- #1

MathLearner123

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**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!