# When do the local homeomorphisms given by a covering map extend to deck transformations?

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?

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 $$X,Y, p$$.