SubLemma 5.4
If I is a closed subinterval of 
and every 
has a neighborhood on which f has a lift, then f has a lift
on I.
Proof
By compactness, we can cover I by closed intervals 
on which f has a lift, and we may assume
for 
.
By induction on j, Sublemma 5.3 lets us extend the
lift on 
to a lift on 
.