Equal area map from sphere to hemisphere The Mercator projection is the unique conformal projection which maps parallels to horizontal lines and meridians to vertical lines. There is an interesting hemispheric analogue of this, which maps parallels to parallels and meridians to meridians, which is called the Gilbert's two-world projection.
Question 1: does there also exist an equal-area map from the sphere to the hemisphere which maps parallels to parallels and meridians to meridians?
Question 2: if so, would either the equal area or conformal map preserving meridians and parallels be unique, possibly given some extra details (such as a central point on the sphere that maps to itself)?
My thinking is that one possibility would be to just naively halve the longitudes and preserve the latitudes, choosing some arbitrary central point to preserve along the equator. It is pretty easy to see this will be equal area, and I thought that this would probably be the unique solution, but then I thought you could also try to halve the latitudes and preserve longitudes, such that one of the two poles is preserved, basically mapping the entire sphere onto the northern or the southern hemisphere. This wouldn't preserve areas as a small circle around the south pole would become a huge ring around the equator, but perhaps there is some way to adjust latitudes such that the result is equal area and meridians and parallels are preserved. That would be three solutions: one equatorial and two polar. Are these the only three?
 A: tl;dr: "Yes" and "no" respectively.

A mapping from a unit sphere to a unit hemisphere naturally cannot preserve area (because the hemisphere has only half the area), but for posterity let's agree to use "equal-area" to mean area-halving.
If we fix an arbitrary hemisphere $H$ on the unit sphere, there exist infinitely many equal-area mappings to $H$. These include:

*

*The longitude-halving mappings mentioned in the question. There are infinitely many of these because we can choose an arbitrary axis $A$ whose endpoints lie on the boundary of $H$, choose an arbitrary "date line" determined by the "poles" (endpoints of $A$), halve longitude with respect to this date line, and finally rotate the image hemisphere about $A$ onto $H$.

*There is another equal-area mapping to $H$: Here let $A$ denote the axis for which $H$ is the northern hemisphere, and let $C$ be the cylinder with axis $A$ and circumscribed about the sphere. Projection away from $A$ is an area-preserving mapping from the sphere to $C$. (Mathematicians sometimes call this fact Archimedes' (hat box) theorem.) There is an obvious equal-area mapping from $C$ to the half-cylinder corresponding to the northern hemisphere; the composition is an equal-area mapping from the sphere to $H$. This is not exactly "latitude-halving", but is qualitatively similar. Technically we can then rotate about $A$, again obtaining infinitely many mappings.

These mappings send latitudes to latitudes and longitudes to longitudes if and only if $A$ is the geographical axis, but if I understand the geographic constraints that still leaves a one-parameter family of ambiguity (the choice of date line) even if $H$ is a fixed hemisphere bounded by two "antipodal" longitudes.
(If it matters, there is no conformal mapping from a once- or twice-punctured sphere to a hemisphere, e.g., because of Liouville's theorem in complex analysis or by the Koebe-Poincaré uniformization theorem.)
