# Homotopy Lifting Problem

$$X, \tilde{X}$$ 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 $$I \times I$$

• 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 $F|_{0\times I}$? – Beckham Myers Aug 18 at 18:41
• 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 $F^{-1}(F(0 \times I))$ – Aneesh Aug 18 at 19:16
• 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 – Aneesh Aug 18 at 19:21

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 $$I \times \{ t_0 \}$$.

• 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? – Aneesh Aug 19 at 19:28
• 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 $l$. – hochs Aug 19 at 19:30
• Ah, I think I see what you mean - I'll leave future comments till after I've finished digesting the proof. – Aneesh Aug 19 at 19:31
• 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 $l/2$? – Aneesh Aug 19 at 19:50
• That should work too. The organization using $l/2$ is from my laziness - but not an important feature of the proof. – hochs Aug 19 at 19:53