Lifting of a path in covering spaces. Does the lifted path lies in single slice? Let $p: E \to B$ be a covering map and $f: [0,1]  \to B$ is a path in $B$ such that $f(0)=b_{0}$. Let $e_{0}$ be a point in $E$ such that $p(e_{0})=b_{0}$. Then there is a unique lifting of $f$ to a path $\tilde{f}$ in $E$ beginning at $e_{0}$. This result is proved in Topology by Munkres. Proof use the idea of partitioning $[0,1]$ into sub intervals using Lebesgue number lemma. Image of each interval $[s_{i},s_{i+1}]$ under $f$ is contained in an open set $U$ which is evenly covered by $p$. Lifting is done by step by step. It is almost similar to induction. $b_{0}$ belongs to an open subset of $B$ which is evenly covered. We define $\tilde{f}(0)=e_{0}$. Clearly this $e_{0}$ lies in a single slice( Slice is another name for sheet)
Assume $\tilde{f}$ is defined in $[0,s_{i}]$. Then we will apply the concept that $ f( [s_{i},s_{i+1}]) $ is contained in $U$ and let $\{V_{\alpha}\}$ be the partition of $ p^{-1}(U)$ into slices. $f(s_{i})$ belongs to one slice and let it be $V_{0}$. Since intervals are connected and $\tilde{f}$ is continuous, image of $[s_{i},s_{i+1}]$ lies in $V_{0}$.
Does this $V_{0}$ is the same slice where $e_{0}$ lies ? Does connectivity of  $[0,s_{i}]$ says this ? But this need not happen as our partition only ensures that image of each sub interval under $f$ lies in one $U$. That $U$ may vary according to interval. 
But if we try to move from first sub interval $[0,s_{1}]$ to second sub interval $[s_{1},s_{2}]$ and so on. Then image of $0$ and image of $s_{1}$ under $\tilde{f}$ lies in same slice  and hence the complete path lies in same slice. 
This affects the lifting of loops at $b_{0}$ as lifting will give a loop again at $e_{0}$. Where did I went wrong in the second argument ?
 A: You do not explain what you mean by a "slice", but I guess it is that what is usually called a sheet. That is, if $U \subset B$ is evenly covered, then $p^{-1}(U)$ is the disjoint union of open subsets $V_\alpha \subset E$ which are mapped by $p$ homeomorphically onto $U$. These $V_\alpha$ are sheets over $U$.
But in general evenly covered sets cannot be arbitarily large. As an example take $exp : \mathbb R \to S^1, exp(t) = e^{it}$. We have $b_0 = 1$. Then any open $U \subsetneqq S^1$ is evenly covered, but $S^1$ is not evenly covered. Now consider the loop $\phi : [0,1] \to S^1, \phi(t) = exp(2\pi t)$. Then you need at least two evenly covered open subsets of $S^1$ to get a partitioning of $[0,1]$ into subintervals which are mapped by $\phi$ into one of these open sets. Let us work with $U_1  = S^1 \setminus \{ -i\}$ and $U_1  = S^1 \setminus \{ i\}$. The sheets over $U_1$ are the intervals $(-\pi/2 +2k\pi, 3\pi/2 + 2k \pi)$ and the sheets over $U_2$ are the intervals $(\pi/2 +2k\pi, 5\pi/2 + 2k \pi)$. You may partition $[0,1]$ into the intervals $[0,1/2]$ and $[1/2,1]$. Each of them is mapped by $\phi$ into one of the $U_i$. If you construct a lift with $e_0 = 0$, then you will see what happens with your construction: You do not get a closed path. The reason is that each sheet over $U_i$ intersects two sheets over $U_j$ when $i \ne j$. In fact, $V_1 = (-\pi/2, 3\pi/2)$ is the sheet over $U_1$ containing $0$, and $1/2$ is lifted to $\pi$. But then $V_2 =  (\pi/2, 5\pi/2)$ is the sheet over $U_2$ containing $\pi$, and $1$ lifts to $2\pi$ which is contained in $V'_1 = (3\pi/2, 7\pi/2)$. You see that $V_2 \cap V_1 \ne \emptyset$ and $V_2 \cap V'_1 \ne \emptyset$.
In fact, in your proof you construct a sequence of sheets $W_i$ over evenly covered $U_i$ such that $f([s_i,s_{i+1}]) \subset U_i$. Then $W_i \cap W_{i+1} \ne \emptyset$ but there is a no guarantee that $W_0 \cap W_n \ne \emptyset$.
