# Thread: Fibre cardinality

1. ## Fibre cardinality

$\displaystyle p: E \to B$ is a covering map. $\displaystyle B$ is connected. Suppose $\displaystyle b_0 \in B$ such that $\displaystyle p^{-1}(b_0)$ has cardinality $\displaystyle k$. Show that $\displaystyle p^{-1}(b)$ has cardinality $\displaystyle k,\, \forall b \in B$.

Why does the base space need to be connected? Here is what I have:

Denote by $\displaystyle card(A)$ the cardinality of $\displaystyle A$.

Let $\displaystyle b \in B$. Let $\displaystyle U_0$ be an open neighborhood of $\displaystyle b_0$ evenly covered by $\displaystyle p$. $\displaystyle p^{-1}(U_0) = \{V_i\}_{i=1}^k$ where the $\displaystyle V_i$ are disjoint. Then each $\displaystyle V_i$ contains a distinct element of $\displaystyle E$ that maps to $\displaystyle b$. That is, $\displaystyle card(p^{-1}(b)) \geq k$.

Now suppose that there is some $\displaystyle b\in B$ such that $\displaystyle card(p^{-1}(b)) > k$ (is strictly greater than $\displaystyle k$). But this implies that there are at least $\displaystyle k+1$ distinct elements in the preimage of $\displaystyle b_0$, contradicting $\displaystyle card(p^{-1}(b_0)) = k$.

That seemed like it went way too quick and way too simply relative to the standard exercises in my course. What gives?

2. Originally Posted by cribby
$\displaystyle p: E \to B$ is a covering map. $\displaystyle B$ is connected. Suppose $\displaystyle b_0 \in B$ such that $\displaystyle p^{-1}(b_0)$ has cardinality $\displaystyle k$. Show that $\displaystyle p^{-1}(b)$ has cardinality $\displaystyle k,\, \forall b \in B$.

Why does the base space need to be connected? Here is what I have:

Denote by $\displaystyle card(A)$ the cardinality of $\displaystyle A$.

Let $\displaystyle b \in B$. Let $\displaystyle U_0$ be an open neighborhood of $\displaystyle b_0$ evenly covered by $\displaystyle p$. $\displaystyle p^{-1}(U_0) = \{V_i\}_{i=1}^k$ where the $\displaystyle V_i$ are disjoint. Then each $\displaystyle V_i$ contains a distinct element of $\displaystyle E$ that maps to $\displaystyle b$. That is, $\displaystyle card(p^{-1}(b)) \geq k$.
Now suppose that there is some $\displaystyle b\in B$ such that $\displaystyle card(p^{-1}(b)) > k$ (is strictly greater than $\displaystyle k$). But this implies that there are at least $\displaystyle k+1$ distinct elements in the preimage of $\displaystyle b_0$, contradicting $\displaystyle card(p^{-1}(b_0)) = k$.
The above part is a bit awkward logically, especially "for some" and its negation in order to establish the contradiction.

My suggestion is to divide B into two subsets for the sake of the contradiction. One of them is the set consisting of $\displaystyle card(p^{-1}(b)) = k$ and the other is the set consisting of $\displaystyle card(p^{-1}(b))) \neq k$. Use the fact that a covering map is an open map and B is connected, which implies that B cannot be the union of two disjoint nonempty open sets. I'll leave it to you to establish the necessary contradiction.

Or,

You can construct a bijective map between $\displaystyle p^{-1}(b_0) \rightarrow p^{-1}(b_1)$ for an arbitrary pair of points $\displaystyle b_0$ and $\displaystyle b_1$ in B. For instance, choose a path f between $\displaystyle b_0$ and $\displaystyle b_1$ and lifts that path to g in E whose initial point is $\displaystyle y_0 \in p^{-1}(b_0)$ and terminal point is $\displaystyle y_1 \in p^{-1}(b_1)$ such that pg=f.

That seemed like it went way too quick and way too simply relative to the standard exercises in my course. What gives?

3. It looks like I ended up with something similar to your first suggestion. I essentially showed that for any two points in an evenly covered open neighborhood having k-cardinality preimage, the preimage of the two points must have cardinality k. Then used this with a set construction similar to your suggestion (sorry, I'm being code-lazy this morning), showed this set must be open and its complement must be open. The connectedness of the base space establishes that the constructed set is, in fact, the entire base space as it is nonempty by assumption.

If its not too much of a bother, however, I would appreciate a little more enlightenment on why connectedness is necessary. Perhaps a counterexample to the statement with a non-connected base space?

Thank you much!

4. Originally Posted by cribby
It looks like I ended up with something similar to your first suggestion. I essentially showed that for any two points in an evenly covered open neighborhood having k-cardinality preimage, the preimage of the two points must have cardinality k. Then used this with a set construction similar to your suggestion (sorry, I'm being code-lazy this morning), showed this set must be open and its complement must be open. The connectedness of the base space establishes that the constructed set is, in fact, the entire base space as it is nonempty by assumption.

If its not too much of a bother, however, I would appreciate a little more enlightenment on why connectedness is necessary. Perhaps a counterexample to the statement with a non-connected base space?

Thank you much!
If $\displaystyle p:E \rightarrow B$ is a covering map, then we assume conventionally E and B are locally path connected and path connected.

However, for the sake of a counterexample of your question, we can consider this covering map:

Let (X', p) be a covering space of X and (Y', q) be a covering space of Y, where p is an m-fold and q is an n-fold covering map for X and Y, m>=1, n>=1, $\displaystyle m \neq n$, respectively.
Then $\displaystyle (X' \coprod Y', f)$ is the covering space of $\displaystyle X \coprod Y$ defined by $\displaystyle f|_{X'}=p, f|_{Y'}=q$.
Now your base space is disconnected (see here) and satisfies the condition for the counterexample.