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: Every 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.