# 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 $\phi(p)=p$.

## 3 Answers

Have you seen homology and degree of continuous mappings? If so, this is pretty short:

Since $f$ is homotopic to the identity, $\deg f = \deg \text{id}_{S^2} = +1$. On the other hand, if a continuous map $g:S^n\to S^n$ has no fixed points, then $\deg g = (-1)^{n+1}$. This shows that $f$ must have a fixed point, since otherwise $\deg f = (-1)^{2+1} = -1$.

Alternatively, you can prove that any map $S^n\to S^n$ with no fixed point is homotopic to the antipodal map. (This is used to prove the statement about $g$ in the previous proof.) Note that if $f$ has no fixed points, then $$h(x, t) = (1-t)f(x)-tx$$ is nonzero, so $h(x,t)/|h(x,t)|$ defines a homotopy from $f$ to the antipodal map. (This proof is in Hatcher, Algebraic Topology, section 2.2.) And the antipodal map is homotopic to the identity if and only if $n$ is odd, so this proves the claim by contraposition.

Alternatively, this follows from the Lefschetz fixed point theorem. As $\phi$ is homotopic to the identity map, the Lefschetz number of $\phi$ is just the Euler characteristic of $S^2$, which is 2. As $2 \neq 0$, it follows that $\phi$ has at least one fixed point.

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.

• I had that thought too, but it breaks down if $\phi(p)$ is antipodal to $p$. – Harald Hanche-Olsen Oct 25 '13 at 17:30
• @HaraldHanche-Olsen Thanks! I tried a second idea, and I invite you to poke a hole in it. :-) – user43208 Oct 25 '13 at 18:02