Unique Partition of Evenly Covered Sets in Algebraic Topology

[JasonRox]

Note: I have many questions and will keep posting new ones as they come up. To find the questions simply scroll down to look for bold segments. Feel free to contribute any other comments relevant to the questions or the topic itself.

Here it is...

Let p:E->B be continuous and surjective. Suppose that U is an open set of B that is evenly covered by p. Show that if U is connected, then the partition of p^(-1)(U) into slices is unique.

Ok, I barely understand what it's asking me to show. Is it saying the partition of p^(-1)(U) is unique? Because I highly doubt it's that because any set can be partitioned many different ways!

I think it's saying that if we have a map p1:E->B that is continuous and surjective and that U is evenly covered by p1, then the partition p1^(-1)(U) will be the same as p^(-1)(U).

Is that it?

 

[matt grime]

What is the definition of 'evenly covered', and what is the definition of 'partitioned into slices'.

 

[JasonRox]

I have no idea what the definition of "partitioned into slices" is.

Being purely honest here. I'll have to wait until I get home to look up the definition since Google isn't helping me here.

I found the solution online and haven't read it. Personally, I don't even want to glance at it to get the definition of whatever it is.

I know that every slice of p^(-1)(U) is connected because they are all homeomorphic to U by restricting p.

 

[ircdan]

since U is evenly covered by p, p^-1( U) = union( V_k) where V_k are mutually disjoint open subsets of V, the V_k are the "slices" and {V_k} is a partition of p^-1( U) into slices

 

[JasonRox]

since U is evenly covered by p, p^-1( U) = union( V_k) where V_k are mutually disjoint open subsets of V, the V_k are the "slices" and {V_k} is a partition of p^-1( U) into slices

Yeah, I already know that. I probably just overcomplicated the question.

I don't see how it's unique because you can partition p^(-1)(U) in other ways. That is by making every point in p^(-1)(U) a set and that union of sets is p^(-1)(U) and all the intersections are disjoint.

 

[ircdan]

Yeah, I already know that. I probably just overcomplicated the question.

I don't see how it's unique because you can partition p^(-1)(U) in other ways. That is by making every point in p^(-1)(U) a set and that union of sets is p^(-1)(U) and all the intersections are disjoint.

those sets you create won't necessarily be slices, you need to show the partition of p^(-1)(U) into slices is unique, not just any partition

just go back to the definition
Let p:E->B be continuous. U subset B is evenly covered by p if p^-1(U) is a union of disjoint open sets in E, call them {V_j}, s.t. for each j the restriction of p to V_j is a homeomorphism of V_j onto U. In this case, the V_j's are called slices.

 

[Nolen Ryba]

Once you see why this is true you're going to kick yourself because it is quite simple. Here's the definition of partition into slices:

partition into slices = partition into disjoint open subsets such that the restriction of p to each open subset is a homeomorphism

 

[mathwonk]

if you define the terms propwrly the reuslt is trivial. i.e. the slices are just conncetde components of the inverse image.

evenlycovered emans covering space.

 

[JasonRox]

Once you see why this is true you're going to kick yourself because it is quite simple. Here's the definition of partition into slices:

partition into slices = partition into disjoint open subsets such that the restriction of p to each open subset is a homeomorphism

Oh boy, I know the feeling now.

I'll keep posting my Algebraic Topology questions here I guess.