Let $S^2$ be the unit sphere in $\mathbb{R}^3$ and $f:S^2\to S^2$ continuous with the property, that it exists a non-empty open subset $U$ of $S^2$ with $f(x)=x$ for every $x\in U$ and $f(x)\notin U$ for every $x\in S^2\setminus U$. Show: $f\simeq \operatorname{id}$
I have to find a continuous function $h: S^2\times [0,1]\to S^2$ with $h(x,0)=f(x)$ and $h(x,1)=\operatorname{id}$ for every $x\in S^2$
I do not really find a way to tackle this problem. Can someone give a hint? Thanks in advance.
Note that we may assume without loss of generality that $U$ is an open disk since they form a basis of the topology. Then $S^2\setminus U$ is also a disk, and hence contractable. Radially contracting this disk to a point, and extending this homotopy to $U $ in the natural way, therefore sends all points of $S^2 \setminus U$ to a single point. This homotopy will be $h_{[0,\frac{1}{2}]}$. For the second half of the homotopy, just let $h (x,t)= (2t-1)x$ on $S^2\setminus U$ thought of as a disk in $\mathbb{C}$, extended in the natural way to $U$.