Constructing a convex polyhedron from a spherical polyhedron Suppose I am given an arbitrary 3-dimensional convex polyhedron $P\subset\Bbb R^3$ that contains the origin.
I can "blow it up" to a spherical polyhedron by projecting all edges and vertices (away from the origin) to the unit sphere (centered at the origin):

What about the other direction?

Question: Given a spherical polyhedron, is there a "convex polyhedron" whose projection is exactly the given spherical polyhedron? And how to construct it explicitly?

For me, a spherical polyedron is a tiling of the 2-sphere where the edges are great circle arcs.
And I know that there is always a convex polyhedron with the same combinatorics as the given spherical polyhedron, but I ask specifically about a convex polyhedron that projects to the given spherical polyhedron.
 A: The answer is negative: not every spherical polyhedron comes from a convex one.
The argument is basically the same I gave over here: the spherical polyhedron can have more degrees of freedom than the convex polyhedron.

The following is a copy of a part of the answer behind the link:

[...] given the combinatorial type of a spherical polyedron, the realization space of that type (i.e. the space of all spherical polyhedra with this combinatorial type) has a local dimension of $2n$, where $n$ is the number of vertices.
What do I mean by that: you can describe your spherical polyhedron basically by drawing some points on the sphere, and stating between which points there should be a line. The line is then uniquely determined as the great circle arc between these points (yes, there is a choice which arc to take, but lets ignore this for now).
So if we placed our points carefully, then none of these arcs intersect, and what we have is a spherical polyhedron.
But note that we can move each point slighly, and the arcs move accordingly. And if we moved the points slightly enough, then the arcs stay disjoint, and the construct stays a spherical polyhedron.
Since each vertex moves on the surface of the 2-sphere, each vertex has two degrees of freedom, and the whole construct has $2n$ degrees of freedom.

Now consider the $7$-sided prism (the argument can certainly be improved, but as presented here, we need an $n$-prism with $n\ge 7$).
This prism has $14$ vertices, and by the argument presented above, the spherical 7-prism has $2\times 14=28$ degrees of freedom.
However, a convex polyhedron has as many degrees of freedom as its dual (because they determine each other uniquely). The dual of the 7-prism is the 7-sided bipyramid, which has $9$ vertices. And the position of these vertices determine the bipyramid uniquely. Each vertex has three dregrees of freedom, and so the 7-prism has at most $3\times 9=27$ degrees of freedom.
In other words, the projection of the convex prism to the spherical one (which is continuous) cannot be surjective, given the larger dimension of the realization space of the image.
A: It depends on whether you require the vertices to stay fixed, or whether you consider isomorphs (with the same structure but different vertex positions on the sphere) to be the "same" polyhedron.
If you want them to stay put, then there is a much simpler proof than the one given in another answer, though it amounts to the same thing:

For a convex polyhedron, all vertices of a given face must be
coplanar,
this condition is not necessary for a spherical polyhedron, and spherical polyhedra with skew faces exist
therefore there are spherical polyhedra which have no
corresponding convex polyhedron.

But if you consider all isomorphs to be essentially the same polyhedron, then it always has a corresponding convex polyhedron, known as its canonical form.
