Uniformization for annuli with boundary Let $(A,g)$ be a compact surface with boundary, diffeomorphic to the standard annulus $\{z\in\mathbb{C}:1\le|z|\le 2\}$, equipped with a smooth metric $g$. 

Does there always exist a conformal (bijective) map
  $\phi:A\to\{z\in\mathbb{C}:1\le|z|\le R\}$, for some $1<R<\infty$?

A possible idea is to use the well-known fact that $A\setminus\partial A$, as a Riemann surface (whose conformal structure is induced by $g$), is biholomorphic to either $\{1<|z|<R\}$ (for some $1<R<\infty$), $\{0<|z|<1\}$ or $\mathbb{C}\setminus\{0\}$. Is there a way to rule out the latter two cases and to prove that the biholomorphism extends nicely to $\partial A$?
 A: Yes. I assume that by "compact surface diffeomorphic to a closed annulus" you mean
a bordered surface. Then by definition of a bordered surface, each boundary component has more than one point. Now an open Riemann surface homeomorphic to
an annulus is conformally equivalent to $\{ z: r<|z|<R\}$, where
$0\leq r<R\leq+\infty$. If each boundary component is not a point, we have
the honest annulus with $r=1,R<\infty$.
That a neighborhood of a puncture is not conformally equivalent to
a neighborhood of the non-degenerate boundary component is a consequence of
the elementary theorem on the removable singularity.
EDIT. Let me add some detail. The punctured annuli are characterized by the property that the extremal
length of the family of all homotopically non-trivial curves is zero.
On the other hand, if we have a bordered Riemann surface, homemorphic to a
non-degenerate annulus, we can estimate this extremal length from below.
To do this, we put on our annulus some conformal metric, in which the
length of the boundary components is not zero.
In your case, the conformal structure is defined by a diffeomorphism. Just pull
back the Euclidean metric on $1\leq |z|\leq R$ by your diffeomorphism.
For the definition and properties of
extremal length, see any of the two books of Ahlfors,
Conformal invariants or Lectures on quasiconformal mappings. 
An alternative proof can be obtained by referring to Kwak's theorem,
which is the generalization of the removable singularity theorem to maps between
Riemann surfaces. (Ann. Math., 90 (1969) 9-22, or S. Lang, Introduction to complex hyperbolic spaces, Springer, 1987.) 
A: An alternate proof that interior of A is not biholomorphic with punctured disc or punctured plane is to notice that that there is a harmonic function on the interior of A which is 0 and 1 on the different boundary components respectively . Such functions do not exist on the punctured disc or the punctured plane. In fact this harmonic function has no critical points and the corresponding holomorphic 1 form has its periods an imaginary multiple of integers .Hence composing the integral of a multiple of the holomorphic one form with the exponential function gives the uniformizing map.
