14 Jack Palmer Sanders
(x,(t-e)/(l-e)) e k(L ). Again p (F,G,e)(x,t) e U. We conclude that
p (E) c W(K,U) and that p is continuous at (f,g,a).
The proof that p is continuous at (f,g,0) and (f,g,l) is
1.5. Proposition. Let A be the subspace of
Map(I,Y) x Map(I,Y) x (0,1] x (0,1] consisting of those (f,g,a,a') such
that f(l) = g(0); if a = 1, then f(t) = f(l) for all t e I; and if
a' = 1, then g(t) = g(0) for all t e I. Define p : A - Map(I,Y) by
p (f,g,l,a') = g; p (f,g,a,l) = f; and for a ^ 1, a' ^ 1, and
x = (a'-aa')/(l-aa'),
p (f,g,a,a')(t) =f(t/x), for 0 t _x;
= g((t-x)/(l-x)), for x £ t _ 1.
Then p is continuous.
Proof. It is sufficient to show that the same function
p : A -*C(I,Y) is continuous, where A c C(I,Y) x C(I,Y) x (0,1] x (0,1]
is A with the relative topology. The function q: (0,1] x (0,1] - I
defined by q(a,a') = (a'-aa')/(1-aa') is continuous at all points except
(1,1). Let A' be the open subset of A consisting of those
(f,g,a,a') e A such that either a ^ 1 or a' ^ 1. Then pJ,/ is the
A' 1Xq ) C(I,Y) x C(I,Y) x I —5_- C(I,Y),
c c
where p is the map of (1.4). Thus p | , is continuous. Now let W(K,U)
be a subbasic open neighborhood of p (f,g,l,l). Since
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