Past exam of the metric spaces and complex analysis course of the University of Oxford / 2021 / 2 / c / Solution

We first show that a continuous injection $j:S^1\to S^1$ must be surjective. Otherwise its image misses a point $q$. Stereographic projection identifies $S^1\setminus \{q\}$ with $\mathbb{R}$, giving a continuous injection $u:S^1\to \mathbb{R}$. The extreme value theorem supplies distinct points $p_{-},p_{+}$ where $u$ attains its minimum $m$ and maximum $M$, with $m<M$ because $u$ is injective.

The two circular arcs from $p_{-}$ to $p_{+}$ each contain a point at which $u=(m+M)/2$, by the intermediate value theorem. Those points are distinct because they lie in the disjoint interiors of the two arcs. This contradicts injectivity, proving the claim about $j$.

Now suppose $F:S^2\to S^1$ were a continuous injection. Its restriction to the equatorial circle $E=\{(x,y,0):x^2+y^2=1\}$ would be a continuous injection from $S^1$ to itself, hence $F(E)=S^1$. The north pole $N=(0,0,1)$ is outside $E$, but $F(N)$ must equal $F(e)$ for some $e\in E$. This contradicts injectivity. Therefore