orientation independent of triangulation I have $2$-dimensional compact connected orientable manifolds... It is known that those can be triangulated in such a way that everything fits nicely: number of triangles is finite, each edge is edge of exactly two triangles etc. 
Orientation on it is given in the following way: we choose an arbitrary simplex and give orientation to it and then follow the rule that two adjacent simplexes must induce opposite orientation on their common edge... 
Suppose we have two triangulations. Is the question about orientation answered in the same way, no matter how we break the space into pieces? 
My primitive ideas were following: 
I think that if we take barycentric subdivisions on any of these... orientations remain unchanged.
What if we take enough times barycentric subdivisions of both triangulations simoultaneously and then take all possible intersections... Would we get another triangulation, i.e. a simplicial complex with diameters of simplexes much smaller? And then, if we were lucky, use the fact that subdivisions don't change  anything.
Is there an elementary way to answer this, as I don't know anything about homology?
My another question is, how could one prove that this is indeed a topologically invariant, i.e. if we know that simplicial complex is orientable then its homeomorphic space (space which we broke into pieces) is orientable. 
 A: The most revealing proof is to show that orientability of your compact, connected, triangulated surface $S$ is equivalent to the 2nd singular homology group $H_2(S;\mathbb{R})$ being isomorphic to $\mathbb{R}$. Since $H_2(S;\mathbb{R})$ is a topological invariant independent of the choice of triangulation, it follows that orientability of $S$ with respect to any two triangulations is equivalent. This kind of proof is carried out in many algebraic topology books.
The method you suggest does not work as stated, because one triangulation can be rather "wild" with respect to the other one. For example it is quite possible for the intersection of the 1-skeleta to be a Cantor set. 
Nonetheless, there is another way to concoct a proof along these lines. First one shows that triangulation #1 can be isotoped to be in general position with respect to triangulation #2. Once that is accomplished, your subdivision argument will work. Carrying this proof out in careful detail, and in particular getting all the general position arguments to work correctly, requires the Schönflies theorem.
