**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.

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