Here is a theorem in Hatcher's algebraic topology.

(Hatcher-Algebraic Topology p.61)

Let $(X,x_0),(C,c_0)$ be topological spaces and $p:(C,c_0)\rightarrow (X,x_0)$ be a covering map.

If $C$ and $X$ are path-conneted, then $|\pi_1(X,x_0):p_*(\pi_1(C,c_0))|$ is the number of sheets of $p$.

So far, I have proven below statement

(Munkres-Topology p.346)

If $C$ is path-connected then $|\pi_1(X,x_0):p_*(\pi_1(C,c_0))|=|p^{-1}(x_0)|$.

I haven't used path-connectivity of $X$ yet. How do I prove Hatcher's theorem via the statement I have proved?


Prove that for any $x_0 \in X$, the set $\{ x \in X : |p^{-1}(x)| = |p^{-1}(x_0)| \}$ is clopen in $X$. It follows that if $X$ is connected, this is the whole set.

Now it's a problem of definitions: the number of sheets of $p$ is defined to be the cardinality $\kappa$ such that $|p^{-1}(x)| = \kappa$ for every $x$. It doesn't always exist: sometimes the cardinalities are not the same everywhere.

By the way I think you missed this part of p.61 of Hatcher's book:

If $p : \tilde X \to X$ is a covering space, then the cardinality of the set $p^{-1}(x)$ is locally constant over $X$. Hence if $X$ is connected, this cardinality is constant as $x$ ranges over all of $X$. It is called the number of sheets of the covering.

  • $\begingroup$ Why the range is $\mathbb{N}$? Why $|p^{-1}|$ is finite? $\endgroup$ – Rubertos Nov 26 '14 at 8:49
  • $\begingroup$ @Rubertos Ah, right. Replace $\mathbb{N}$ by $\mathbb{N} \cup \{\infty\}$ in this case (with the discrete topology). Or if you don't want to deal with that, take some $x_0 \in X$, let $k_0 = |p^{-1}(x_0)|$, and prove that $\{ x \in X : |p^{-1}(x)| = k_0 \}$ is clopen. $\endgroup$ – Najib Idrissi Nov 26 '14 at 8:51
  • $\begingroup$ As a subspace of the extended real, the subspace topology on $\mathbb{N}\cup\{\infty\}$ is not the discrete topology. Do you mean to equip $\mathbb{N}\cup\{\infty\}$ the discrete topology independently? $\endgroup$ – Rubertos Nov 26 '14 at 8:57
  • $\begingroup$ Yes, I mean to equip $\mathbb{N} \cup \{\infty\}$ with the discrete topology independently. But really, what you want to prove is that $x \mapsto |p^{-1}(x)|$ is constant on connected components, that's just it -- the range doesn't matter much (in fact I don't think there's a reasonable "range" to consider: $|p^{-1}(x)|$ could be any infinite cardinality, I'm just collapsing everything as "$\infty$"). $\endgroup$ – Najib Idrissi Nov 26 '14 at 8:57
  • $\begingroup$ I just proved that $X\rightarrow \mathbb{N}\cup\{\infty\}$ is continuous and as you said, this shows that $f(X)$ is a singleton. If $f(X)=\{n\}$ then it is fine, but when $f(X)=\{\infty\}$, then we cannot compare cardinalities of $p^{-1}(x)$ and $p^{-1}(y)$ $\endgroup$ – Rubertos Nov 26 '14 at 9:12

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