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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?

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  • $\begingroup$ Don't you mean $h\colon U\to M$? $\endgroup$ – Daniel Bernoulli Dec 5 '16 at 6:44
  • $\begingroup$ Yes, indeed. Thanks you $\endgroup$ – user394438 Dec 6 '16 at 2:45
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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.

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