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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)

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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.

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  • $\begingroup$ Can I also prove it as follow? Let be $U$ connected, so every $U_i$ is connected as well and also $\phi (U_i)$ for every $\phi \in \mathrm{Deck}(p)$. But $\phi (U_i) \subset p^{-1}(U) = \bigsqcup U_j$ so $\exists j: \phi (U_i) \subset U_j$ for connectedness. In the same way $\exists k: \phi^{-1}(U_j) \subset U_k$. We necessarily have $k = i$ so $U_j \subset \phi(U_i)$. So $\phi(U_i) = U_j$. Is this correct? $\endgroup$ – Marco All-in Nervo Feb 16 at 11:03

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