I'm just learning algebraic topology and have hit a problem I can't do. Lets say we have two topological spaces $X$ and $Y$ where $X$ is connected, and a continuous function $f:X \to Y$. Let $p:Y^\prime\to Y$ be a covering map of $Y$. Say $f_1^\prime$ and $f_2^\prime$ are two lifts of $f$. I want to prove that if $f_1^\prime(x_0)=f_2^\prime(x_0)$ for some $x_0$ then $f_1^\prime(x)=f_2^\prime(x)$ for all $x$. Could someone point me in the right direction?
Different Lifts of the Same Function
2 Answers
I think we need some more hypotheses. Taking $X = \{0, 1\}$ will already give a lot of counterexamples. So I will assume that $X$ is connected. The idea is to show that the subset of $X$ on which your lifts agree is open and closed.
So take $x \in X$, and an evenly covered neighborhood $U$ of $f(x)$. Let $U_1'$ and $U_2'$ be the slices above $U$ containing $f_1'(x)$ and $f_2'(x)$, respectively. Argue that there is an open neighborhood $V$ of $x$ such that $f_1'(V) \subset U_1'$ and $f_2'(V) \subset U_2'$, and that the two lifts either agree or disagree on all of $V$. Let me know if I should say more!
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0You were right. My book actually said that it had to be connected, I just forgot to mention it in the question. What you said doesn't really help me. I'm very new to everything topological and have no idea where to go from there. If you could just make a slightly more complete argument I might get it. – 2012-07-07
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0@Parakee If you've gotten to this point, then it wouldn't hurt to put that work in the question, so that I don't tell you things that you already know. Let's see. Can you find a $V$ like this, for starters? This would just use that the $f_i'$ are continuous. – 2012-07-07
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1Once you've done this, suppose that $f_1'(x) = f_2'(x)$. Then $U_1' = U_2'$ (why?), and recall that (1) $p$ restricted to $U_1'$ is injective and (2) $f_1'$ and $f_2'$ are lifts of the same function $f$. Use this to show that $f_1'(z) = f_2'(z)$ for all $z \in V$. Does that help? Cheers, – 2012-07-07
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0Thank you. That helped. I get it now. – 2012-07-07
It seems to me that this question of lifting maps in the theory of covering spaces is best handled by using the relation between covering maps of spaces and covering morphisms of groupoids, since then lifting of maps is related to lifting of morphisms. This idea is handled in my book "Topology and groupoids" (T&G) Chapter 10, which is available from amazon. The point is that applying the fundamental groupoid functor $\pi_1$ to a covering map of spaces gives what is called a covering morphism of groupoids, which is a translation to groupoids of the idea of unique path lifting. The main result is then that in this way we get an equivalence of categories between covering maps of $X$ and covering morphisms of $\pi_1X$, for suitably locally nice spaces $X$. So the lifting problem translates into the equivalent following result, which is 10.3.3 of T&G:
Let $p : \widetilde{G}, \tilde{x} \to G, x$ be a covering morphism, and $f : F, z \to G, x$ a morphism such that $F$ is connected. Then $f$ lifts to a morphism $\tilde{f} : F,z \to \widetilde{G}, \tilde{x}$ if and only if the characteristic group of $f$ is contained in that of $p$; and if this lifting exists, then it is unique.
There is also the result that covering morphisms of a groupoid $G$ are equivalent to actions of $G$ on sets, which is quite an old result in groupoid theory (see Higgins' book: "Categories and Groupoids"). But the translation of the above into actions may not be quite as clear.
I can give more details if needed.