Question about existence of path-lifting property

[MathLearner123]

I'm reading "Complex Made Simple" by David C. Ullrich and here i have a problem with the proof of a theorem:

Theorem
Suppose that $p : X \to Y$ is a covering map. If $\gamma : [0,1] \to Y$ is continuous, $x_0 \in X$ and $p(x_0) = \gamma(0)$ then there exists a unique continuous function $\tilde{\gamma} : [0,1] \to X$ such that $\tilde{\gamma}(0) = x_0$ and $p \circ \tilde{\gamma} = \gamma$.

This is the proof that author gives:

To prove the existence, let $A$ be the set of $t_0 \in [0,1]$ such that there exists a continuous $\tilde{\gamma} : [0, t_0] \to X$ with $\tilde{\gamma(0)} = x_0$ and $p(\tilde{\gamma}(t)) = \gamma(t)$ for all $t \in [0, t_0]$. We need to show that $A$ is both open and closed.

I don't know how to show that $A$ is both open and closed, and as a mention I never took a Algebraic Topology class before. I tried to read some books, but can't understand very good what happens there and in the book mentioned by the author (A Basic Course in Algebraic Topology by William S. Massey) I don't find this proof for existence (that one from the book I don't understand). Thanks!

 

[fresh_42]

$A$ is a subset of $[0,1].$ The only subsets of $[0,1]$ that are closed and open at the same time are $\emptyset$ and $[0,1]$ because $[0,1]$ is a connected set. So if we can show that a) $A\neq \emptyset$, b) $A$ is open, and c) $A$ is closed, then $A=[0,1].$ If so, we can chose $t_0=1$ and we are done. I assume that $0\in A$ proves a). Remains to show b) and c).

 

[WWGD]

Often, you show openness by proving if property holds at some $t$, then it holds as well in a neighborhood of $t$. Closedness can be shown either by showing the complement is open or through sequences, so that $S$ is closed if for every sequence $s_n \in S$, if $S_n \rightarrow s$, then $s \in S$, i.e. the limit$s$ of a sequence of points in $S$ is itself in $S$.

 

[MathLearner123]

I think I proved the openness. I will write it here as soon as I get home. I know that I can use sequences to prove the closedness but I don't know actually how to apply this on my exercise. Can you help me for closedness please?

 

[Office_Shredder]

I think I proved the openness. I will write it here as soon as I get home. I know that I can use sequences to prove the closedness but I don't know actually how to apply this on my exercise. Can you help me for closedness please?

Here's a sketch I would do, lots of details left to you
1.) if $t_0\in A$ then $t\in A \forall t < t_0$.
2.) if $t_k\to t$ is a sequence with $t_k\in A$ for all k, then $t\in A$ unless $t_k< t$ for all k by (1)
3.) you can have pick a strictly increasing subsequence (unless $t=t_k$ eventually)
4.) use the paths given by that subsequence to build a path that stretches all the way to $t$.

A classic technique is to take an arbitrary sequence and restrict it to a sequence that exactly reflects the interesting case you have to cover.

 

[mathwonk]

I like a little trick called a "lebesgue number", that he invented to clarify some of Riemann's integration concepts. If one has an open cover of a compact subset C of a metric space, there is a positive number e>0, such that the e-ball around any point of C is entirely contained in one of the open sets of the cover.

Then one pulls back to the closed unit interval, the open cover of Y by the "evenly covered" open sets that define the covering map. Then any subdivision of the unit interval into short enough subintervals will be such that each subinterval maps entirely into an evenly covered set in Y.

Then you just define the lift on one of these subintervals at a time. I.e. if a subinterval maps entirely into an evenly covered set of Y, the lift to X is uniquely defined by knowing where its initial point goes. just continue for every subinterval.
And this also proves uniqueness at the same time.

the point is that defining the lift on a subinterval [0,t1] is trivial if the subinterval maps entirely into an evenly covered set in Y. But [0,1] is a union of a finite chain of such subintervals, 0 ≤ t1 ≤ t2 ≤ ...≤ 1. so first define the lift on [0,t1], then on [t1,t2],.....

 

[mathwonk]

here's a proof of the lebesgue number: define a function d(x) on C as "how far inside the cover" is the point x. I.e. d(x) is the largest positive number e such that the open ball around x of radius e is entirely inside one set of the open cover. This function has positive values and is continuous, hence has a positive lower bound on the compact set C, the lebesgue number. (If you have trouble showing d is continuous, you can reduce the cover first to a finite subcover, and then d is the maximum, over each open set, of the distance from x to the outside of that open set, which is continuous by the triangle inequality.)

 

[mathwonk]

or if you want to stick closer to the author's proof, basically you want to show that if 0 < t1 < 1, and the lift of gamma is defined on every subinterval [0,t0] with t0<t1, then the lift can also be defined on [0,t1]. For that, choose an open set U containing t1, and evenly covered. Then for some t0<t1, the image under gamma of the interval [t0,t1] is contained in U, and the already defined lift on [0,t0] takes t0 into a unique component V of the preimage of U in X. Since V≈U, and gamma maps [t0,t1] in to U, it is easy to extend the lift of gamma mapping [t0,t1] into V, via [t0,t1]-->U-->V.

 

[MathLearner123]

Thanks for all help! I think I got it!