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Let $p: Y \rightarrow X$ be a (surjective) covering map and let $G$ be the deck group of $p$. Let $U$ open in $X$ be such that the disjoint open sets $V, W \subseteq Y$ are both homeomorphic to $U$ through $p$. Then $(p \rvert_V)^{-1} \circ (p \rvert_W) : W \rightarrow V$ is a homeomorphism which also preserves fibers.

Are there any conditions on the spaces which guarantee that this homeomorphism is the restriction of some deck transformation? One obvious condition would be that $G$ has to act transitively on fibers if we want this to happen for any $V$ and $W$.

But is this enough? If not, does this at least happen if the spaces have "good" properties, for example local path connectedness, or are even topological manifolds?

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It is not even true for "nice" spaces. As an example consider $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$.

Take $V = (0,1) \cup (1,2)$ and $W = (1,2) \cup (2,3)$. Both are mapped by $p$ homeomorphically onto $U = S^1 \setminus \{-1, 1\}$. Then $p_{W,V} = (p \rvert_V)^{-1} \circ (p \rvert_W) : W \rightarrow V$ is the identity on $(1,2)$, but on $(2,3)$ it has the form $p_{W,V}(t) = t - 2$. Therefore it is not the restriction of some deck transformation. Note that in this example $U$ is evenly covered.

The same phenomenon will occur for any covering $p : Y \to X$ with more than one sheet if $U$ is evenly covered, but not connected. See Covering projections: What are the sheets over an evenly covered set? for a discussion.

Thus the minimal requirement will be that $U$ is a connected evenly covered open set. In that case the set $p^{-1}(U)$ decomposes uniquely into sheets which are mapped by $p$ homeomorphically onto $U$. These sheets are the components of $p^{-1}(U)$. Thus for any two sheets $V, W$ there exists exactly one homeomorphism $\phi : W \to V$ such that $p(\phi(x)) = p(x)$. We have $\phi = p_{W,V}$.

Then if $G$ acts transitively on fibers, it is true. To see that, pick $y \in V$ and let $y' = p_{W,V} (y)$. There exists a deck transformation $h$ such that $h(y) = y'$. Clearly $h$ maps each component of $p^{-1}(U)$ homeomorphically onto some component of $p^{-1}(U)$. Since $V, W$ are such components and $h(V) = W$, we are done.

That $G$ acts transitively on fibers can be assured by suitable assumptions on $X,Y, p$.

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