Why are the sheets the path connected components?

Question

In Bredon's Topology and Geometry we find the following definition for covering spaces and maps:

A map $p: X \to Y$ is called a covering map (and $X$ is called a covering space of $Y$) if $X$ and $Y$ are Hausdorff, arcwise connected, and locally arcwise connected, and if each point $y \in Y$ has an arcwise connected neighborhood $U$ such that $p^{-1}(U)$ is a nonempty disjoint union of sets $U_\alpha$ (which are the arc components of $p^{-1}(U)$), on which $p\mid_{U_\alpha}$ is a homeomorpism $U_\alpha \to U$. Such sets $U$ will be called elementary, or evenly covered.

Why does it follow in the above setting that the sheets are actually the arcwise connected components of $p^{-1}(U)$?

Answer 1

Since $p^{-1}(U) - U_\alpha$ is a union of open subsets of $p^{-1}(U)$, it is an open subset of $p^{-1}(U)$. Also, $U_\alpha$ is an open subset of $p^{-1}(U)$. It follows that $U_\alpha$ is a disjoint union of connected components of $p^{-1}(Y)$, and therefore also a union of arcwise connected components.

However, $U_\alpha$ is homeomorphic to $U$ and is therefore arcwise connected. It follows that $U_\alpha$ is an arcwise connected component.

Answer 2

That $p^{-1}(U)$ is a nonempty disjoint union of sets $U_α$ means that there is a nonempty index set $A$ such that

1. $p^{-1}(U) = \bigcup_{\alpha \in A} U_\alpha$
2. $U_\alpha \cap U_\beta = \emptyset$ for $\alpha \ne \beta$
3. A subset $S \subset p^{-1}(U)$ is open in $p^{-1}(U)$ if and only if $S \cap U_\alpha$ is open in $U_\alpha$ for all $\alpha \in A$. Here the spaces $p^{-1}(U)$ and $U_\alpha$ carry of course the relative topology inherited from $X$.

By 3. and 2. all sheets $U_\alpha$ over $U$ are open in $p^{-1}(U)$, and therefore also open in $X$ because $p^{-1}(U)$ is an open subset of $X$.

We conclude that each nonempty connected $C \subset p^{-1}(U)$ is contained in a (by 2. unique) sheet $U_{\alpha_C}$. Since $C \ne \emptyset$, we have $C \cap U_{\alpha_C} \ne \emptyset$ for at least one $\alpha_C$. Clearly $C' = C \cap U_{\alpha_C}$ is open in $C$. But also $C'' = C \cap \bigcup_{\alpha \ne \alpha_C}U_\alpha$ is open in $C$. Since $C = C' \cup C''$, $C' \cap C'' = \emptyset$ and $C' \ne \emptyset$, the connectedness of $C$ shows that $C'' = \emptyset$. Hence $C \subset U_{\alpha_C}$.

Each sheet $U_\alpha$ over $U$ is arc connected because it is homeomorphic to $U$. Let $P$ denote the arc component of $p^{-1}(U)$ such that $U_\alpha \subset P$. Since $P$ is connected, it is contained in a unique sheet $U_{\alpha_P}$. Thus $U_\alpha \subset U_{\alpha_P}$. By 2. we see that $\alpha = \alpha_P$. Thus $U_\alpha = P$, i.e. $U_\alpha$ is an arc component of $p^{-1}(U)$. Since the sheets over $U$ partition $p^{-1}(U)$, they are in fact nothing else than the arc components of $p^{-1}(U)$.