Lemma 27 Revisited. Suppose is a covering map. Then there is a covering map such that is the fibre product of and .
Proof. Let be the fibre product of and . There is a map given by . Let be the fibre product of and ; i.e.
There is a map given by . This is a covering map and injective, so it is a homeomorphism.
Let be continuous, be a covering map and , choices of basepoint. We have already seen that a choice of such that determines an elevation of to at . Fix such a . The pre-image of in is in bijection with the set of cosets
This raises the question, when do two cosets determine the same elevation?
Exercise 24. and determine the same elevation if and only if
that is, the set of elevations of to is in bijection with .
Let and be graphs of spaces and suppose we have the following data.
(a) A combinatorial map given by and .
(b) Covering maps for each .
(c) Covering maps for each , such that whenever adjoins .
This determines a continuous map . When is really a covering map?
Theorem 17. is a covering map if
(i) for all adjoining , the edge map is an elevation of ; and
(ii) wherever adjoining and , every elevation to arises as an edge map of .
Proof (sketch). It’s enough to consider our local model : and . and be covering maps defining and a map . By Lemma 27, is a covering map if and only if is a fibre product with respect to some covering:
Every map in the diagram is injective, so (for each component)
and it follows that is the fibre product of and . The result follows.