# Surface covered by hyperbolic plane admits a hyperbolic metric

Let $$S$$ be a surface. Is it true that if $$S$$ is covered by the hyperbolic plane (or a subset thereof) then it admits a Riemannian metric of constant negative curvature? How does the metric (or multiple metrics) arise?

Is the converse true? Again, how do we construct the covering(s)?

Relevant references would be much appreciated.

The statement is not true. The universal covering of $$T^2$$ is homeomorphic to $$\mathbb R^2$$, and $$\mathbb R^2$$ is homeomorphic to $$\mathbb H^2$$, so there exists a universal covering map $$\mathbb H^2 \to T^2$$. But $$T^2$$ does not have a hyperbolic metric, by the Gauss-Bonnet theorem.
The correct statement is that if there exists a covering map $$f : \mathbb H^2 \to S$$ such that the deck transformation action of $$\pi_1(S)$$ on $$\mathbb H^2$$ is an action by isometries of $$\mathbb H^2$$ then $$S$$ admits a Riemannian metric of constant negative curvature.
The proof is to take any open subset $$U \subset S$$ which is evenly covered by $$f$$, choose an open subset $$\tilde U \subset \mathbb H^2$$ such that $$f$$ maps $$\tilde U$$ to $$U$$ by a homeomorphism, and define the Riemannian metric on $$U$$ as the pushforward via the map $$f$$ of the Riemannian metric on $$\tilde U$$. The hypothesis that the deck action is by isometries is needed here to show that the Riemannian metric on $$U$$ is well-defined independent of the choice of $$\tilde U$$.
The converse is only true with an additional hypothesis as well, namely that the Riemannian metric on $$S$$ is geodesically complete. The proof takes some work, but you can find it carefully written up in textbooks on hyperbolic geometry.