Continuity in the lifting lemma for a covering map

Question

Suppose $f: M \rightarrow N$ is a covering map between topological spaces and $g: U \rightarrow N$ be a continuous function with U simply connected. Then for a base point $z_0 \in U$ and any other $z_1 \in U$ we pick a path $\gamma$ connecting $z_0$ and $z_1$, and then define $h: U \rightarrow M$ in terms of the unique lift of the curve $g^o\gamma$. This can be done by the lifting lemma, and we know $f^oh = g$. But is this enough to guarantee continuity of h? Or do we need that N and M are connected and locally path connected?

Answer 1

I believe continuity of $h$ is automatic.

Proof: Let $u\in U$. Then $g(u)\in N$, so there is an open neighborhood of $g(u)$, $V\subseteq N$, with the property that $f^{-1}(V)$ is a disjoint union of open subsets of $M$, each of which is mapped by $f$ homeomorphically onto $V$.

Let $W\subseteq f^{-1}(V)$ denote the unique open subset of $M$ with the property that 1) $h(u)\in W$ and 2) $f|_W:W\rightarrow V$ is a homeomorphism.

Then, near $u\in U$, $h = (f|_W)^{-1}\circ g $ is a composition of continuous functions, so is continuous.

Answer 2

It is not enough. As an example take the Warsaw circle $N$ which is simply connected. It has non-trival coverings $p : M \to N$ with non-pathconnected $M$. See my answers to Classification of covering spaces for spaces that are not locally path connected: counterexamples?, especially to point 3.

Then take $U = N$ and $g = id$. There does not exist a continuous lift $h : N \to M$. Such a map would be a section of $p$, and this is possible only when $p$ is a homeomorphism. See my answer to If a covering map has a section, is it a $1$-fold cover?.