Abel's theorem in problems and solutions - School of Mathematics

Solutions 205
Since by hypothesis the point the centre of the circle C, is a root
of the equation then There thus exists a value
such that the disc of radius equal to with centre at the point
has no intersection with the curve (Figure 114). Consider now a
different complex number and consider another mapping
Let be the image of the circle C under the mapping Since
the curve is obtained from the curve displacing
it by the vector (cf., 246). If the length of the vector is
smaller than then the curve is displaced with respect to by so
small an amount that along one turns around the point as many
times as along (Equivalently, one may imagine, conversely, that the
point be displaced instead of the curve, see Figure 114). Since the
curve turns times around the point the curve will turn
times around the point as well. Following the same reasoning
as before, we obtain that inside the disc D there are roots of the
equation (taking into account their multiplicities).
345. Let be an arbitrary point, different from and
We thus have 5 different images of the point under the mapping
Let these images be If a continuous curve C
starts from the point then at every point at least
a continuous image of the curve C under the mapping starts. If
two continuous images of the curve C were starting from the point
then the curve C should have at least six continuous images. This is
not possible because an equation of degree 5 cannot have more than five
roots. Consequently the point is not a point of non-uniqueness of the
function
Consider now 5 discs with a certain radius with
centres at the points Choose sufficiently small so that these discs
be disjoint. By virtue of the result of Problem 344 there exists a disc
with centre at the point such that for every point inside this
disc there exists at least (and consequently only) one image in each one
of the discs on the plane. If C is a continuous curve
which lies entirely in the disc all images of its points lie in the discs
But thus a continuous image of the curve C under the
mapping cannot jump from one disc to another, and each one of the
images of C lies entirely in one of the discs If the curve
C, which lies entirely in the disc begins and ends at the point then
the end points of its continuous image are images of the point under