Question about lifting of a function

[MathLearner123]

Theorem 1
Suppose that $p : X \to Y$ is a covering map. Suppose $\gamma_0, \gamma_1 : [0, 1] \to Y$ are continuous, $x_0 \in X$ and $p(x_0) = \gamma_0(0) = \gamma_1(0)$. Fie $\tilde{\gamma}_0$ and $\tilde{\gamma}_1$ be the continuous functions mapping [0,1] to X such that $\tilde{\gamma}_j(0) = x_0$ and $p \circ \tilde{\gamma}_j = \gamma_j$ for $j = 0, 1$. If $\gamma_0$ and $\gamma_1$ are path-homotopic then $\tilde{\gamma}_0(1) = \tilde{\gamma}_1(1)$.


Theorem 2
Suppose that $p:X \to Y$ is a covering map, $D$ is a path-connected, locally path-connected and simply connected topological space and $f:D \to Y$ is continuous. Suppose that $a \in D$. Fix $x_0 \in X$ with $p(x_0) = f(a)$. There exists a unique continuous function $\tilde{f} : D \to X$ such that $f(a) = x_0$ and $p \circ \tilde{f} = f$.

I don't understand very well the proof of Theorem 2:

Given $b \in D$, let $\gamma:[0,1] \to D$ be continuous with $\gamma(0) = a$ and $\gamma(1) = b$. Let $\tilde{\gamma}$ be a lifting of $f \circ \gamma$ with $\tilde{\gamma}(0) = x_0$.

Ok. Until there I understand that he constructs the unique path-lifting.

Define $\tilde{f}(b) = \tilde{\gamma}(1)$. Since $D$ is simply connected, Theorem 1 shows that $f$ is well defined.

Here I don't understand anything. If you can help me to understand.. from where is that $\tilde{f}$ function already defined? Thanks!

 

[pasmith]

The conditions of Theorem 2 require that (1) there exists a path between any two points of D (ie. a continuous \gamma: [0,1] \to D such that \gamma(0) is one of the two points and \gamma(1) is the other) and (2) that any two paths in D between those points are path-homotopic.

Since the continuous images of path-homotopic paths are path-homotopic, it follows that if \gamma' : [0,1] \to D is another path with \gamma'(0) = a and \gamma'(1) = b then f \circ \gamma and f \circ \gamma' are path-homotopic, so that by Theorem 1 the end points of the lifts of these paths are the same; this value therefore does not depend on the specific choice of path in D from a to b but only on the end point b. Thus the definition \tilde f(b) \equiv \tilde{\gamma}(1) makes sense: it doesn't depend on \gamma, but only on the fact that \gamma(0) = a and \gamma(1) = b. Since we are guaranteed that there will exist a path from a to any point in D and that any two such paths are path-homotopic, in this way we can define a function on the whole of D.

 

[WWGD]

@mathlearner: Use that simply-connectedness of $D$ (asumed in the problem) implies that $\pi_1(D)=\{e\}$, i.e., trivial. This means any two paths in $D$ are homotopically equivalent. Basically, there is one path between two points up to homotopy.