Lifting of a path in covering spaces. Does the lifted path lies in single sheet? Including the proof outline for those that can't access Munkres or don't want to look.
On page 342 Munkres proves a path $f:[0,1]\rightarrow B$ starting at $b_o = p(e_0)$ has a unique lifting to a path $\tilde{f}$ in $E$ starting at $e_0$ by taking a cover of B consisting of evenly covered (by p) open sets U of B. He pulls them back to $[0,1]$, subdivides $[0,1]$ into intervals $[s_i,s_{i+1}]$ using Lebesgue number lemma to ensure that for every such subinterval, it's image $F([s_i,s_{i+1}])$ is contained in some $U\subset B$. To define $\tilde{f}$ he defines $\tilde{f}(0)= e_0$ first. Then,  $f([s_i,s_{i+1}])\subset U$ and we have $p^{-1}(U)$ is partitioned by disjoint $\{V_\alpha\}$. So  $\tilde{f}(s_i)$ is in one sheet call it $V_0$. Then for all $s$ in that interval he defines $\tilde{f}(s): = (p|_{V_0})^{-1}(f(s))$. He says continue in this manner. Since each subinterval $[s_j,s_{j+1}]\subset [0,1]$ is connected and $\tilde{f}$ is continuous on the subinterval (b/c p|V_{0} a homeomorphism onto U) we know the path lifting upstairs $\tilde{f}([s_j,s_{j+1}])$ lies in one sheet. But doesn't [0,1] itself being connected imply the entire lifting is contained in one sheet? Or do we not get that because we defined $\tilde{f}$ in pieces (gluing lemma)? I know there's another post about this but I don't think the top answer addressed the poster's question fully. If they did, can someone explain it to me? Thanks! I also didn't have enough credit or karma or w.e. to comment.
 A: It is better to consider an example than to ask on an abstract level why the entire lifting shouldn't be contained in one sheet.
Given a path $\gamma : I \to B$ in $B$, the minimal requirement for its lift $\tilde \gamma : I \to E$ being contained in a single sheet is that $\gamma$ is contained in an evenly covered subset of $B$ - otherwise it does not make any sense to speak about sheets. Conversely, if you have a path $\gamma$ which is contained in an evenly covered open subset $U \subset B$, you know  that $\tilde \gamma$ is contained in a single sheet over $U$. That is the reason for partioning $I$ in small subintervals: Pasting together the lifts of the paths on these subintervals gives a lift of the whole path. If you do that with a  closed path in $B$, there is no reason to expect that its lift is a closed path in $E$ which would be the case if the lift is contained in a single sheet.
Now see my answer to Lifting of a path in covering spaces. Does the lifted path lies in single slice? Take the standard covering $p : \mathbb R \to S^1, p(t) = e^{2\pi it}$, and the (closed) path $\gamma(t)= e^{2\pi it}$ in $S^1$ which wraps once around $S^1$. This is not contained in any evenly covered subset of $S^1$, thus it cannot have a lift contained in a single sheet. And, by the way, the lift is no closed path, it has the form $\tilde \gamma(t) = t$. This fact alone shows that that $\tilde \gamma$ cannot be contained in a sheet.
Edited:
I guess your doubts come from Munkres' construction. Let us introduce the following ad-hoc notation: Given a path $\gamma$ in $B$, a lifting structure for $\gamma$ consists of

*

*a partition of $I = [0,1]$ in subintervals $J_i = [s_{i-1},s_i]$, $i=1,\ldots,n$, with $s_0 =0$ and $s_n =1$


*a family of evenly covered open $U_i \subset B$, $i=1,\ldots,n$, such that $\gamma(J_i) \subset U_i$.
Based on any such lifting structure, Munkres constructs a lift of $\gamma$ by  subsequently lifting $\gamma_i = \gamma \mid_{J_i}$ to paths $\tilde \gamma_i$ in $E$ such that $\tilde \gamma_1(0) = e_0$ and $\tilde \gamma_{i+1}(s_i) = \tilde \gamma_i(s_i)$ for $i=1,\ldots,n-1$.  It seems that you want to know if it is possible that the total lift $\tilde \gamma$ is contained in a single sheet of $p$. Yes, it is possible, but not in general as my above example shows.
Once we have proved the existence of a lift, it is also clear that $\gamma$ has a unique lift starting at $e_0$. Thus the result of Munkres' lifting construction does not depend on the choice of a lifting structure. Now there is a lifting structure with a mininal $n = n_{min}$. It may happen that $n_{min}=1$, and then $\tilde \gamma$ is trivially contained in a single sheet of $p$. But if $n_{min} > 1$ which means that $\gamma$ is not contained in any single evenly covered open $U \subset B$, the lift $\tilde \gamma$ can never be contained in a single sheet of $p$ simply because any sheet projects to an evenly covered set in $B$.
Therefore, even if we start with a lifting structure with $n > 1$, it may be possible that $n_{min}=1$ in which case the lift is contained in a single sheet. In other words, if $n > 1$, we may simply have overlooked that there is another lifting structure with $n = 1$.
A: You cannot say that the whole path lifts to to one sheet because you do not know that this path lies in one of the open sets $U$ that you used to cover $B$. Remember from the definition of the covering space $p^{-1}(U)$ is the union of disjoint open sets for sufficiently small open sets $U$. The sets that you use to cover $B$ must have this property in order for the argument to work.
So the reason that $\tilde{f}([s_i,s_{i+1}]) $ lies in one sheet is exactly because $p^{-1}(U)$ is a disjoint union of open sets.
A: Yes, if I have a lifting of a path to the covering space, I should be able to construct a lifting of an open neighborhood of the path to the covering space ("a single sheet covering" of that open neighborhood). But the argument in Munkres is supposed to work starting from a general open cover, where I might have to use 2 incompatible lifts on open sets that overlap, in order to get my lifting of the path.
