S}^2$ is a mapping, homotopic to the identity map.  Show that there is a fixed point $\phi(p)=p$.

Here is an idea: for each point $p \in S^2$, let $v_p$ be the result of orthogonally projecting $\phi(p)-p \in \mathbb{R}^3$ onto the tangent plane at $p$. If $\phi$ has no fixed point, this would given an everywhere nonzero continuous section of the projection $T(S^2) \to S^2$, which we know can't exist ("hairy ball theorem"). 
Edit: Of course the idea above is botched, as Harald pointed out. A possibly flaky second idea is to consider a homotopy $H(t, -)$ from $\phi$ to the identity. Assume that $\phi$ has no fixed point. For each $p$, let $t(p)$ be the least $t$ such that $\|H(t, p) - p\| = (1/2)\|\phi(p)-p\|$. Then define $\psi(p) = H(t(p), p)$. This $\psi$ has no fixed point either, and $\psi(p)$ is never antipodal to $p$. The remaining thing to check is that $p \mapsto t(p)$ is continuous. 
