# Munkres Lemma 54.1 (Unique lifting)

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?

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).
• After some more reflection, I realize that this argument is not actually sound. It is not automatic that $\overline{f}$ is continuous. In fact, lifts are typically continuous by definition.
'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.