We have shown that the graph of f contains at least one component whose inverse is the entire interval [0, T], and whose multiplicity is odd. There must be an odd number of such components, which will be called complete components. The remaining (incomplete) components all have an even number of ordinary points at any argument, and are defined only on a proper sub-interval of [0, T].

We must now show that on some component of the graph there exist two points for which the corresponding diagonal points in the C-plane are on opposite sides of C. We again consider a fixed point P at **f and a variable point Q at **f on C. We erect a square with PQ as a side and with free corners **f and **f adjacent to P and Q respectively. As s varies from zero to T, the values of s for which **f and **f cross C will be denoted by **f and **f respectively. We have **f, plus tangent points. These s-values are just the ordinary values of **f.

The values **f are the ordinary values at **f of a multi valued function g (t) which has components corresponding to those of f (t).

We first define a function b (t) as follows: given the set of squares such that each has three corners on C and vertex at t, b (t) is the corresponding set of positive parametric differences between t and the backward corner points. The functions f and b have exactly the same multiplicity at every argument t. Now with P fixed at **f, **f-values occur when the corner **f crosses C, and are among the values of s such that **f. The roots of this equation are just the ordinates of the intersections of the graph of b with a straight line of unit slope through **f in the b-plane (the plane of the graph of b). We define these values as **f, and define g (t) in the same way for each t. Thus we obtain g (t) by introducing an oblique g (t) -- axis in the b-plane.