Proof of injective correspondence on covering space

Question

Let $ \Gamma $ $:\hat {X} \rightarrow X$ be a covering space and $ x \in \hat {X}.$$\quad$$ \Gamma (x) = x \in X$. Let $ F = \Gamma ^{-1} (x)$. We define the function $\phi : \pi_{1}(X,x) \rightarrow F$ by $\phi (\{p\}) = \hat {p}(1)$, where $\hat {p}$ is the lifted path. $\qquad $ $\qquad$ $\qquad $ $\qquad$ $\quad $ $\quad$$\quad $ $\quad$$\quad $ $\qquad$Theorem$ -$ If $\hat{X}$ is simply connected, then $\phi$ is a $1-1$ correspondence. $\qquad$$\qquad $ $\quad$$\quad $ $\quad$$\quad $ $\quad$ Proof $-$ If $\hat {X}$ is simply connected, we will produce an inverse $\varphi : F \rightarrow \pi_{1}(X, x)$. For $f \in F$, choose a path $p$ from $a$ to $f$. Since $\hat {X}$ is simply connected, any two such choices are homotopic keeping the end points fixed. Thus $\{\Gamma \circ p\}$ is a well defined element 0f $\pi_{1} (X, x)$. Clearly $\varphi \circ \phi = 1$, since we may choose the original path to define $\varphi$. On the other hand. $\phi \circ \varphi = 1$ since, given $f \in F$ and a path $p$, $p$ is a lifting of $\pi \circ p$.$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\qquad$$\qquad $ $\qquad$$\qquad $ $\qquad$$\qquad $ $\qquad$$\quad $ $\quad$$\quad $ $\quad$$\quad $ $\qquad$ In the above proof, I am not being able to understand the reasoning behind this argument "Clearly $\varphi \circ \phi = 1$, since we may choose the original path to define $\varphi$. On the other hand. $\phi \circ \varphi = 1$ since, given $f \in F$ and a path $p$, $p$ is a lifting of $\pi \circ p$."

Answer 1

I'll call the basepoint of $\hat{X}$ as $\hat{x}$.

The map $\phi$ takes a loop, lifts it uniquely to a path starting from $\hat{x}$ and then looks at the endpoint. $\varphi$ starts with a point, takes a path from $\hat{x}$ to that point, then projects it down to a loop in $X$.

To show $\varphi\circ\phi = 1$, start with a loop $\gamma$ in $X$. It lifts to some path $\tilde{\gamma}$ and $\phi(\gamma) = \tilde{\gamma}(1)$. To find $\varphi(\tilde{\gamma}(1))$ you need to take a path from $\hat{x}$ to $\tilde{\gamma}(1)$ and project it down. Think about what works as such a path.

$\tilde{\gamma}$ is such a path and projects down to $\gamma$

To show $\phi\circ\varphi = 1$, start with a point $q$ in the fibre and take a path $\tilde{\gamma}$ from $\hat{x}$ to $q$. Then $\varphi(q) = \Gamma\circ\tilde{\gamma}$. To find $\phi(\Gamma\circ\tilde{\gamma})$ you need to lift this loop to $\hat{X}$ and look at the end point. Think about what works as a lift of this loop.

Since $\tilde{\gamma}$ projects to this loop and starts at $\hat{x}$, it is a suitable lift of $\Gamma\circ\tilde{\gamma}$, and has end point $q$