Uniformization theorem: Equivalence between simply connected Riemann surfaces and closed Riemannian 2-manifolds? Recently, I read a proof of the uniformization theorem using the Ricci flow on closed Riemannian 2-manifolds. It is implied by some sources that this is equivalent to the uniformization theorem for simply connected Riemann surfaces for example Wikipedia says:
"The uniformization theorem also has an equivalent statement in terms of closed Riemannian 2-manifolds: each such manifold has a conformally equivalent Riemannian metric with constant curvature."
I am unable to see why this is true. Could someone please direct me to a proof? 
Remark:
I did find an earlier question that addresses part of this issue, but didn't get the answer I was looking for:
Uniformization of metrics vs. uniformization of Riemann surfaces
 A: One cannot trust Wikipedia too much on these matters: Anybody can edit a Wikipedia article. The Wikipedia article is sloppy about distinguishing compact and noncompact surfaces. The Uniformization Theorem has the following equivalent forms:
A. Every simply-connected Riemann surface is biholomorphic to $S^2$ or to ${\mathbb C}$ or to the unit disk $\Delta$ in ${\mathbb C}$. 
B. Every connected Riemann surface $X$ is biholomorphic either to $S^2= {\mathbb C}P^1$ (with its standard complex structure), or to the quotient  of $U={\mathbb C}$ or of $U=\Delta$ by a group $\Gamma$ of linear-fractional transformations of $U$ acting on $U$ freely and properly discontinuously. 
C. Every connected Riemannian surface $(S,g)$ admits a positive smooth function $\lambda$ such that $(S, \lambda g)$ is a complete Riemannian manifold of constant curvature. (Note that the surface $S$ is not required to be oriented.) 
Remark. i. In all three formulations, the surface $S$ is not required to be compact. The compactness assumption made by the Wikipedia article is totally unnecessary. Compactness is used in some of the proofs but not in other proofs. 
ii. There is no way to prove the UT for general noncompact surfaces from the UT for compact surfaces. 
Equivalence of the three statements (A, B and C) is not hard to establish. The key facts are the existence of a conformal Riemannian metric on every Riemann surface and that every group $\Gamma$ as in Part B, acts on $U$ isometrically with respect to the Euclidean or hyperbolic metric respectively. 
Regarding the Ricci flow, what is true is that there is a proof of the Uniformization Theorem (UT) for compact (closed) Riemann surfaces via the Ricci Flow (RF). This result by itself does not imply the full UT. 
In order to prove the UT for noncompact surfaces using RF one would have to work much harder than in the compact case and I am unaware of such a proof in the literature. Even short-term existence of the flow becomes a problem. See for instance 
Xiaorui Zhu, Ricci Flow on Open Surface, J. Math. Sci. Univ. Tokyo 20 (2013), 435–444. 
for some partial results on proving UT via RF on open surfaces. 
Of course, if your (say, simply connected) Riemann surface is compact, then RF indeed does the job: First equip your surface $X$ with an arbitrary conformal Riemannian metric $g_0$ (i.e. a metric which in the local holomorphic coordinates of $X$ has the form $\rho_k(z)|dz|^2$). For a proof of existence of such a metric see my answer here. 
Then apply the normalized RF to $g_0$. This, in finite time, converges to a constant curvature metric $g_T$ (the curvature has to be positive). For surfaces, RF preserves the conformal class of the metric. Hence, $g_T$ is still a conformal Riemannian metric on $X$. After rescaling, $g_T$ has curvature $1$. Now, use the theorem (due to Killing and Hopf in all dimensions) that all compact simply-connected surfaces of curvature $1$ are isometric to each other. Hence, $(X,g_T)$ is isometric to the standard unit sphere $S^2$. The isometry $f: X\to S^2$ has to be conformal in the sense of Riemannian geometry, hence (after changing orientation if needed) is conformal in the sense of complex analysis. qed   
