Null-homotopic loop around puncture implies that the surface is a sphere

general-topologyhomotopy-theoryriemann-surfaces

  • Let $X$ be a Riemann surface with a point $x\in X$.

  • Let $U\subseteq X$ be a neighborhood of $x$ with a homeomorphism $\phi:U\to D$ to the unit disc $D$.

  • Let $p\in U$ be any point other than $x$ and $\gamma$ a loop in $U$ based at $x$ such that $[\gamma]$ generates $\pi_1(U-\{p\},x)$.

I want to show that if $\gamma$ is null-homotopic in $X-\{p\}$, then $X$ is isomorphic to the sphere.

My idea: any component of the preimage of $U-\{p\}$ in the universal covering of $X-\{p\}$ is certainly a punctured disc if $\gamma$ is null-homotopic. But the universal covering is simply-connected and the only simply-connected Riemann surface that "naturally contains" a punctured disc is the plane since the plane is the sphere punctured at infinity. Then $X-\{p\}$ can only be the plane and thus $X$ is the sphere. But this is too vague… is there a simple and clean way to do this? Maybe one needs homology for this (which I am not familiar with yet)?

Best Answer

Theorem. $\quad$ Let $X$ be a Riemann surface with a point $p\in X$. Let $\phi:U\xrightarrow{\sim}\mathcal{D}$ be a normalized chart at $p$. Let $x\in U$ be any point other than $p$ and $\gamma$ a loop in $U$ based at $x$ such that $[\gamma]$ generates $\pi_1(U-\{p\},x)$. Then if $\gamma$ is nullhomotopic in $X-\{p\}$, it holds that $X\cong\mathbb{C}^\infty$.

Proof. $\quad$ Let $\mathcal{F}:\mathcal{U}\to X-\{p\}$ be a universal unbranched covering such that $\mathcal{U}\in\{\mathbb{C}^\infty,\mathbb{C},\mathcal{H}\}$. Let $A$ be a connected component of $\mathcal{F}^{-1}(U-\{p\})$. It holds that $\mathcal{F}|_{A}:A\to U-\{p\}$ is an unbranched path-connected covering. Since $\gamma$ is nullhomotopic in $X-\{p\}$, the homotopy lifting property implies that any lift of $\gamma$ to $A$ is also nullhomotopic in $\mathcal{U}$ and therefore a loop. Hence, it follows that ${\mathcal{F}|_{A}}_\ast(\pi_1(A,q))=\pi_1(U-\{p\},x)$ for any $q\in{\mathcal{F}|_{A}}^{-1}(\{x\})$. Thus, $\mathcal{F}|_{A}:A\to U-\{p\}$ is isomorphic to $\operatorname{Id}:U-\{p\}\to U-\{p\}$ and so it is a finite covering.

Now, we do an operation known as the "filling of holes". The basic idea is Theorem 8.4. in Forster's Lectures on Riemann Surfaces. However, the theorem also works for non-proper coverings with the weaker assumption that all restrictions $\mathcal{F}|_{A}:A\to U-\{p\}$ as constructed above are finite coverings (just look at Forster's proof and you'll see). Then, there exists a branched covering $\mathcal{G}: \mathcal{V}\to X$ such that $\mathcal{V}-{\mathcal{G}}^{-1}(\{p\})=\mathcal{U}$ and $\mathcal{G}|_{\mathcal{U}}=\mathcal{F}$. There exists only one Riemann surface $Y$ with a discrete subset $B\subseteq Y$ satisfying $Y-B\in\{\mathbb{C}^\infty,\mathbb{C},\mathcal{H}\}$, that is $Y=\mathbb{C}^\infty$ with $B=\{\infty\}$. Therefore, $\mathcal{V}=\mathbb{C}^\infty$ and $\mathcal{G}:\mathbb{C}^\infty\to X$ is an unbranched covering. It holds that $|{\mathcal{G}}^{-1}(\{p\})|=|\{\infty\}|=1$ and the fact that $\mathcal{F}|_{A}$ is injective implies that $\operatorname{mult}_{\infty}(\mathcal{F})=1$. It follows that $\mathcal{G}:\mathbb{C}^\infty\to X$ is a covering of degree 1 and, consequently, $\mathcal{G}$ is a biholomorphism.

Related Question