Orientation of closed combinatorical surfaces I have some problems with the equivalence of definitions of orientation. I know two definitions of orientation, namely:


*

*A surface is orientable if it contains no 1-sided curves (a 1-sided curve in a combinatorical surface $F$ is a curve such if $p^{-1}(C)$ has just one component with $p:F'\rightarrow F$ and $F'$ the surface we obtain by cutting $F$ along $C$)


Also there exists the following lemma: A one-sided curve is always non-seperating (non-seperating means that the number of components of $F'$ has the same as $F$). Can we use this fact? And why is that true? From my point of view one can prove this easy with contradiction,true?!
Now the second definition:


*

*An orientation of a closed combinatorial surface F is an assignment of a clockwise or anticlockwise \circulation" to each face (really an ordering of its vertices, considered up to cyclic permutation), such that at any edge, the circulations coming from the two incident faces are in opposition. 


Now i want to show directly that a surface has an orientation if and only if it is orientable in the sense that it contains no 1-sided curves.
Can someone help me with this question? Thank you!
 A: Assume $F$ is oriented in the sense of the second definition.
Let $C$ be a curve.
Wlog.(!) the curve runs along edges.
The map $p^{-1}(C)\to\{-1,+1\}$ that maps a point to $+1$ if the "current" edge at that point is in forward direction of $C$ and to $-1$ if it is in reverse, is well defined (must be checked at vertices) and continuous. 
As each edge belongs to two faces and is differently oriented for these, ther exist both edges yielding $+1$ and edges yielding $-1$, hence $p^{-1}(C)$ has (at least) two connected components. Hence no one-sided curve exists.
The other direction is a bit more complicated if one wants to be formally exact, so the following is rather a sketch of an idea:
Assume $F$ is oriented in the sense of the first definition.
Select an arbitrary face and orient its edges.
Extend this face by face to the whole surface by checking faces having at least one edge already oriented.
If ever a conflict arises (i.e. the next step would orient an edge differently from an already given orientation by its other face), we find a onesided curve as follows: There is a sequence of faces between the conflicting faces that runs within alrady oriented faces. 
This also determines a sequence of edges (those that are shared by subsequent faces in this sequence). Join the midpoints of these edges to obtain a closed curve (i.e. starting and ending at the conflicting edge). Using the orientation given to the edges crossed along its way, we get a notion of "left" and "right" of the curve, which gets "flipped" by completely traversing the curve, thus showing that $p^{-1}(C)$ is connected.
