When is a local homeomorphism a semicovering map?

In a previous post “What is a Semicovering Map?,” I gave an introduction to semicovering maps. A semicovering is a slight generalization of covering map that becomes particularly relevant when you’re dealing with locally complicated spaces. In particular, a semicovering $p:E\to X$ is a local homeomrphism with unique lifting of all paths (for every path $\alpha:[0,1]\to X$ and point $e\in E$ with $p(e)=\alpha(0)$ , there is a unique lift $\beta:[0,1]\to E$ with $\beta(0)=e$ ). For context, I recommend taking a look at the introductory post (or the original papers [1] and [3]) first to gain intuition for the subtle difference between covering maps and semicovering maps.

A natural question to ask is: how much can we weaken the definition of “semicovering map?” More precisely, is there a weaker condition that allows us to promote a local homeomorphism to a semicovering? I really enjoy the logical investigations that these kinds of questions lead to.

This question comes from the paper [4], which gives an answer similar to the one I’ll give in this post. My interest in this question was piqued by questions of some non-Archimedian geometers, who identified a geometric analogue of semicoverings. I shared some of my suspicions with these geometers and this post is just me writing down the details. Certainly, the authors of [4] deserve the credit for originally answering this question.

Definition: We say a map $p:E\to X$ has the endpoint-lifting property whenever $\alpha:[0,1]\to X$ is a path and $\beta:[0,1)\to E$ is a map satisfying $p\circ \beta=\alpha|_{[0,1)}$ , $\beta$ extends (not necessarily uniquely) to a path $\beta:[0,1]\to E$ such that $p\circ\beta=\alpha$ .

The endpoint-lifting property should be thought of as a completeness-type property of $E$ relative to $p$ : if you have a map $[0,1)\to E$ and it agrees with the start of a true path in $X$ , then you can complete it to a true path $[0,1]\to E$ . There are some conditions on $E$ (without referring to $p$ ) that imply this, but these include very strong compactness conditions and do not improve known results.

Example: Let $T=\{0\}\times[-1,1]\cup\{(x,\sin(1/x)\mid 0<x\leq 1\}$ be the closed topologist sine curve and let $p:T\to [0,1]$ be the projection onto the x-axis. This map does not have the endpoint-lifting property because we can define continuous map $\beta:[0,1)\to T$ by $\beta(t)=(1-t,\sin(\frac{1}{1-t}))$ . Now $\alpha:[0,1]\to [0,1]$ , $\alpha(t)=1-t$ is a path and agrees with $p\circ\beta$ on $[0,1)$ . However, we cannot assign a value to $\beta(0)$ so that $\beta$ is continuous at $0$ . You can have both spaces path conecnted if you take the projection of the Warsaw circle onto the ordinary circle.

Throughout this post, we’ll assume the space $X$ is non-empty and path connected. With this assumption, a map $p:E\to X$ with the endpoint lifting property (or stronger property) will always be surjective.

Theorem: Suppose convergent sequences in $E$ have unique limits and $p:E\to X$ is a local homeomorphism. Then the following are equivalent:

$p$ is a semicovering map,
$p$ is a Hurewicz fibration,
$p$ has the endpoint-lifting property.

Note: the condition that convergent sequences in $E$ have unique limits is formally weaker than being Hausdorff. Sometimes a space with this property is called a “US-space.”

Proof of Theorem. The proof of 1. $\Leftrightarrow$ 2. is Theorem 7.5 in [2]. I’d like to focus on the equivalence of 1. and 3 because this part is, in my view, a little more fun. By definition, a semicovering has “unique lifts of all paths rel. basepoint.” Clearly, this implies the endpoint-lifting property. The direction 1. $\Rightarrow$ 3. follows. The rest of this post will be to prove 3. $\Rightarrow$ 1.

Suppose $p$ is a local homeomorphism with the endpoint-lifting property. We must show that $p$ has unique lifting of all paths. Let $\alpha:[0,1]\to X$ be a path starting at $\alpha(0)=x$ and $p(e)=x$ . We must find a unique path $\beta:[0,1]\to E$ such that $\beta(0)=e$ and $p\circ \beta=\alpha$ . We proceed similar to how you might prove that closed intervals are compact, that is, by defining a convenient set and analyzing the supremum. Let $$A=\{t\in (0,1]\mid \exists \beta_t:[0,t]\to E\text{ s.t. }\beta_t(0)=e\text{ and }p\circ\beta_t=\alpha|_{[0,t]}\}$$

Existence of path lifts: To show that lifts exist, we must show that $1\in A$ . Since $p$ is a local homeomorphism, we can find an open neighborhood $U$ of $e$ that $p$ maps homeomorphically onto the neighborhood $p(U)$ of $x$ . Find $s>0$ such that $\alpha([0,s])\subseteq p(U)$ . The formula $\beta_s(t)=p|_{U}^{-1}(\alpha(t))$ gives a lift $\beta_s$ latex of $\alpha|_{[0,s]}$ . Therefore $s\in A$ and we have $\sup(A)>0$ . Moreover, the definition of $A$ also ensures that $A$ is an interval of the form $[0,t)$ or latex $[0,t]$ for $0<t\leq 1$ . Now the assumption that $p$ has the endpoint-lifting property tells us that if if we knew $\alpha|_{[0,t)}$ could be lifted, then we can also lift $\alpha|_{[0,t]}$ . Hence, $A$ must actually be a closed interval of the form $[0,v]$ where $v=\max(A)$ .

But if $v<1$ , we can only lift $\alpha|_{[0,v]}$ to a path $\beta:[0,v]\to E$ and no more. But $p$ is still a local homeomorphism. So we can find an open neighborhood $U'$ that $p$ maps homeomorphically onto the open set $p(U')$ . Since $\alpha(v)\in p(U')$ , we can find $v<w$ such that $\alpha([v,w])\subseteq p(U')$ . Like before we define $\beta(t)=p|_{U'}^{-1}(\alpha(t))$ when $t\in [v,w]$ . This extends $\beta$ to a map $\beta:[0,w]\to E$ that is a lift of $\alpha|_{[0,w]}$ .

However $w>max(A)$ ; a contradiction. Since $v>0$ and $v<1$ is false, we must have $v=\max(A)=1$ . This proves existence.

Uniqueness of path-lifts: To prove uniqueness, suppose $\beta_1,\beta_2:[0,1]\to E$ both satisfy $\beta_i(0)=e$ and $p\circ\beta_i=\alpha$ . Let $B=\{t\in (0,1]\mid \beta_{1}|_{[0,t]}=\beta_{2}|_{[0,t]}\}.$

Using the neighborhood $U$ of $e$ (as above) that is mapped homeomorphically to $p(U)$ , we can see that we must have $\beta_{1}|_{[0,t]}=\beta_{2}|_{[0,t]}$ for at least some small $t>0$ . Hence, $B\neq \emptyset$ . Just like $A$ , the set $B$ must be an interval containing $0$ . What if $B=[0,s)$ for some $0<s\leq 1$ ? Then $\beta_{1}|_{[0,s)}=\beta_{2}|_{[0,s)}$ but $\beta_1(s)\neq \beta_2(s)$ . Picking a convergent increasing sequence $s_1<s_2<s_3<\cdots \to s$ in $[0,s]$ , we see that $\{\beta_1(s_n)\}$ and $\{\beta_2(s_n)\}$ are equal but converge to distinct limits points $\beta_1(s)$ and $\beta_2(s)$ .

However, we have assumed that convergent sequences in $E$ have unique limits. Therefore, $B$ must be a closed interval.

Now we must show $v=\max(B)=1$ . Suppose that $v<1$ . Then $\beta_{1}|_{[0,v]}=\beta_{2}|_{[0,v]}$ . Take a neighborhood $U'$ that maps homeomorphically on the neighborhood $p(U')$ of $\alpha(v)$ . Because $p|_{U'}:U'\to p(U')$ is injective, $\beta_1([v,w])$ and $\beta_2([v,w])$ lie in $U'$ for some $v<w$ , and $p(\beta_1(t))=p(\beta_2(t))$ for all $v\leq t\leq w$ , we must have $\beta_{1}|_{[v,w]}=\beta_{2}|_{[v,w]}$ . Thus $w\in B$ ; a contradiction that $v$ is the maximum. We conclude that $v=1$ , i.e. $\beta_1=\beta_2$ . $\square$

References:

[1] J. Brazas, Semicoverings: a generalization of covering space theory. Homology Homotopy Appl. 14, (2012) 33–63. Open Access.

[2] J. Brazas, A. Mitra, On maps with continuous path lifting. Preprint. 2020. https://arxiv.org/abs/2006.03667

[3] H. Fischer, A. Zastrow, A core-free semicovering of the Hawaiian Earring. Topology Appl. 160, (2013) 1957–1967. Open Access.

[4] M. Kowkabi, B. Mashayekhy, H. Torabi, When is a local homeomorphism a semicovering map? Acta Mathematica Vietnamica, 42, (2017) 653-663. https://arxiv.org/abs/1602.07260