$f:S^2\to S^2$ is homotopic to $\operatorname{id}$

Question

Let $f:S^2\to S^2$ be a continuous with the property that it exists a nonempty,open $U\subseteq S^2$ with $f(x)=x$ for every $x\in U$ and $f(x)\notin U$ for $x\in S^2\setminus U$.

Prove that $f$ is homotopic to $\operatorname{id}: S^2\to S^2$.

My first idea was to take the homotopy

$H: S^2\times [0,1]\to S^2$ given by $(x,t)\mapsto \dfrac{tx+(1-t)f(x)}{\|tx+(1-t)f(x)\|}$.

For $x\in U$ this works. For $x\in S^2\setminus U$ one has to argue that $tx+(1-t)f(x)\neq 0$ for every $t\in [0,1]$

If $tx+(1-t)f(x)=0$ for some $t$, then the direct line from $x$ to $f(x)$ has to intersect the origin, and therefor $f(x)=-x$, so $f(x)$ and $x$ are antipodal.

I tried to argue that this can not happen, but I did not succeed, and I do not think that this homotopy will be enough for a proof.

I have found this statement in a textbook:

For even $n$ has every continuous $f: S^n\to S^n$ a fixpoint or an antipodal point. Someone knows if this is exclusive, or inclusive? So can such a continuous function have a fixpoint and an antipodal point?

I would like to know if my initial homotopy actually works, and it is worth trying to find the missing details, or if I should search for an other homotopy.

Answer 1

Your homotopy does not work in general. If $U$ is "small", then $f(x) = -x$ is indeed possible for $x \in S^2 \setminus U$.

Choose $x_0 \in U$ and let $T$ denote the tangential plane to $S^2$ at $x_0$, i.e. $T = \{ x_0 + y \in \mathbb R^3 \mid \langle x_0, y \rangle = 0 \}$. For $r \in (-1,1)$ define $T_r = \{ rx_0 + y \in \mathbb R^3 \mid \langle x_0, y \rangle = 0 \}$. This is a plane which is parallel to $T$ and contains the point $rx_0$. It separates $S^2$ in two closed spherical caps $C^+$ and $C^-$ such that $x_0 \in C^+$ and $-x_0 \in C^-$. Both caps are homeomorphic to the disk $D^2$. The intersection $S = C^+ \cap C^-$ is a circle which is the common boundary of the two caps.

For $r$ sufficiently close to $1$ we have $C^+ \subset U$. Clearly $f$ is identity on $C^+$ and in particular the identity on the $S$. By construction we have $f(C^-) \subset C^-$: If $x \in C^- \cap U$, then $f(x) = x \in C^-$ and if $x \in C^- \cap (S^2 \setminus U)$, then $f(x) \in S^2 \setminus U \subset S^2 \setminus C^+ \subset C^-$.

Using the identification $h : C^- \stackrel{\approx}{\to} D^2$ we now see that $f : C^- \to C^-$ is homotopic to the identity rel. $S$ which implies that $f$ is homotopic to $id_{S^2}$. In fact, consider $g = h^{-1}f h : D^2 \to D^2$ which is the identity on $S^1$ and take the "linear homotopy" on $D^2$ from $g$ to $id$. This gives the desired homotopy on $C^-$.

Answer 2

You can do this directly by construction of a homotopy as the other answers show.

A map $f \colon \mathbb{S}^2 \to \mathbb{S}^2$ can have antipodal and fixed points at the same time. For this view $\mathbb{S}^2$ as embedded in $\mathbb{C} \times \mathbb{R} \cong \mathbb{R}^3$ and put $f(z,x) = (z^2,x)$. Then $f(0,1) = (0,1)$ is a fixed point and $f(-1,0) = (1,0)$ shows that $f$ has antipodal points.

Here a homological proof of the original statement: $\require{AMScd}$ Pick $x \in U$ and an $1 > \varepsilon > 0$ such that the open ball $U' := B_\varepsilon(x)$ is in $U$, where $\mathbb{S}^2$ is equipped with the standard metric. Let $V' = \mathbb{S}^2 \setminus \overline B_{\varepsilon / 2} (x)$. Clearly $f(U') \subset U'$ and $f(V') \subset V'$ and $f$ restricts to the identity on $U' \cap V'$. The fact that $\mathbb{S}^2 = U' \cup V'$ is an open cover yields a Mayer-Vietoris sequence which yields a commutative diagram $\require{AMScd}$ \begin{CD} H_2(U') \oplus H_2(V') @>>> H_2(\mathbb{S}) @>\partial>> H_1(U' \cap V') @>>> H_1(U') \oplus H_1(V')\\ @V V H_2(f|_{U'}) \oplus H_2(f|_{V'}) V @VV H_2(f) V @VV H_2(f|_{U' \cap V'}) = \operatorname{id}_{H_2(U' \cap V')} V @V VH_1(f|_{U'}) \oplus H_1(f|_{V'}) V \\ H_2(U') \oplus H_2(V') @>>> H_2(\mathbb{S}) @>\partial>> H_1(U' \cap V') @>>> H_1(U') \oplus H_1(V') \end{CD} with exact rows. Now $\partial$ is an isomorphism as $U'$ and $V'$ are contractible. Hence, $H_2(f) = \operatorname{id}_{H_2(\mathbb{S}^2)}$. By the Hurewicz theorem $\pi_2(f) = \operatorname{id}_{\pi_2(\mathbb{S}^2)}$, hence $f = f \circ \operatorname{id}_{\mathbb{S}^2}$ is homotopic to $\operatorname{id}_{\mathbb{S}^2}$ by definition of $\pi_2(f)$.