Theorem 1Suppose 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 2Suppose 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!