To show that a quotient geometry is a projective space we need the concept of isomorphism.
Def: Let G = (
,I) and G' = (
', I') be rank 2 geometries. G and G' are isomorphic if there exists a map
:

' such that
(
) =
' and
(
) =
' with
restricted to these domains being a bijection, and
P
and B
,
(P) I'
(B).
is called an isomorphism.An automorphism is an isomorphism from G onto itself. If the blocks of G are called lines, then an automorphism is usually called a collineation.
Example: The points of a 3-dimensional projective space P are given by the 15 non-zero binary vectors of length 4 written as row vectors. The planes of this space are the same vectors written as column vectors. A point and a plane are incident if their dot product is 0 (mod 2). A line is the set of points incident with two distinct planes. Thus, the plane (0,0,0,1)T = {0010,0100,0110,1000,1010,1100,1110} and the plane 1000T = {0001,0010,0011,0100,0101,0110,0111}, and so, the set of points {0010,0100,0110} form a line. Incidence of points and lines, and lines and planes is given by set inclusion. Let Q be the point 0001, then the set of planes containing Q is {0010T,0100T,0110T,1000T,1010T, 1100T,1110T} and the set of lines containing Q is {a ={0001,1000,1001}, b={0001,0100,0101}, c={0001,0010,0011}, d={0001,1100,1101}, e={0001,1010,1011}, f={0001,0110,0111}, g={0001,1110,1111}}. We form the quotient geometry P/Q:
| POINTS | LINES Through POINT |
|---|---|
| a | 0010T, 0100T, 0110T |
| b | 1000T, 0010T, 1010T |
| c | 1000T, 0100T, 1100T |
| d | 0010T, 1100T, 1110T |
| e | 0100T, 1010T, 1110T |
| f | 1000T, 0110T, 1110T |
| g | 1100T, 0110T, 1010T |
Example: Continuing with the above example. Define a map
on the points of P by
(x,y,z,w) = (w,y,z,x). Notice that
maps the points of 0001T to the points of 1000T bijectively. One then checks to see that the points of a line of 0001T are mapped to the points of a line of 1000T. For instance, the points of the line {0010,1100,1110} which is the intersection of 0001T and 1100T, are mapped to {0010,0101,0111} which lie on the line which is the intersection of 1000T and 0101T. This
is therefore an isomorphism between the planes 0001T and 1000T.
Proposition 1.4.2: Let P be a d-dimensional projective space, Q a point of P. Then there exists a hyperplane of P that does not pass through Q.
Pf: By Lemma 1.3.8, one can extend Q to a basis of P, {Q,P1,P2,...Pd}. The subspace H = <P1,P2,...,Pd> is spanned by d independent points. Hence H has dimension d-1 and is therefore a hyperplane. Since the original basis is an independent set, Q
H.
Theorem 1.4.1: Let P be a d-dimensional projective space, Q a point of P. Then the quotient geometry P/Q is a projective space of dimension d-1.