To see the whole surface of a cube it is sufficient to have two viewpoints above two opposite vertices of a cube. We can see three faces of a cube from the first viewpoint and other three faces from the second viewpoint.
Suppose each face is marked by numbers from 1 to 6 on outside of faces and from 7 to 12 on inside of faces. Also suppose faces are not transparent.
By rotating the cube we can only see numbers from 1 to 6 from the two viewpoints because faces are not transparent.
Analogously now consider we have a tesseract in 4D.
Its 24 faces are marked with numbers from 1 to 24 on one side of faces and with numbers from 25 to 48 on opposite sides of faces. Faces are not transparent.
Now my question is:
If the tesseract is rotating (in all possible angles of rotation about all three planes) in 4D, can an observer in 3D see all 48 numbers or only 24 numbers?
(Observer has the same two viewpoints as with a cube.)
Seeing the animation on the Wikipedia link, my intuition tells me that an observer can see both sides of the faces even though they were not transparent.
The "faces" of a 3-cube (i.e., the things you get by setting one coordinate to $\pm 1$) are cubes, and there are eight of them. They each have an "inside" and an "outside" and of the 16 numbered sides, you can see only 8 from your two exterior viewpoints.
The 2D-components of a 3-cube are sort of analogous to edges of a 2-cube in 3-space. There are indeed 24 of those. But they don't have an "inside" and an outside, just as a line-segment in 3-space doesn't have "sides".
But just as the edges of a cube in 3-space, if you think of them as very thin tubes, have 1/4 inside the cube and 3/4 outside, so you could label them with two different colors, the 2D-components of a 3-cube in 4-space can also be labelled similarly, and indeed, you can "see" only 24 of the 48 labels. By "See", I mean that you can pick a point on one of the labelled parts of the tube, and connect it by a line segment to the viewpoint, with no portion of the interior of the line segment intersecting the 3-cube.
It's often easiest to understand these things by working up from lower dimensions. Let's start with a square (a "1-cube in 2-space"). 4 vertices (which are 0-dimensional --- they're just points) and 4 edges (each 1-dimensional, a line-segment). The line-segments could be labelled with an "inside" and an "outside", but if we draw the top row of this figure:
you can see what I've done: I've colored the edges blue, the vertices red, and I've made a smaller and larger copy of the square, and filled in the "corners" with red arcs. These arcs are kind of "fattened up" versions of the vertices. And now, where we used to have one edge with two sides, we've got two parallel edges; where we used to have a vertex (where it was hard to talk about "sides") we've got two parallel arcs, serving as proxies for the "inside" and "outside". From the exterior of the square, we can see the four "outside" blue edges and red arcs, but none of the "inside" ones.
When we move to a cube (a 2-cube in 3-space) we have 6 vertices (zero-dimensional) 12 edges (1-dimensional) 6 faces (2-dimensional)
I've drawn only three of the faces, to save myself time and drawing problems. Again I "blow up" the shape and make two copies of each top-dimensional thing; I've drawn the "inside" copy in light blue. Now for each edge in the left-most picture, I've drawn a quarter-tube in green, and for each vertex in the left-most picture, I've ... skipped drawing a 1/8-sphere that would fill in the little triangular hole you see. But in this "fattened" version, you can see the outside of the vertex, of three edges, and of three faces. Actually, if I'd drawn them, you'd see part of the outside of 6 more edges. So 3 of the edges are visible ONLY from this vantage point, 3 are visible ONLY from the opposite vantage point, and 6 are partly visible from both. But NONE of the interior edges are visible, etc.
When you get to 4-space, there are
12 vertices 24 "edges" 24 "faces" 12 "solid" faces (3-dimensional things)
And you can blow up things in much the same way, and you'll find that the outsides of all these things are visible (given two opposing viewpoints on opposite ends of a main diagonal of the hypercube), but that the "insides" are not.