# A continuous map $f:S^2\rightarrow S^2$ such that $f(x)\neq f(-x)$ for all $x$ is surjective

For all $$x\in S^2$$, let $$-x$$ denote its antipode. Let $$f:S^2\rightarrow S^2$$ be a continuous map such that $$f(x)\neq f(-x)$$ for all $$x\in S^2$$. Show that $$f$$ must be surjective.

I'm working through an old practice exam, and I've been stuck on this question for hours.

I think this has something to do with the Borsuk Ulam Theorem, or maybe Brouwer degree but I am not sure how to construct a function $$S^2\rightarrow \mathbb{R}$$ that makes $$f$$ not being surjective a contradiction, or that contradicts anything about degrees.

I've found a somewhat similar question here, but not sure how to adapt it to this problem.

I know that if $$f$$ is not surjective, $$\exists y\in S^2$$ such that $$y\notin f(S^2)$$, so I can construct a well-defined map $$g:S^2\rightarrow S^2,\hspace{2cm} g(x)=\frac{f(x)-y}{|f(x)-y|}$$ that's homotopic to $$f$$. Not sure how that helps.

I also thought of the map $$h:S^2\rightarrow S^2,\hspace{2cm}h(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|}$$ which we can well-define, and is homotopic to both $$f$$ and $$f\circ A$$ where $$A:x\rightarrow -x$$.

I also know that if $$f$$ is not surjective then its degree is $$0$$, and that the degree of homotopic maps is equal, so the degree of $$g$$, $$h$$ and $$f\circ A$$ is zero, but not sure if there's anything wrong with that either.

The most elementary answer possible would be most appreciated.

Remember that $$S^2 \cong \mathbb{R}^2 \cup \{ \infty \}$$ (this is stereographic projection).
So (towards a contradiction) say we're given a map $$f : S^2 \to S^2$$ isn't surjective. Well without loss of generality we can say it misses $$\infty$$, so that composing with stereographic projection gets us a map $$f' : S^2 \to \mathbb{R}^2$$.
Now, since $$\forall x . f(x) \neq f(-x)$$, we see that the same must be true for $$f'$$...