Show that a path $\beta: I \rightarrow B $ lifts to a path $\bar{\beta}: I \rightarrow E .$

Question

The question is:

Let $p: E \rightarrow B $ be a fibration. Show that a path $\beta: I \rightarrow B $ with $\beta(0) = b_{0}$ and $e_{0} \in p^{-1}(b_{0})$ lifts to a path $\bar{\beta}: I \rightarrow E $, i.e., $p \circ \bar{\beta} = \beta,$ with $\bar{\beta} (0) = e_{0}.$

Now, I have read this question Show that $\phi$ is a fibration. and I claim that I understand it which means that I understand how to prove that something is a fibration by proving the HLP for it. but still I am unable to answer the question I mentioned above. and I got a hint that the proof is by direct application of fibration definition and that we should define a homotopy that gives us something homotopic to a point. but still I am unable to write a rigorous and thorough proof for it. so could anyone help me in doing so?

Answer 1

First, prove the following:

1. A point $x$ of a space $X$ corresponds to a (continuous) function $\mathbf{x} : \ast \to X$ that sends $*$ to $X$.

2. Paths $\gamma : I \to X$ from $x$ to $y$ are in bijective correspondence to homotopies $H : \mathbf{x} \simeq \mathbf{y}$ with $\mathbf{x},\mathbf{y}: \ast \to X$ the functions associated to $x$ and $y$, where $H$ is defined via $H(*,t ) = \gamma(t)$.

From now on I will make an abuse of notation and write $x$ for both a point $x \in X$ and the associated function $\ast \to X$.

Fix a path $\gamma : I \to B$ from $b_0$ to $b_1$. This gives a homotopy $H : b_0 \simeq b_1$. Since $e_0 \in p^{-1}(b_0)$, its associated map $e_0 : \ast \to E$ satisfies

$$ p \circ e_0 = b_0 = H_0. $$

Now, as $p$ is a fibration, we have a homotopy $G : \ast \times I \to E$ such that $pG = H$ and $G_0 = e_0$. If we define $\widetilde{\gamma}(t) := G(\ast,t)$, then

$$ \widetilde{\gamma}(0) = G_0(\ast) = e_0(\ast) = e_0 \in E $$

and

$$ p \circ \widetilde{\gamma}(t) = p \circ G(\ast,t) = H(\ast,t) = \gamma(t). $$