Differentiation 385open. In what follows we shall use the term open function in the sense
of strongly open.A function f: A --+ B is said to be light if it is not constant on any
continuum (a compact connected set). In other words, f is light iffor each
b EB its inverse image f-^1 (b) is totally disconnected (Definition 2.31).
M. S. Stoi1ow [113, 114] was the first to recognize that lightness and
openness of the mapping defined by a nonconstant analytic function are
the fundamental topological properties of such functions, meaning that all
other topological properties of this class of functions are derivable from
those two properties. Furthermore, he also proved that every light and
open function on a two-dimensional manifold is topologically equivalent to
some analytic function.
In 1952, Eggleston and Ursell [39] provided rather elementary proofs
for the lightness and openness of analytic mappings by making use of the
topological index. A simplified version of Eggleston and Ursell proof was
presented by Whyburn in [128] and in the first edition of [129]. We refer the
reader to those. papers and to Whyburn's book for a detailed discussion of
this topic.
6.22 CONTINUITY OF THE FIRST DERIVATIVE OF
AN ANALYTIC FUNCTIONSuppose that f is defined in some open set A C C. The existence of f'
everywhere in A implies the continuity off in A (Theorem 6.1). The fact
that f' is also continuous in A was for a long time derived, with recourse to
complex integration theory, by proving that the existence of f' in A implies
that of f 11 in A, as well as the existence of all successive derivatives f^111 ,
J(iv), ••• , so that all these functions are necessarily continuous in the sameset. The desirability of finding a direct proof of the continuity of f^1 has
been felt for some time when early in 1959, R. L. Plunkett [93] provided
such a proof by purely topological methods, to which others followed by
E. H. Connell [24] in 1961 (see also [25]), A. H. Read [100] in 1961, and G.
T. Whyburn [130] in 1962 (see also [129]).
We proceed to give an account of this important result based on
Connell's papers.
Theorem 6.32 Let A be a bounded open set, and suppose that f is
continuous on A and open on A. If B is a component of the complement
of f(A - A), then the condition f(A) n B =/:- 0 implies that f(A) :J B.
Proof J(A) n B is open in B and J(A) n B = J(A) n B is closed in B.
Since B is connected, it follows that J(A) contains B.