(Covering space) Does Deck move whole sheets in whole sheets?

Question

Let $p: E \to B$ be a covering space. Let be $b \in B$, $U \in I(b)$ an evenly covered neighborhood of $b$, $U_i$ the sheets over $U$ such that $p^{-1}(U) = \coprod{U_i}$. If $\phi \in Deck(p)$, is true that for every $i$ it exists $j$ such that $\phi(U_i) = U_j$, i.e. every automorphism moves a whole sheet in another whole sheet? (maybe with some other assumptions)

Answer 1

Yes, assuming $U$ is connected, because the deck transformations permute fibers. (If you leave out connectedness, $\phi$ will permute the corresponding connected components.)

Suppose $p^{-1}(U) = \bigsqcup_{i\in I} U_i$ with each $U_i$ mapped homeomorphically to $U$ by $p$. For each $i\in I$ and $x\in U$, let $x_i$ be the preimage under $p$ of $x$ in $U_i$. Now fix $i_0 \in I$. Consider the function $f_{i_0}:U\to I$ defined by sending $x\in U$ to the unique $j_0\in I$ such that $\phi(x_{i_0}) = x_{j_0}$. This map is continuous (giving $I$ the discrete topology), so it is locally constant (and thus constant, by connectedness) on $U$, and so we know $\phi(U_{i_0}) \subset U_{j_0}$ for some $j_0\in I$, but it must then be that $\phi(U_{i_0}) = U_{j_0}$, since each contains exactly one preimage of each $x\in U$ and $\phi$ permutes these.