The PoincareKlein Map
 The map k is given by k(x,y) = (2x/(1+x^{2}+y^{2}),2y/1+x^{2}+y^{2})).
 The map k maps the hline with boundary points AB to the
intersection of the euclidean line AB with D.
 The inverse is
k^{1}(x,y) =
(x/1+(1x^{2}y^{2})^{½},y/1+(1x^{2}y^{2})^{½}).
Proof
Suppose that (x,y) is in D. This corresponds to P(x,y,0) in R^{3}.
In vector notation, the line ZP has equation tp+(1t)z = 0,
i.e. each point on ZP has the form (tx,ty,(1t)), for some t.
This meets S where (tx)^{2}+(ty)^{2}+(1t)^{2} = 1, which simplifies to
t^{2}(x^{2}+y^{2}+1) = 2t. The solution t = 0 corresponds to Z. The
other solution, t = 2/(1+x^{2}+y^{2}), corresponds to P'. Thus we have
P' is
(2x/(1+x^{2}+y^{2}),2y/(1+x^{2}+y^{2}),
12/(1+x^{2}+y^{2})).
We obtain P" by replacing the zcoordinate by 0.
Thus, we have k(x,y) = (2x/(1+x^{2}+y^{2}),2y/(1+x^{2}+y^{2})).
For part (2), we show that the points which map to a line segment
lie on an hline. The point k(x,y) in D is on the line L : ax+by+c = 0
if and ony if 2ax/(1+x^{2}+y^{2})+2by/(1+x^{2}+y^{2})+c = 0,
i.e. 2ax+2by+c((1+x^{2}+y^{2})) = 0.
If c = 0, this reduces to ax+by = 0, i.e.
If c ≠ 0, it is becomes x^{2}+y^{2}2(a/c)x2(b/c)y+1 = 0.
The distance of O from the line L is d = c/(a^{2}+b^{2})^{½}
Thus, L cuts D if and only if d < 1, i.e.
(a/c)^{2}+(b/c)^{2} > 1.
Using the
Lemma
The circle C : x^{2}+y^{2}2ax2by+c = 0 is orthogonal to the circle
C : x^{2}+y^{2} = 1
if and only if c = 1 and a^{2}+b^{2} > 1.
we see that the points on L lie on a circle C orthogonal to C.
Since k maps D to D, it maps the hline
C_{n}D to the segment L_{n}D.
Finally, the map k is defined on the boundary C, and fixes each
point of C 
if P is on C, then P = P' = P" in the definition of k.
Thus, the boundary points ofthe hline are the endpoints of the
segment.
For part (3), we apply the projections in reverse order.
If P" is the point (x,y,0), the orthogonal projection maps it to the
point P'(x,y,z), where x^{2}+y^{2}+z^{2} = 1 and z < 0.
so z = (1x^{2}y^{2})^{½}.
For brevity, we will use z in our calculations. Then any point ZP'
has the form (tx,ty,tz+1t). This cuts the xyplane when t = 1/(1+z),
i.e. at the point (x/(1+z),y/(1+z)), as required.

