Let $E$ and $X$ are topological spaces and $p:E \rightarrow X$ be a covering map. Whys are all the fibers homeomorphic?
Asked
Active
Viewed 891 times
2
-
In you definition of covering, is $X$ connected ? – Olórin Mar 13 '15 at 21:37
-
Let's assume it is arc-wise connected. – Babai Mar 13 '15 at 21:38
-
1A standard argument (works e.g. also for the dimension of a manifold) is that the continous assignment $x\in X \mapsto #p^{-1}(x)$ is locally constant. Hence it is constant on connected components. – Daniel Valenzuela Mar 14 '15 at 06:16
2 Answers
7
They're not necessarily; you need to assume $X$ is connected.
If $X$ is connected, consider the equivalence relation given by $x \sim y$ if $f^{-1}(x)$ is homeomorphic to $f^{-1}(y)$. (For covering maps, this is the same as the equivalence relation given by saying that $|f^{-1}(x)| = |f^{-1}(y)|$.) Show that equivalence classes are open using the definition of a covering space. Now if there is more than one equivalence class, you can use these to write $X$ as a disjoint union of nonempty open sets; contradiction, since we assumed $X$ connected.
1
if we assume $X$ is connected then basically fiber of a single point is a discrete set...and fiber of two distinct point have same cardinality ... so any bijection is actually a homeomorphism.
Anubhav Mukherjee
- 6,708
-
-
you can prove it in many ways... see this http://math.stackexchange.com/questions/1169319/about-path-connected-covering-spaces/1169354#1169354 – Anubhav Mukherjee Mar 13 '15 at 21:56