I'm struggling with the following question, hopefully someone can help me.
Let $f, g: \mathbb{S}^{2n}\rightarrow\mathbb{S}^{2n}$ be two contionuous mappings. Show that if $\deg(f) + \deg(g) \neq0$, then there exists a $x\in\mathbb{S}^{2n}$ such that $f(x)=g(x)$.
I don't see how to even start, honestly. I'm aware of some theorems related to this question, like for example;
it is known that for these continuous mappings from $\mathbb{S}^{n}$ to itself if the mapping $f$ has a fixed point there is a formula for the degree namely $\deg(f)=(-1)^{n+2}$ same for if it has an antifix point in which case it should equal $1$ i.e. $\deg(f)=1$ this implies that for spheres of even dimension there must either be a fixed point or an antifixed point.
If the degree of such different mappings is equal they are homotopically equivalent.
I think I should be able to use these facts to solve the mentioned question, but I don't see it. Any help in the right direction would be greatly appreciated!
Thanks!
We use the following fact:
Lemma. Let $a, b \colon \mathbb{S}^k \to \mathbb{S}^k$ be two continuous maps. If $a(x)$ and $b(x)$ are never antipodal, then $a \simeq b$.
Proof. Embed $\mathbb{S}^k \subseteq \mathbb{R}^{k+1}$. Since $a$ and $b$ are never antipodal, we may consider the linear homotopy $$H \colon [0,1] \times \mathbb{S}^k \to \mathbb{R}^{k+1} - \mathbf{0}; \quad \quad H_t(x) = (1-t)a(x) + tb(x)$$ from $a$ to $b$. Projecting $H$ back onto the sphere yields the desired homotopy.
Onto the main problem. Suppose that $f(x) \neq g(x)$ for all $x \in \mathbb{S}^{2n}$. Then $f$ and $-g$ are never antipodal, and hence homotopic. It follows that $\deg(f) = \deg(-g)$. On an even-dimensional sphere, the antipodal map $-\, \colon \mathbb{S}^{2n} \to \mathbb{S}^{2n}$ has degree $-1$, yielding $$\deg(f) + \deg(g) = \deg(f) - \deg(-g) = 0.$$