If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $f(\mathbb{S}^n)\neq \mathbb{S}^n$, where $\mathbb{S}^n=\left \{x\in\mathbb{R}^{n+1}:\|x\|=1\right \}$. Prove that $f$ has a fixed point.
This is an exercise on the chapter about Brouwer's fixed-point theorem in a Differential Topology book (Chinese version). I think we may need to extend $f$ to a map from $D^{n+1}=\left \{x\in\mathbb{R}^{n+1}:\|x\|\leq 1\right \}$ to itself, so that we can apply Brouwer's fixed-point theorem to get some useful conclusions. However, I don't know how to achieve the goal.
Another thought: If $y_0\in \mathbb{S}^n\setminus f(\mathbb{S}^n)$, using stereographic projection, there is a homeomorphism $\varphi :\mathbb{S}^n\setminus \{y_0\}\to \mathbb{R}^n$. Hence $\varphi \circ f:\mathbb{S}^n\to \mathbb{R}^n$ is a continuous function. However, I don't know how to proceed from this.
I'm not good at geometric stuffs. If my question is silly, forgive me please. Any help would be appreciated!
Since $f(\mathbb S^n)$ is a closed subset of $\mathbb S^n$ and is not the whole $\mathbb S^n$, there is a subset $D$ of $\mathbb S^n$ homeomorphic to the closed unit disc $\mathbb D^n$ so that $f(\mathbb S^n) \subset D$. Thus $f|_D : D\to D$ is a continuous map and by Brouwer's fixed point theorem it has a fixed point $x_0$. This fixed point is of course also a fixed point of $f$.
Remark on how to choose $D$: Assume that $y_0\in \mathbb S^n$ is not in the image of $f$. Let $g : \mathbb S^n \to \mathbb R$ be the function $g(x) = x\cdot y_0$ (usual dot product). Then $g\circ f: \mathbb S^n \to \mathbb R$ is a continuous function so that $g\circ f (y) <1$ for all $y\in \mathbb S^n$. Since $\mathbb S^n$ is compact, there is $\delta \in (0,1)$ so that $g\circ f (y) \le 1-\delta$ for all $y\in \mathbb S^n$. Now choose $D = \{ y\in \mathbb S^n : g(y) \le 1-\delta\}$.