### Extending a Projective Map from Four Points

Suppose we know where the 4 corners of a square go under a projection.
Can we locate the image of any other point, x, from this?
Here’s how, but we need a topological crutch.
Assume a coordinate system where the coordinates of the corners of the square are A=(0, 0), B=(0, 1), C=(1, 0) & D=(1, 1).
The projection maps A, B, C & D to A', B', C' and D'.
We adopt the notation that juxtaposition of two point expressions is a line expression and conversely.
We use lower case letters for lines.
We use enough parens to be unambiguous.

We locate the image of the square’s center, F=(1/2, 1/2)=(A'D')(B'C'), as the intersection of the (image of the) two diagonals: A'D' and B'C'.
Line AB is parallel to CD and thus A'B' intersects C'D' on the image of the line at infinity (lai).
Likewise A'C' intersects B'D' at E' on the lai and we can draw the lai since we have two points on it.
i = ((A'B')(C'D'))((A'C')(B'D')).
We now draw ‘vertical’ and ‘horizontal’ bisectors of the square.
v = F((A'B')(C'D')) and h = F((A'C')(B'D')).
The vertical bisector is parallel to AB and its image goes thru E' and the image of the center.
Likewise for the horizontal bisector.
By further bisections we can continue this process and fill in an ever finer image of the coordinate system in the original square and thus locate the image of arbitrary points within the square.
The image of the adjacent unit square can be drawn by drawing a line ‘parallel’ to a diagonal AD thru point B.
That intersects the line AC at the image of the point (−1, 0).
The mesh can thus be extended to cover the entire plane.

Many years ago Dick Tracy was concerned with a photo of a hotel window that had been taken from some room in another hotel across the street.
A mathematician was invoked to determine from which room the photo had been taken.