I am reading through parts of Fomenko and Fuchs' Homotopical Topology. It's a great book for someone with some pre-existing exposure to algebraic topology, but there are omitted details and the occasional typo. I'm attempting to understand the argument for the theorem in Chapter 9.6. There are shortcuts I can use involving the exact sequence of a fibration, but I'd like to understand the specific argument given in the book to avoid the risk of circular reasoning.
Let $p: E \to B$ be a Serre fibration, and suppose $x_0, x_1 \in B$ are in the same path component of the base. To show $p^{-1}(x_0)$ and $p^{-1}(x_1)$ are weakly equivalent, we will construct a natural bijection from $\pi(X,p^{-1}(x_0))$ to $\pi(X,p^{-1}(x_1))$ (homotopy classes of maps $X \to p^{-1}(x_i)$) for an arbitrary CW complex $X$. Specifically, I am seeking to understand why this bijection is well-defined.
Choose a path $s$ from $x_0$ to $x_1$. Let $f: X \to p^{-1}(x_0) \subseteq E$ be an arbitrary continuous map. Define the homotopy $\Phi: X \times I \to B$ by $\Phi(x,t) = s(t)$. It goes from the constant map $X \to x_0$ to the constant map $X \to x_1$. Then $f$ is a lifting of $\Phi_0$, so by the homotopy lifting property, we have some $\widetilde{\Phi}: X \times I \to E$ lifting $\Phi$. The restriction $\widetilde{\Phi}|_{X \times 1}$ is a map $X \to p^{-1}(x_1)$, so denote this map by $g$. The bijection we want is $[f] \mapsto [g]$. We wish to show this map is well defined with respect to the choice of homotopy lifting and the homotopy class of the path $s$.
Let $s'$ be a path from $x_0$ to $x_1$ path-homotopic to $s$. Construct $\Phi', \widetilde{\Phi'}$, and $g'$ as above. We want a homotopy from $g$ to $g'$. Let $S: [-1,1] \times I \to B$ be a homotopy from the path $s^{-1}s'$ (the reverse of $s$ followed by $s'$) to the constant path at $x_1$. Define a homotopy $\Psi: (X \times [-1,1]) \times I \to B$ by $\Psi((x,u),t) = S(u,t)$. Let $\widetilde{\Psi_0}: X \times [-1,1] \to E$ be defined by $\widetilde{\Psi_0}(x,u) = \widetilde{\Phi}(x,-u)$ for $u \leq 0$ and $\widetilde{\Psi_0}(x,u) = \widetilde{\Phi'}(x,u)$ for $u \geq 0$. Notice that $\widetilde{\Psi_0}$ lifts $\Psi_0$. Apply the homotopy lifting property to get a homotopy $\widetilde{\Psi}$ lifting $\Psi$. Then $\widetilde{\Psi}$ restricted to $X \times ((-1 \times I) \cup ([-1,1] \times 1) \cup (1 \times I))$ is a homotopy from $g$ to $g'$ as desired.
My issue is why do we have to go through the trouble of constructing $\widetilde{\Psi}$? Is $\widetilde{\Psi_0}$ not the required homotopy?
In general $\widetilde\Psi_0$ will be a homotopy of maps $X\rightarrow E$ rather than $X\rightarrow p^{-1}(x_1)$ as we require for the proof.