TITLE: Simplicial complexes vs. geometric realization of abstract simplicial complexes
QUESTION [7 upvotes]: A finite abstract simplicial complex is a pair $D=(S,D)$ where $S$ is a finite set and $D$ is a non-empty subset of the power set of $S$ closed under the subset operation, e.g. $(\{a,b,c\},\{\emptyset,\{a\},\{b\},\{c\},\{a,b\}\})$.
For $n\geq 0$ the topological space $\Delta^n=\{(x_0,...,x_n)\in\mathbb{R}^{n+1}\mid x_i\geq 0, \sum x_i =1\}$ is called the standard $n$-simplex. A topological space homeomorphic to the standard $n$-simplex is called an $n$-simplex. For $n\geq 1$ the $n+1$ faces of any $n$-simplex are $n-1$ simplices.
A finite topological simplicial complex is a pair $(X,F)$ where $X$ is a topological space and $F=(F_1,...,F_m)$ is a finite sequence of embeddings $F_k:\Delta^{i_k}\to X$ such that

$X=\cup_k F_k(\Delta^{i_k})$
$F_k\neq F_l$ if $k\neq l$ 
for every $1\leq k\leq m$ with $i_k\geq 1$ and for every face $A$ of the $i_k$-simplex $F_k(\Delta^{i_k})$ there is a $1\leq l\leq m$ with $F_l(\Delta^{i_l})=A$
for every $1\leq k\neq k'\leq m$ the simplex $F_k(\Delta^{i_k})\cap F_{k'}(\Delta^{i_{k'}})$ is a face of each of them.

I hope, the definitions are correct.
There is the notation of a geometric realization of a finite abstract simplicial complex: Let $D=(S,D)$ be a finite abstract simplicial complex. Then choose a total order on $S$, w.l.o.g. $S=\{1,...,M\}$. The colimit of the functor sending an element $\{0,...,n\}$ of the poset $D$ (considered as a category) to $\Delta^n$ is the geometric realization $|D|$ of $D$.
If I am not mistaken there are finite topological simplicial complexes which are not the geometric realization of a finite abstract simplicial complex. This is because the choice of the total order determines an orientation of the realization. I think the projective plane for example is not in the image of the realization functor.
My question is: Is there a reasonable notation of a geometric realization for abstract simplicial complexes which has exactly the topological simplicial complexes as its image or do I have a wrong understanding somehow?
I have realized that the original question does not make sense. Please let me ask if this is the right way to understand the situation:
A finite triangulation of a space is the same as a "finite topological simplicial complex". Every finite triangulation is the realization of a finite abstract simplicial complex. The realization of a finite abstract simplicial complex comes with a "direction" of each 1-simplex such that the neighbouring edges are pointing in the same directions (they are glued together in this way). The triangulation is orientable if and only if one can permute these "directions" of the 1-simplices such that all the neighbouring edges are pointing in opposite directions. How can I see that this condition gives the right concept of orientability? Why "opposite"?

REPLY [3 votes]: To answer just a small part of your question, three small picture-examples that can help see why "edges match up in opposite directions" is the right condition for "orientability":

Analogy with how vertices match up when you glue paths together along vertices.  Each path has one vertex oriented as "source", and another oriented "target".  To glue paths together in a consistently oriented way, we do it like this: $ \quad \cdot \longrightarrow \cdot \longrightarrow \cdot \quad $ where a "source" vertex is matched to a "target" vertex; in other words, a vertex should have opposite orientations in the edges that it joins.  (Matching them in the same orientation would give $ \quad \cdot \longrightarrow \cdot \longleftarrow \cdot \qquad $ ...ouch!)
(Working from the picture in David Sayer's example.) Each triangle in his picture has the edges in its boundary oriented clockwise, which makes them match up consistently at their vertices, in the sense of the previous point.  When we paste the triangles together along an edge, that edge is no longer part of the boundary of the resulting quadrilateral.  So looking at just the resulting boundary, we want that still to be consistently oriented, with each vertex occurring as a source of one edge and a target of another.  Having the glued edge matched in opposite directions is what will make that work...  contemplate how the vertices of the "gluing edge" occur in the remaining boundary edges, as sources/targets, and what will make sure that they still cancel out.
Now, imagine you're standing somewhere on the surface of either triangle; the notion of orientation you can see, locally, is the idea of "clockwise around you".  What should it mean to say the edges are consistent with this?  It seems reasonable to say that as you walk up to any edge of the triangle, its direction should agree with your idea of "clockwise around you".  So let's you're on the right-hand triangle and you sidle up to the "gluing edge"; it's on your left, and it points forward — that's good, that's clockwise!  Now you step over it; but now it's on your right, so to still agree with your orientation, it should point backwards, seen from this side... and it does!  Awesome!

As I see it, these (and a few other such examples) are the intuitively natural seeds from which the various abstractions and generalisations of orientation grow, and whenever I have trouble seeing why some choice is made the way it is, chasing it back to something like this usually resolves the question.
Looking outward: the ideas in 1 and 2 are the first steps down the path of combinatorial approaches to orientation, where the orientaion is on entire simplices, as seen in simplicial complexes like you sketch.   No. 3 points down a different path, to more geometric notions of local orientation at a point, which can be expressed in terms of  local homology groups, among other ways; Hatcher's excellent free book has a nice treatment of this, with really good examples/exercises, in Section 3.3 under "Orientations and Homology".