Real and Complex Analysis (Rudin)

326 REAL AND COMPLEX ANALYSIS
Suppose u satisfies these conditions. Then (2) shows that 'If covers Yu' and
therefore Theorem 16.11 shows that both g, and gu are obtained by continuation
of (f, D) along this same chain 'If. Hence g, = gu.
Thus each tel is covered by a segment J, such that gu = g, for all u E
I ('\ J,. Since I is compact, I is covered by finitely many J,; and since I is
connected, we see in a finite number of steps that g1 = go.
IIII
Our next item is an intuitively obvious topological fact.
16.14 Theorem Suppose r 0 and r 1 are curves in a topological space X, with
common initial point IX and common end point fJ. If X is simply connected, then
there exists a one-parameter family {y,} (O::s;; t ::s;; 1) of curves from IX to fJ in X,
such that Yo = ro and Y1 = r 1•
PROOF Let [0, n] be the parameter interval of r 0 and r l' Then
(0 ::s;; s ::S;; n)
(n ::S;; s ::S;; 2n)
(1)
defines a closed curve in X. Since X is simply connected, r is null-homotopic
in X. Hence there is a continuous H: [0, 2n] x [0, 1] -+ X such that
H(s, 0) = r(s), H(s, 1) = c E X, H(O, t) = H(2n, t).
(2)
If : 0 -+ X is defined by
(rei~ = H(fJ, 1 - r)
(0 ::S;; r ::S;; 1, 0 ::S;; fJ ::S;; 2n),
(2) implies that is continuous. Put
y,(fJ) = [(1 - t)e i9 + te- i '] (0 ::S;; fJ ::S;; n, 0 ::S;; t ::S;; 1).
Since (ei~ = H(fJ, 0) = r(fJ), it follows that
y,(O) = (1) = r(0) = IX (0 ::S;; t ::S;; 1),
YAn) = ( -1) = r(n) = fJ (0 ::S;; t ::S;; 1),
(0 ::S;; fJ ::S;; n)
and
Y1(fJ) = (e-i~ = (ei (21t-9)) = r(2n - fJ) = r 1 (fJ)
This completes the proof.
(0 ::S;; fJ ::S;; n).
IIII
The Monodromy Theorem
The preceding considerations have essentially proved the following important
theorem.