# Question about lifting of a function

• A
• MathLearner123
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!

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 $D$.

MathLearner123
@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.

Last edited:
MathLearner123

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