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I'm doing a course on Topology and have been trying to make sense of the outline of a proof my professor gave during a lecture which went over my head at the time. I have attempted to fill in the gaps from my notes but wanted to ask whether I have done so correctly as I'm a not sure about my arguments.

I want to prove the following (as it was stated in my lecture):

Suppose $p:X'\to X$ is a covering space, then if $[\gamma]\in \pi_1(X,x_0)$ is such that $\gamma$ lifts to a non-closed path, then $\gamma$ is not null homotopic.

I seek to prove the contrapositive, that is, $\gamma$ null homotopic implies it lifts to a closed loop.

My Attempt

Fix $[\gamma]\in \pi_1(X.x_0)$ null homotopic. Let $F: I \times I \to X$ is this null homotopy so that $F(0,t)=\gamma(t)$ for all $t\in I$ and $F(1,t)=x_0 \in X$ for all $t \in I$ where $x_0$ is just some single point. Also, $F(s,0)=x_0=F(s,1)$ for all $s\in I$ as the base point is unchanged in the null homotopy.

The lifting lemma I have seen states the following:

Given a covering space $p:X'\to X$ and $f:I^n\to X$, then for any $x\in p^{-1}(f(0))$, there exists a unique lifting $\tilde{f}:I^n \to X'$ such that $\tilde{f}(0)=x$.

Pick any $x_0'\in p^{-1}(F(0,0))$. By the lifting lemma, $\exists ! \tilde{F}:I\times I \to X' $ with the property that $\tilde{F}(0,0)=x_0'$. Now, by definition of the lifting $p\circ \tilde{F}=F$. Whence, for any $s\in I$, $(p\circ \tilde{F})(s,0)=F(s,0)=x_0$ for all $s \in I$. Therefore, $\tilde{F}(s,0)\in p^{-1}(x_0)$ for all $s \in I$ so $\tilde{F}(I\times \{0\})\subset p^{-1}(x_0)$.

As $\tilde{F}$ is continuous and $[0,1]\times \{0\}$ is connected, $\tilde{F}(I\times \{0\})$ is connected. However, by definition of the covering map, $\exists U^{nbhd} \owns x_0$, and $\exists h_U :U\times \Delta \to X'$ with $\Delta$ a discrete space such that the projection $\pi_U:U\times \Delta \to U$ satisfies $\pi_U=p\circ h_U$. But then $p^{-1}\circ \pi_U = h_U$ so $(p^{-1}\circ \pi_U)(\{x_0\}\times \Delta) = h_U(\{x_0\}\times \Delta)$ so that $p^{-1}(x_0)= h_U(\{x_0\}\times \Delta)$. But then, as $\{x_0\}\times \Delta$ is a discrete space and $h_U$ is continuous, if $p^{-1}(x_0)$ has cardinality greater than $1$, it is necessarily disconnected. So, then since $\tilde{F}(s,0)\in p^{-1}(x_0)$, $p^{-1}(x_0)$ has exactly one element. So then $\tilde{F}(s,0)=x_0'$ for all $s\in I$. Likewise, we can show that $\tilde{F}(s,1)=x_0'=\tilde{F}(1,t)$ for all $s,t\in I$. So then $\tilde{F}$ is a homotopy between the lift of $\gamma$, given by $\tilde{\gamma}(t)=\tilde{F}(0,t)$ and the constant loop $\tilde{F}(1,t)=x_0'$ as $t$ moves over $I$.

I am the most uncertain about the argument involving the $h_U$ function. I think the argument should be obvious but it feels unintuitive the way I have done it. Also, doesn't this proof give us the stronger result that null homotopic paths lift to null homotopic paths?

Paul Frost
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Joseph Basford
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  • I don't see why "null homotopic paths lift to nullhomotopic paths" would be stronger. In fact, it's easier to prove: A nullhomotopy of a path lifts to a homotopy which at $t = 1$ lifts a constant path, so by discreteness of fibres it, too, must be constant. – Ben Steffan Oct 31 '24 at 17:02
  • @BenSteffan The definition of null homotopy we use in the course is that a based loop is null homotopic if it is path-homotopic to a constant. I thought that meant any null homotopy is a closed loop in our definitions, but I'm still getting used to working with the machinery. – Joseph Basford Oct 31 '24 at 17:26
  • What is $I^n $? The usual lifting lemma works with $n=1$. – Paul Frost Oct 31 '24 at 17:44
  • @PaulFrost I meant $\underbrace{[0,1]\times [0,1] \times \dots \times [0,1]}_{\text{n times}}$ and this was just what I had copied down from the lecture. For $n=2$ it is the lifting of the homotopy of some paths, but I don't know about the case $n\geq 3$. – Joseph Basford Oct 31 '24 at 18:03

1 Answers1

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Your concern is justified. $p^{-1}$ is no (single-valued) function, thus $p^{-1} \circ \pi_U = h_U$ does not make sense. Your argument would require that $p$ is a bijection locally.

However, you collected the necessary ingredients.

The set $A = I \times \{1\} \cup \{0,1\} \times I$ is connected and mapped by $F$ to $x_0$. Hence $\tilde F(A)$ is connected and contained in $p^{-1}(x_0)$, thus contained in the component of $x'_0 =\tilde F(0,0)$ in $p^{-1}(x_0)$. But $p^{-1}(x_0)$ is discrete, thus the component of $x'_0$ is $\{x'_0\}$. This shows that $\tilde F$ maps $A$ to $x'_0$ and proves that the lift of $\gamma$ is a null-homotopic closed loop.

Paul Frost
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  • I see, thank you. I am still struggling to understand why exactly the fibres under $p$ are discrete. This is what I was trying to show with my clumsy argument with $h_U$. Would you mind spelling out for me why these are discrete? – Joseph Basford Nov 01 '24 at 18:46
  • Which definition of a covering space do you use? – Paul Frost Nov 01 '24 at 21:07
  • For us, a covering space of $X$ is a set $X'$ and map $p:X'\to X$ such that for any $x\in X$, there is an open neighbourhood $U\owns x$, along with a discrete set $\Delta$ and map $h_U: \Delta \times U \to X'$ with $p\circ h_U = \pi_U$, where $\pi_U$ is the projection from $\Delta \times U$ to $U$. – Joseph Basford Nov 03 '24 at 17:46
  • @JosephBasford Certainly you have to require that $h_u$ is a homeomorphism (otherwise any surjective local homeomorphism would be a covering; to see this take $\Delta$ to be a one-point space). But if you know that $h_U$ is a homeomorphism, then you know that the discrete space ${x} \times \Delta$ is homeomorphic to $p^{-1}(x)$. This means that the fiber $p^{-1}(x)$ is discrete. – Paul Frost Nov 06 '24 at 09:03
  • I see, that is very clear now, thank you! – Joseph Basford Nov 06 '24 at 09:10