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Let $\Pi$ be a hyperbolic polygon such that $S_{\Pi}$ is a hyperbolic surface (see below) and let $\Gamma$ be the group generated by the edge pairing isometries of $\Pi$. Consider the orbit space $\mathbb{H}^2/\Gamma$ with the quotient metric defined as for $S_{\Pi}$ (see Wikipedia).

Is $S_{\Pi}=\mathbb{H}^2/\Gamma$ true only for complete $S_{\Pi}$? Or is it true for all $S_{\Pi}$ but if $S_{\Pi}$ is not complete then $\Gamma$ is not fixed point free and/or not discontinuous?

Extract from "Geometry of Surfaces" by Stillwell:

Definition: A hyperbolic polygon $\Pi$ is a region of $\mathbb{H}^2$ bounded by a simple polygonal path of finitely many $\mathbb{H}^2$-line segments and segments of $\partial \mathbb{H}^2$, called the proper and improper edges of $\Pi$, respectively. The endpoints of edges are called vertices of $\Pi$, with those in $\mathbb{H}^2$ being called proper. An edge pairing of $\Pi$ is a partition of the proper edges into pairs $\{e,e'\}$ of equal length together with an $\mathbb{H}^2$-isometry $g_{e,e'}:e \rightarrow e'$ for each pair. Points $w\in e$ and $g_{e,e'}(w)=w'\in e'$ are said to be indentified by the edge pairing. We also say that $w$ is identified with $w''$ if $w$ is identified with $w'$ and $w'$ is identified with $w''$. Such a chain of identifications can occur with vertices and we call a maximal set $\{v_1,...,v_k\}$ of identified vertices a vertex cycle. An edge pairing of $\Pi$ defines an identification space $S_{\Pi}$.

Theorem: $S_{\Pi}$ is a hyperbolic surface when the angles of each vertex cycle sum to $2\pi$.

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    You have mis-understood and/or mis-copied the theorem. There is a missing and very crucial hypothesis: the angle sum of every vertex cycle must equal $2\pi$. – 2017-01-18
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    You're right! I forgot to write that down. This is essential for the proof ! – 2017-01-18

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What's true is that $S_\Pi$ is isometric to $\mathbb{H}^2/\Gamma$ if and only if $S_\Pi$ is complete. Stillwell makes that point in the remark at the end of Section 5.5. The "only if" direction is easy, because of the evident completeness of $\mathbb{H}^2/\Gamma$. The "if" direction is harder and is carried out in more advanced texts.

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    Thank you for the answer. Then I guess that the idea of constructing a non-complete surface $S_{\Pi}$ as an orbit space of a $\Gamma\subset Iso(\mathbb{H}^2)$ that is not discontinuous and/or fixed point free, doesn't make any sense? – 2017-01-18
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    The orbit space certainly makes sense as a topological space, using the quotient topology on the decomposition of $\mathbb{H}^2$ into orbits. The trouble is that it makes no sense to think of this quotient space as a surface unless some very strong additional hypothesis is assumed, such as that $\Gamma$ is properly discontinuous and fixed point free. – 2017-01-19