Why is covering map a fibration? Incorrect proof?

Question

This is on page 66, Tom Dieck's

A covering $p:E \rightarrow B$ is a fibration.

What concerns me is its proof.

The proof is claims is suffices to consider local lifts - how?

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On page 63, (Prop 3.1.3) there is a uniquely lifting from $f:X \rightarrow B$ to $F:X \rightarrow E$, give that $X$ is connected.

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Suppose we can lift locally. (Definition below) Then we may piece up all the local fibrations by uniqueness. But $X$ is not necessarily connected here. Only $I$ is.

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So how does the piecing work?

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Local lifiting: on $V \times I \subseteq X \times I$, with respect to an initial condition $a:V \times \{0 \} \rightarrow E$, $h:V \times I \rightarrow B$, then exists $H$ ,
$$ H: V \times I \rightarrow E, pH=h, Hi=a.$$

Answer 1

$X$ is irrelevant here. Given $a : X \to E$ and a homotopy $H : X \times I \to B$ such that $p \circ a = H_0$ we find an open covering $\{ V_\alpha \}$ of $X$ and local lifts $G^\alpha : V_\alpha \times I \to E$ of $H \mid_{V_\alpha \times I}$ such that $a \mid_{V_\alpha} = G^\alpha_0$. We have to show that $G^\alpha$ and $ G^\beta$ agree on $(V_\alpha \times I) \cap (V_\beta \times I)$. This is the same as showing that $G^\alpha \mid_{\{ x \} \times I} = G^\beta \mid_{\{ x \} \times I}$ for all $x \in V_\alpha \cap V_\beta$.

But Proposition 3.1.3 applies for $X' = \{ x \} \times I$, $f = H \mid_{\{ x \} \times I} \to B$ and $F_0 = G^\alpha \mid_{\{ x \} \times I}, F_1 = G^\beta \mid_{\{ x \} \times I}$ because $F_0(x,0) = G^\alpha(x,0) = a(x) = G^\beta(x,0) = F_1(x,0)$.

Note that what we have used here is known as unique path lifting.