Cardinality of fibers same for covering maps

[PsychonautQQ]

I'm having trouble following one part of a proof.

Proposition: For any covering map $p: X-->Y$, the cardinality of the fibers $p^{-1}(q)$ is the same for all fibers

Proof: If U is any evenly coverd open set in $X$, each component of $p^{-1}(q)$ contains exactly one point of each fiber. Thus, for any $q,q' contained in U$, there are one-to-one correspondences between $p^{-1}(q)$ <---> {components of $p^{-1}(U)$} <--> $p^{-1}(q')$,

which shows that the number of components is constant on U.

next is the part I'm confused about:
"It follows that the set of points $q' \in X$ such that $p^{-1}(q')$ has the same cardinality as $p^{-1}(q)$ is open."

why is this?

 

[andrewkirk]

The covering map needs to go from Y to X, not X to Y as written above, in order for the post to make sense. Otherwise statements like 'evenly covered set in X' don't make sense.

I can't make anything of the above proof. Further, the theorem will not even be true unless we require Y to be path connected (some, but not all, authors require that as part of the definition of a covering map). Otherwise we could set $p:(X=A\dot\cup B)\to(Y =C\dot\cup D)$ where $p$ gives a 1:1 covering of A on C and a $\aleph_0$ covering of B on D.

Assuming that Y is path connected, take two points $a,b\in X$ and a path $\gamma$ between them such that $\gamma(0)=a,\gamma(1)=b$. Let $y= \inf\{x\in[0,1]\ :\ |p^{-1}(\gamma(x))|=|p^{-1}(\gamma(a))|\}$ and set $c=\gamma(y)$.

Then we should be able to prove $|p^{-1}(c)|=|p^{-1}(a)|$ and $c=b$ by considering an open nbd of $c$ that is evenly covered with some cardinality.

 

[PsychonautQQ]

10/10 for deciphering my garbage and giving a great response anyway

 

[Infrared]

I think the given proof makes sense, but I'll try to add some detail.

For $x\in X$, there exists $U$ a neighborhood of $x$ such that $p^{-1}(U)=\bigsqcup_{\lambda\in\Lambda}U_\lambda$ where each $U_\lambda$ is homeomorphic to $U$ and taken homeomorphically to $U$ by restricting $p$. So for each element of $q\in U$, the cardinality of $p^{-1}(q)$ is the same as that of $p^{-1}(x)$ (both are in one-to-one correspondence with $\Lambda$).

So each element of $X$ has a neighborhood consisting of points whose fibers all have equal cardinality. This means that the set of points whose fiber has cardinality $|\Lambda|$ is open. But its complement is also open by the same argument. If $X$ is open, this is enough to conclude that every point has fiber of cardinality $|\Lambda|$ (I don't think we need path-connectedness).

 

[WWGD]

Similar to Andrew Kirk's argument, assuming path-connectedness of the base, given any points a,b, consider a path joining a with b. A path ( viewed as a continuous map $f:[0,1] \rightarrow X$) is compact, and, of course, also connected . Take an open cover of the path of evenly-covered neighborhoods about the points in the path and then consider a finite subcover to show that points in consecutively-numbered open sets have fibers of the same cardinality (consider the overlap between consecutively-numbered open sets in the finite subcover), so that the fibers over $f(0), f(1)$ have the same cardinality. This then applies for any $a,b$.

 

[andrewkirk]

A path ( viewed as a continuous map $f:[0,1] \rightarrow X$) is compact...

If I am correctly recalling some work I did several years back on the Koch Snowflake, that is (very surprisingly - to me at least) not necessarily the case. The boundary $\partial K$ of the Koch Snowflake $K$ - a fractal shape - can be parameterised by a continuous, surjective map $\gamma:[0,1]\to\partial K$. But IIRC the path image $\partial K$, which has infinite length, is not compact in the subspace topology of $\mathbb R^2$. It is compact in the order topology imposed by the parametrisation, but that is different from the subspace topology, which is the one that is applicable in this problem.

For anyone interested, the analysis of the snowflake is here (formatting is not great unfortunately. It dates from before my rediscovery of LaTeX). It doesn't prove that the boundary is non-compact, but my memory tells me that I did convince myself of that, I just can't find a written proof. The intuition for why it's not compact is that it's infinitely long under the metric of $\mathbb R^2$.

For the covering map case, we can try to circumvent the non-compactness of the path image by taking an open cover of [0,1] rather than its image. But then we cannot be confident that the images of the sets in our finite, open subcover are themselves open - as a path is not required to be an open map.

On another note - the proof above by @Infrared looks to me like it works, in which case, as Infrared says, path-connectedness of X is not needed, only connectedness.

 

[WWGD]

If I am correctly recalling some work I did several years back on the Koch Snowflake, that is (very surprisingly - to me at least) not necessarily the case. The boundary $\partial K$ of the Koch Snowflake $K$ - a fractal shape - can be parameterised by a continuous, surjective map $\gamma:[0,1]\to\partial K$. But IIRC the path image $\partial K$, which has infinite length, is not compact in the subspace topology of $\mathbb R^2$. It is compact in the order topology imposed by the parametrisation, but that is different from the subspace topology, which is the one that is applicable in this problem.

For anyone interested, the analysis of the snowflake is here (formatting is not great unfortunately. It dates from before my rediscovery of LaTeX). It doesn't prove that the boundary is non-compact, but my memory tells me that I did convince myself of that, I just can't find a written proof. The intuition for why it's not compact is that it's infinitely long under the metric of $\mathbb R^2$.

For the covering map case, we can try to circumvent the non-compactness of the path image by taking an open cover of [0,1] rather than its image. But then we cannot be confident that the images of the sets in our finite, open subcover are themselves open - as a path is not required to be an open map.

On another note - the proof above by @Infrared looks to me like it works, in which case, as Infrared says, path-connectedness of X is not needed, only connectedness.

Still, I don't think it would be hard to show that a path that is compact in the subspace topology exists in spaces with minimal requirements; we don't need to have every path be compact in this topology, only that a compact path exists between any two points.EDIT: I would even say that compact paths are dense and that any non-compact path can be perturbed slightly into being compact; just make the map continuous in the subspace topology. EDIT2: Will try to provide a proof of these claims ASAP.

 

[Infrared]

This above discussion confuses me a bit. Isn't $\gamma([0,1])$ the continuous image of a compact set and therefore compact? Compact sets don't have to have finite length (even when this notion makes sense).

 

[WWGD]

This above discussion confuses me a bit. Isn't $\gamma([0,1])$ the continuous image of a compact set and therefore compact? Compact sets don't have to have finite length (even when this notion makes sense).

I agree, I think Andrew is saying it is not compact in the subspace topology of X, but, AFAIK, a path is a continuous map into the space, so continuous in the subspace topology. Maybe he can explain. EDIT: I guess length spaces are the most general spaces where we can speak of length. Maybe he is referring to non-self-intersecting curves, so that my argument of breaking the curve down into pairs of open sets that intersect only each other, i.e., $O_j$ intersects only $\{O_{j-1}, O_{j+1}\}$, but not $O_{j+k}$ for $k \neq \pm 1$ can go through.

 

[WWGD]

I think the given proof makes sense, but I'll try to add some detail.

For $x\in X$, there exists $U$ a neighborhood of $x$ such that $p^{-1}(U)=\bigsqcup_{\lambda\in\Lambda}U_\lambda$ where each $U_\lambda$ is homeomorphic to $U$ and taken homeomorphically to $U$ by restricting $p$. So for each element of $q\in U$, the cardinality of $p^{-1}(q)$ is the same as that of $p^{-1}(x)$ (both are in one-to-one correspondence with $\Lambda$).

So each element of $X$ has a neighborhood consisting of points whose fibers all have equal cardinality. This means that the set of points whose fiber has cardinality $|\Lambda|$ is open. But its complement is also open by the same argument. If $X$ is open, this is enough to conclude that every point has fiber of cardinality $|\Lambda|$ (I don't think we need path-connectedness).

Do you mean If $X$ is connected in the next-to-last line?

 

[lavinia]

Do you mean If $X$ is connected in the next-to-last line?

Right. Infrared's point is that the set of points with a different cardinality must also be open. and these two sets are by definition disjoint. This can't happen if the space is connected.

 

[andrewkirk]

I agree, I think Andrew is saying it is not compact in the subspace topology of X, but, AFAIK, a path is a continuous map into the space, so continuous in the subspace topology.

My mistake - my memory has been playing tricks on me. The path image is indeed compact in the subspace topology, despite its infinite length. There is a proof here about continuous images of compact sets, that looks valid to me. It must have been some other property of the Koch boundary that I was misremembering.

So @WWGD's proof in post 5 stands.

 

[mathwonk]

locally constant integer valued functions on connected spaces are constant. this works at least for finite cardinality fibers.

 

[WWGD]

locally constant integer valued functions on connected spaces are constant. this works at least for finite cardinality fibers.

Beat me to it, I was about to say this. Still, need to show continuity, which shouldn't be too hard.

 

[Infrared]

Beat me to it, I was about to say this. Still, need to show continuity, which shouldn't be too hard.

You don't need continuity, nor do you need the codomain to be discrete. Any locally constant function on a connected space is constant (though locally constant implies continuous).

 

[WWGD]

You don't need continuity, nor do you need the codomain to be discrete. Any locally constant function on a connected space is constant (though locally constant implies continuous).

Ah, my bad, did not read the locally constant part, only the integer-valued part.

 

[PsychonautQQ]

So each element of $X$ has a neighborhood consisting of points whose fibers all have equal cardinality. This means that the set of points whose fiber has cardinality $|\Lambda|$ is open.

Can you explain why all points who's fibers have cardinality $abs(/lambda)$ is open?

 

[Infrared]

Can you explain why all points who's fibers have cardinality $abs(/lambda)$ is open?

Use the following fact: If $X$ is a space and $U\subset X$ is a subspace such that for every element $x\in U$, there exists an open set $U_x\subset U$ that contains $x$, then $U$ is open.

Proof: Writing $U=\bigcup_{x\in U} U_x$ shows $U$ to be a union of open sets, and hence open.

It might be a good idea to review such facts from point-set topology- this stuff doesn't stop being important when you start doing more advanced material :)