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

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)

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.
• 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? – Marco All-in Nervo Feb 16 at 11:03