Self-sewing two punctures of the sphere gives genus one

The set $S^2$ is the inverse image of $\{1\}$ under the continuous polynomial $x^2+y^2+t^2$, so it is closed; it is also bounded. Thus [F5] makes it compact. Distinct points have disjoint relative balls of radius less than half their Euclidean distance, so it is Hausdorff. To check second countability, let $U$ be a relative open neighborhood of $p\in S^2$ and choose $ϵ>0$ with $B(p,ϵ)\cap S^2\subseteq U$. By density in [F8], for each of the three coordinates choose a rational within $ϵ\mathrm{/}8$ of the corresponding coordinate of $p$; the resulting $a\in {\mathbb{Q}}^3$ satisfies $\mathrm{\parallel }a-p\mathrm{\parallel }<\sqrt{3}ϵ\mathrm{/}8<ϵ\mathrm{/}4$. Since $\mathrm{\parallel }a-p\mathrm{\parallel }<ϵ-\mathrm{\parallel }a-p\mathrm{\parallel }$, density in [F8] gives $\rho \in {\mathbb{Q}}_{>0}$ with $\mathrm{\parallel }a-p\mathrm{\parallel }<\rho <ϵ-\mathrm{\parallel }a-p\mathrm{\parallel }$. The rational ball $B(a,\rho )\cap S^2$ contains $p$; if $x\in B(a,\rho )$, then $\mathrm{\parallel }x-p\mathrm{\parallel }\le \mathrm{\parallel }x-a\mathrm{\parallel }+\mathrm{\parallel }a-p\mathrm{\parallel }<\rho +\mathrm{\parallel }a-p\mathrm{\parallel }<ϵ$, so the ball lies in $U$. The collection of these balls is countable by [F9]: enumerate $\mathbb{Q}$ and its positive subset by natural indices, then enumerate the resulting finite tuples in ${\mathbb{N}}^4$ by increasing coordinate sum. It is therefore a countable base. The sphere is path-connected: if $p,p^{\mathrm{\prime }}\in S^2$ and $p^{\mathrm{\prime }}\mathrm{\ne }-p$, normalize the segment $(1-u)p+u{p}^{\mathrm{\prime }}$ for $u\in [0,1]$, which never vanishes; if $p^{\mathrm{\prime }}=-p$, take the least $k\in \{1,2,3\}$ for which the standard basis vector $e_k$ is not parallel to $p$, and put $v=(e_k-⟨e_k,p⟩p)\mathrm{/}\mathrm{\parallel }{e}_k-⟨e_k,p⟩p\mathrm{\parallel }$. Such $k$ exists because the three basis vectors span ${\mathbb{R}}^3$ while $\mathbb{R}p$ is one-dimensional; this $v$ is a unit vector perpendicular to $p$. Concatenating the normalized segments from $p$ to $v$ and from $v$ to $p^{\mathrm{\prime }}$ gives a path. Thus [F6] makes $S^2$ connected. The inverse formulas in the Example show that both stereographic maps are charts covering the sphere; on the overlap their transition is $w=1\mathrm{/}z$, holomorphic with holomorphic inverse. They form a holomorphic atlas satisfying [F1], so the starting object is a compact connected Riemann surface.

On the overlap, $\mathrm{\mid }w\mathrm{\mid }=\mathrm{\mid }z{\mathrm{\mid }}^{-1}$. Thus a point cannot satisfy both $\mathrm{\mid }z\mathrm{\mid }<r$ and $\mathrm{\mid }w\mathrm{\mid }<s$, since that would give $1\mathrm{/}s<\mathrm{\mid }z\mathrm{\mid }<r$ while $r<1\mathrm{/}s$. The complement of the two open disks is the closed annulus $A_0=\{r\le \mathrm{\mid }z\mathrm{\mid }\le 1\mathrm{/}s\}$. The nonzero plumbing fixture is the closed annulus $A_q=\{q\mathrm{/}s\le \mathrm{\mid }z\mathrm{\mid }\le r\}$, because $q>0$ and $w=q\mathrm{/}z$. Both are compact, and the fixture boundaries are attached to the corresponding boundaries of $A_0$ using equal $z$ at the first seam and equal $w$ at the second. By [F2]–[F3], the transition across the second coordinate description is the holomorphic map $w=q\mathrm{/}z$ with inverse $z=q\mathrm{/}w$ on neighborhoods where both coordinates are nonzero; the first seam uses the same coordinate $z$. Thus the quotient charts form a holomorphic atlas. The two annuli are connected and are glued along both boundary circles, so the resulting surface is compact, connected, and has no boundary.

Parameterize each annulus by its angular coordinate and a radial interval. At the $z$ seam the angular coordinates agree. At the $w$ seam, equality of $w$ and the relations $z_{A_0}=1\mathrm{/}w$ and $z_{A_q}=q\mathrm{/}w$ make the angular coordinates agree as well, because $q$ is positive real. Hence the annulus angular coordinates glue to one circle coordinate. The two radial intervals, with their corresponding endpoints identified, are explicitly the semicircles $u\mapsto (\cos \pi u,\sin \pi u)$ and $u\mapsto (\cos \pi u,-\sin \pi u)$. These maps agree at the identified endpoints, have disjoint interiors, and cover $S^1$; they induce a continuous bijection from the compact quotient of the intervals to the Hausdorff space $S^1$, hence a homeomorphism. Call this interval quotient $Q$. Each annulus is homeomorphic to $S^1$ times its radial interval. Since the seam identifications preserve the angular coordinate and identify exactly the corresponding radial endpoints, the map from the disjoint union $S^1\times (I_0\bigsqcup I_q)$ to $S^1\times Q$ descends to a continuous bijection from the sewn quotient to $S^1\times Q$. The sewn quotient is compact by step 2.1, and $S^1\times Q$ is Hausdorff, so this bijection is a homeomorphism. Since $Q\cong S^1$, the sewn surface is homeomorphic to $S^1\times S^1$. As a subspace of ${\mathbb{R}}^4$, it is Hausdorff. Rational balls form a countable base for ${\mathbb{R}}^4$: density [F8] gives a rational center within $ϵ\mathrm{/}4$ of any point in a given open ball, and then a rational radius strictly between that distance and the remaining radius; countability follows from [F9]. The subspace therefore is second-countable, so the charts from step 2.1 make it a Riemann surface.

By [F4] and [A1], the sewn compact connected Riemann surface, being homeomorphic to $T^2$, has genus $g=1$.