Munkres Lemma 54.1 (Unique lifting)

Question

For the uniqueness part of the proof, I am having some trouble understanding why the set $\overline{\overline{f}}([s_{i},s_{i+1}])$ has to be connected in $E$. It is clear $\overline{f}([s_{i}, s_{i+1}])$ is connected in $E$, since it is a continuous image of an interval, but there is no reason why $\overline{\overline{f}}$ has to be continuous on the interval $[s_{i}, s_{i+1}]$. Are liftings automatically assumed to be continuous?

Answer 1

Let's prove lifts are continuous. Let $\overline{f}:[0, 1] \rightarrow E$ be a lift of a path $f:[0, 1]\rightarrow B$. Let $s\in [0, 1]$ and let $U \subseteq E$ be an open neighbourhood of $\overline{f}(s)$ such that $p|_U : U \rightarrow B$ is a homeomorphism. Let $V=f^{-1}(p(U))$ such that $\overline{f}|_V = (p_U)^{-1} \circ f$. Note that $\overline{f}|_V$ is continuous as the composition of continuous maps. Since $V$ is an open neighbourhood of $s$, this implies $\overline{f}$ is continuous at $s$ (a map of topological spaces is continuous at a point iff its restriction to some open neighbourhood of that point is continuous).

This really works for any lift, not just a lift of a path.

Answer 2

No, we do not assume liftings to be continuous. But the statement of Lemma 54.1 states

'a unique lifting to a path $\overline{f}$ in $E$ beginning at $e_0$'. Paths are continuous, so $\overline{\overline{f}}$ is assumed to be in the second half of the proof.