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$X, \tilde{X}$ are hausdorff, path connected spaces

$p:\tilde{X} \rightarrow X$ is a local homeomorphism

Under these assumptions we have the following lemma:

If $f:Y \rightarrow X$ is continuous, and $Y$ is connected, f has at most one lift with respect to $p$, that is if $f_1$, $f_2$ are both lifts with respect to $p$, then $f_1 = f_2$ or there is no point in $Y$ where $f_1 = f_2$


I want to prove the following:

Let $a$, $b$ be homotopic paths in $X$ (so that in particular $a$, $b$ have the same starting and endpoints), via the homotopy $F:I \times I \rightarrow X$, and suppose that $f_t = F(.,t)$ each have lifts $\tilde{f_t}$ with respect to $p$ starting at some $\tilde{x_0} \in p^{-1}(a(0))$.

($I$ is $[0,1]$)

Then the paths $\tilde{f_t}$ are homotopic in $\tilde{X}$ and in particular have a common starting and endpoint.

Attempt:

So far I have been able to construct a lifted homotopy, $\tilde{F}:[0,\epsilon] \times I \rightarrow \tilde{X}$ via considering a neighbourhood of $\tilde{x_0}$, $U$ where $p$ is a homeomorphism, then reasoning that $F^{-1}(p(U))$ is open in $I \times I$ and contains ${0} \times I$, so that $[0,\epsilon] \times I \subset F^{-1}(p(U))$ and $F([0,\epsilon] \times I) \subset p(U)$ for some $\epsilon > 0$. Then I defined $\tilde{F} = p \restriction_{U}^{-1} \circ F \restriction_{[0,\epsilon]\times I}$ a lift of $F$ on the desired domain. We also have that $\tilde{F} \restriction_{{0}\times I} = \tilde{x_0}$

So I know that the set of all $\epsilon \in I$ such that a lift as constructed above exists is nonempty, and that each lift corresponding to such an $\epsilon$ is unique, so that any other lift is either a continuation or a restriction of this lift (that is any other lift on some domain $[0,\epsilon_1] \times I$ must agree with lifts on domains that are contained within this domain i.e. $\epsilon_2 < \epsilon_1$) , which follows by our lemma.

I have also been able to show that given such a lift, if $\epsilon < 1$, we can always construct a continuation of the lift, that is we can find a lift $\tilde{F'}:[0,\epsilon + h] \times I \rightarrow \tilde{X}$, where $h$ depends on $\epsilon$, but don't have control over how much bigger the continuation of the lift is ($h$ is dependent on $\epsilon$). Which implies that if we can show $\sup A \in A$ , then $1 \in A$ and we are done.

I am stuck at this point, as I cannot show that $\sup A \in A$ necessarily.. I would be extremely grateful for any help

EDIT: I should add I am talking about homotopy of paths all throughout this question, no "free" homotopies where endpoints of each path in the homotopy are different - In this case any lifted homotopy will have to be a homotopy of paths by virtue of the fact that $p$ is a local homeomorphism, and so the preimage of an endpoint is a discrete set.. so this isn't a major concern, what is the primary trouble is how to construct a valid lifted homotopy on all of $I \times I$

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  • $\begingroup$ You remark that $F^{-1}(p(U))$ for some neighborhood $U\ni\tilde x_0$ contains $0\times I$. Why is this true? I there any reason a neighborhood on which p restricts to a homeomorphism should contain the entire lift of the path $F|_{0\times I}$? $\endgroup$ – Beckham Myers Aug 18 at 18:41
  • $\begingroup$ Well, first note $\tilde{x_0} \in U$, now we have that $p(x_0) = F(0 \times I)$, as $F$ is a homotopy of paths and thus all $f_s(t) = F(t,s)$ have the same start and endpoints... now since $p(x_0) \in p(U)$, it should be clear that $F^{-1}(p(U))$ contains $0 \times I$, as $F^{-1}(p(U))$ contains $F^{-1}(p(x_0))$, which is $F^{-1}(F(0 \times I))$ $\endgroup$ – Aneesh Aug 18 at 19:16
  • $\begingroup$ I suppose it sounds strange when you say "should contain the entire lift of the path ..", but because $F(0 \times I)$ is really just a singleton, its not too odd $\endgroup$ – Aneesh Aug 18 at 19:21
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If $\tilde{F}(., t) = \tilde{f}_t$ denotes the lifting then it suffices to check that $\tilde{F}$ is continuous into $\tilde{X}$.

Do it "horizontally" locally instead of "vertically" as you've been trying to do.

That is, pick a point $(0, t_0) \in I \times I$ and show that there exists an open set $W \subseteq I \times I$ such that $I \times \{ t_0 \} \subseteq W \subseteq I \times I$ such that $\tilde{F}|_{W}$ is continuous. We can now find a "universal" length (which you have been struggling with) doing it horizontally as follows:

Note that for each $\tilde{x} \in \tilde{f}_{t_0}(I \times \{ t_0 \}) = \tilde{F}(I \times \{ t_0 \}) = K$, there exists an open subset $U_{\tilde{x}}$ containing $\tilde{x}$ such that $p(U_{\tilde{x}})$ is open and $p|_{U_{\tilde{x}}}$ is a homeomorphism. Cover $I \times \{ t_0 \} \subseteq \cup_{\tilde{x} \in K} \tilde{f}_{t_0}^{-1}(U_{\tilde{x}})$ and use the Lebesgue covering lemma to find the length $l > 0$ such that if $J$ is an subinterval of $I$ of length at most $l$ then $J \times \{ t_0 \}$ is contained in some $\tilde{f}_{t_0}^{-1}(U_{\tilde{x}})$. We may as well assume that $l = 1/n$ for some positive integer $n$.

Now the rest is induction with argument similar to yours. First, $\tilde{F}$ is continuous when restricted to some small "vertical" neighborhood of the form $\{ 0 \} \times I_{t_0}$ containing $(0, t_0)$. Pick some $U_{\tilde{x}}$ (as above paragraph) such that $\tilde{F}([0, l] \times \{ t_0 \}) \subseteq U_{\tilde{x}}$. We may cut $I_{t_0}$ if necessary (by using continuity on the "vertical" nbdh $\{ 0 \} \times I_{t_0}$) so that $\tilde{F}(\{ 0 \} \times I_{t_0}) \subseteq U_{\tilde{x}}$. We have $[0, l] \times \{ t_0 \} \in F^{-1}(p(U_{\tilde{x}}))$, so by compactness of $[0,l] \times \{ t_0 \}$ we can fatten $[0,l] \times \{ t_0 \} \subseteq [0,l] \times J_{t_0} \subseteq F^{-1}(p(U_{\tilde{x}}))$ for some neighborhood $J_{t_0}$ of $t_0$. We may cut this further to assume that $J_{t_0} \subseteq I_{t_0}$ so that $\tilde{F}$ is continuous on the first vertical strip $\{0 \} \times J_{t_0}$. Now $\tilde{F} = (p|_{U_{\tilde{x}}})^{-1} \circ F$ on $[0,l] \times J_{t_0}$ by the "unique lifting" lemma that you cited (because both functions start at the same points on the vertical $\{ 0 \} \times J_{t_0}$. The RHS is clearly a continuous expression, so expresses $\tilde{F}$ as a continuous function on a neighborhood of $[0,l/2] \times J_{t_0}'$ where $J_{t_0}'$ is a compact interval containing $t_0$ of length $1/2$ that of $J_{t_0}$.

Now one simply inducts (go from continuity of $\tilde{F}$ on the strip $\{l \} \times J_{t_0}$ to continuity on $[l, l + l/2] \times T_{t_0}$ for some nondegenerate interval $T_{t_0} \subseteq J_{t_0}$ that could be smaller). Since this stops in finite steps (in at most $2n$ steps to be exact), this "cutting" of the intervals around $t_0$ doesn't pose any problem. The point is that our $l$ doesn't depend on "where" we are on the horizontal strip $I \times \{ t_0 \}$.

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    $\begingroup$ Apologies if I'm being hasty here - I haven't finished reading your entire solution, but when you say "First, $\tilde{F}$ is continuous when restricted to some small "vertical" neighbourhood of the form ${0} \times I_{t_0}$", surely since $\tilde{F}(0,t) = \tilde{x_0}$, by construction, we have in fact that $\tilde{F}$ is continuous when restricted to ${0} \times I$ , was this why you said $\tilde{F}$ is continuous when restricted to some small "vertical" neighbourhood of this form? Or were you using the fact that the neighbourhood was small somehow , perhaps later in the proof? $\endgroup$ – Aneesh Aug 19 at 19:28
  • $\begingroup$ What you say is true, but ultimately one wants to get an induction argument going. For the next step, for example, you need to work with continuity of $\tilde{F}$ on a strip $\{ l/2 \} \times I_{t_0}$, because $\tilde{F}$ may not be continuous on $\{ l/2 \} \times I$. The "vertical" slices for $t_0$ will decrease, but the point is that it'll end in finite steps because of universality of $l$. $\endgroup$ – hochs Aug 19 at 19:30
  • $\begingroup$ Ah, I think I see what you mean - I'll leave future comments till after I've finished digesting the proof. $\endgroup$ – Aneesh Aug 19 at 19:31
  • $\begingroup$ Given we have continuity of $\tilde{F}$ on the strip ${l} \times J_{t_0}$, isn't there some $U_{\tilde{x}}$ s.t. $\tilde{F}([l,2l]) \times t_0) \subset U_{\tilde{x}}$, and then proceeding as in paragraph 2, we should get a larger lift $\tilde{F}$ defined on $[0,2l] \times T_{t_0}$ for some smaller interval $T_{t_0}$ right? Why do we have to use steps of $l/2$? $\endgroup$ – Aneesh Aug 19 at 19:50
  • $\begingroup$ That should work too. The organization using $l/2$ is from my laziness - but not an important feature of the proof. $\endgroup$ – hochs Aug 19 at 19:53

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