Lemma 27 Revisited. Suppose \tau : X' \to X is a covering map. Then there is a covering map \sigma: Y'\to Y such that X' is the fibre product of \sigma and d.

Proof. Let Y' = \left\{ (\xi',\eta) \in X' \times Y : \tau (\xi) = i (\eta) \right\} be the fibre product of \tau and i. There is a map d' : X' \to Y' given by \xi' \mapsto (\xi' , d \circ \tau (\xi')).  Let \hat{X} be the fibre product of \sigma and d; i.e.

\hat{X} = \left\{ (\xi,\eta') \in X \times Y' : d(\xi) = \sigma(\eta') \right\}.


There is a map X' \to \hat{X} given by \xi' \mapsto (\tau(\xi') , d'(\xi')).  This is a covering map and injective, so it is a homeomorphism.

Let f: X \to Y be continuous, Y' \to Y be a covering map and x \in X, y= f(x) \in Y choices of basepoint. We have already seen that a choice of y' \in Y' such that y' \mapsto y determines an elevation of f to Y' at y'.  Fix such a y'. The pre-image of y in Y' is in bijection with the set of cosets

\pi_1(Y',y') \backslash \pi_1(Y,y)

This raises the question, when do two cosets determine the same elevation?

Exercise 24. \pi_1(Y',y')g_1 and \pi_1(Y',y')g_2 determine the same elevation if and only if

\pi_1(Y',y') g_1 f_* \pi_1(X,x) = \pi_1(Y',y') g_2 f_* \pi_1(X,x);

that is, the set of elevations of f to Y' is in bijection with \pi_1(Y') \backslash \pi_1(Y) / f_*(\pi_1(X)).

Let \mathfrak{X} and \mathfrak{X}' be graphs of spaces and suppose we have the following data.

(a) A combinatorial map \Xi' \to \Xi given by e' \mapsto e and v' \mapsto v.

(b) Covering maps \varphi_{v'} : X_{v'} \to X_v for each v' \in V(\Xi').

(c) Covering maps \varphi_{e'} : X_{e'} \to X_{e} for each e' \in E (\Xi'), such that \varphi_{v'}\circ\partial_{e'} = \partial_e\varphi_{e'} whenever e' adjoins v'.

This determines a continuous map \Phi : X_{\mathfrak{X}'} \to X_{\mathfrak{X}}.  When is \Phi really a covering map?

Theorem 17. \Phi is a covering map if

(i) for all e' \in E(\Xi') adjoining v' \in V(\Xi'), the edge map \partial_{e'}^{\pm} : X_{e'} \to X_{v'} is an elevation of \partial_{e}^{\pm} : X_{e} \to X_{v}; and

(ii) wherever e \in E(\Xi) adjoining v \in V(\Xi) and v' \in V(\Xi'), every elevation \partial_{e}^{\pm} : X_{e} \to X_{v} to X_{v'} arises as an edge map of \mathfrak{X}'.

Proof (sketch). It’s enough to consider our local model : X_{\mathfrak{X}} = X and Y = X_v. \varphi_{v'} : X_{v'} \to X_v and \varphi_{e'} : X_{e'} \to X_{e} be covering maps defining \mathfrak{X}' and a map \Phi : X_{\mathfrak{X}'} \to X_{\mathfrak{X}}. By Lemma 27, \Phi is a covering map if and only if X_{\mathfrak{X}'} is a fibre product with respect to some covering:

fig2note0401Every map in the diagram is \pi_1 injective, so (for each component)

\pi_1(Y') \cong \pi_1(X_{\mathfrak{X}'}) \cong \pi_1(X_{v'})

and it follows that X_{\mathfrak{X}'} is the fibre product of d and \varphi_{v'}. The result follows.