In defining the map \(\overline{F}\) in the proof of Lemma 54.2, why were we so careful about the order in which we considered the small rectangles?
Solution
Lemma 54.2 states: for a covering map \(p: E \to B\) with \(p(e_0) = b_0\) and a continuous \(F: I \times I \to B\) with \(F(0, 0) = b_0\), there is a unique lifting \(\overline{F}: I \times I \to E\) with \(\overline{F}(0, 0) = e_0\). The proof subdivides \(I \times I\) into rectangles \(R\), each mapped by \(F\) into an evenly covered open set \(U\), and extends \(\overline{F}\) one rectangle at a time.
1. When \(\overline{F}\) is already defined on a connected subset \(C\) of \(R\), it extends uniquely and continuously to \(R\).
\(p^{-1}(U)\) is a disjoint union of open slices \(V_\alpha\). The connected set \(\overline{F}(C)\) lies in \(p^{-1}(U)\), so it lies in a single slice \(V_0\). Define \(\overline{F} = (p|_{V_0})^{-1} \circ F\) on \(R\). This is continuous and agrees with the given values on \(C\), since \(p \circ \overline{F} = F\) there and \(p|_{V_0}\) is injective.
2. Taking the rectangles row by row, from left to right within a row, the part of each rectangle on which \(\overline{F}\) is already defined is the union of its left and bottom edges, which is connected.
At the step for a rectangle \(R\), the rectangles already treated are those in lower rows and those to the left in the same row, together with the left and bottom edges of \(I \times I\), where \(\overline{F}\) was first defined by path lifting. Their union meets \(R\) exactly in the left and bottom edges of \(R\).
3. Q.E.D.
With another order, the part of a rectangle on which \(\overline{F}\) is already defined can be disconnected, for example its left and right edges. The values of \(\overline{F}\) on the two pieces could then lie in different slices over \(U\), and no continuous lift over \(R\) would agree with both. The order of step 2 makes that set connected at every step, so step 1 applies to every rectangle and the pieces fit together into a continuous \(\overline{F}\).