Proof of Theorem B

Theorem B
A projective transformation may be expressed as a composite of
at most three perspectivities.

Proof
Suppose that t is a projective transformation.
Say t^-1 maps (X,Y,Z,U) to (A,B,C,D), so t maps (A,B,C,D) to (X,Y,Z,U).
Note that, as no three of (X,Y,Z,U) are collinear, the same is true of (A,B,C,D).
We show that there is a composite of three perspectivities with the same
effect on (A,B,C,D). By the Fundamental Theorem of Projective Geometry,
this composite is equal to t.

By Lemma 1, we can choose a plane Π which cuts the p-points A, B, C
at A', B', C' such that ΔA'B'C' is acute angled.
By Lemma 2, we can choose a Points P, Q, R on the x-, y- and z-axes
such that ΔPQR is (euclidean)congruent to ΔA'B'C'. Let Π' be the plane
through P,Q,R.
Now consider the following transformations:
Let R be a rotation which maps Π to a plane Π" parallel to Π'.
Let P be the projection from Π" to Π along their common normal.
Together, these map Π rigidly to Π', so it maps ΔA'B'C to ΔP'Q'R'
which is euclidean congruent to ΔA'B'C', and hence to ΔPQR.
Now let T be the transformation corresponding to the euclidean
transformation which maps ΔP'Q'R' to ΔPQR.
By the remarks after the proof of Theorem A, the transformation
A = ToPoR corresponds to a single perspectivity.

As a projective transformation, this maps (A,B,C) to (X,Y,Z). Suppose that
it maps D to E=[k,l,m]. Since no three of (A,B,C,E) are collinear, k,l,m ≠ 0.
From Lemma 3, we have perspectivities:
B(x,y,z) = ((m/k)x,y,z), and
C(x,y,z) = (x,(m/l)y,z).
Note that B and C fix X, Y and Z,
and that CoB maps {k,l,m] to {m,m,m] = U.
Then CoBoA maps (A,B,C,D) to (X,Y,Z,U), so equals t.

Part 3 - Some preliminary results

Theorems A and B