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If I construct a composition of mappings that map the upper half of the unit disk conformally to the entire unit disk, then this mapping is a Riemann mapping, by the Riemann Mapping Theorem, since $D^+$ is simply connected and not the entire complex plane.

My question is: is a conformal mapping from the upper half plane to the (open) unit disk considered a Riemann mapping? What about a conformal mapping from, say, the first quadrant, to the unit disk? Naively, I'd say that these are simply connected regions that are not the entire complex plane.

My guess is that, because these pre-image regions include the point at infinity, they are somehow equivalent to being the entire complex plane - and perhaps this is more of just knowing what the Riemann sphere is? (Thus, these are not Riemann mappings.)

And, by this logic, then a mapping of a vertical or horizontal strip to the unit disk is not a Riemann mapping either.

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this mapping is a Riemann mapping, by the Riemann Mapping Theorem,

The Riemann Mapping Theorem does not say how the map is called. It says that it exists. How to call it is a matter of conventions and definitions, not theorems. The convention I am familiar with is (as on Riemann mapping theorem wiki page):

Riemann mapping = a holomorphic bijective map of a simply-connected domain onto the unit disk.

Riemann mapping theorem is a statement that such a thing exists for every simply-connected domain in $\mathbb C$, except $\mathbb C$ itself.

What about a conformal mapping from, say, the first quadrant, to the unit disk? Naively, I'd say that these are simply connected regions that are not the entire complex plane.

Correct.

these pre-image regions include the point at infinity

Incorrect. They are unbounded, but they do not include the point at infinity because their complement is also unbounded. In terms of the Riemann sphere, $\infty$ is one of their boundary points.

a mapping of a vertical or horizontal strip to the unit disk is not a Riemann mapping either.

It is. These are simply-connected domains of the complex plane. They do not contain the point at infinity.

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  • $\begingroup$ Ok, thank you so much for clearing that up for me, @Behaviour. Have a great night :) $\endgroup$
    – User001
    Dec 9, 2014 at 4:41
  • $\begingroup$ I have a simple follow-up question, if you wouldn't mind answering, @Behaviour: If I consider the upper half plane, is the only "boundary" of this region the real axis? Are there any more boundaries of this region? Thanks... $\endgroup$
    – User001
    Dec 14, 2014 at 0:53
  • $\begingroup$ Boundary is taken with respect to some ambient space. With respect to the plane $\mathbb C$, the boundary is the real line. With respect to the Riemann sphere $\widehat{\mathbb C}$ it is $\mathbb R\cup \{\infty\}$. $\endgroup$
    – user147263
    Dec 14, 2014 at 0:58
  • $\begingroup$ Ok, got it. Thanks so much, @Behaviour. Have a great night :) $\endgroup$
    – User001
    Dec 14, 2014 at 1:12
  • $\begingroup$ Sorry, one more question, @Behaviour :( if I add the point at infinity to, say, the upper half plane, or to a quadrant, both of which are unbounded regions, are these regions still considered simply connected sets? There is no gap / hole between the upper half plane and the point at infinity, right? Thanks... $\endgroup$
    – User001
    Dec 14, 2014 at 8:01

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