Understanding orientation of simplicial complex I am reading a basic course in homology theory from book of paul alexandroff.How can one prove the following:"a closed surface is orientable if and only if one can orient triangles of any of it's triangulations in such a way that the oriented complex thus arising has boundary zero".
 A: Ah...a consistent choice of surface normals. Well, Alexandroff's book is talking about generalized triangulated surfaces, which may not be embedded (or immersed) in 3-space. But for those that are, I think I can make a pretty decent argument. Let me assume that the surface is at least somewhat smooth, like a soccer ball that's been divided up into pentagons and hexagons: the polygons aren't quite "flat", but the underlying surface is smooth enough to have a normal vector at every point. 
So suppose you have such a smooth surface in 3-space, and you have a triangulation of the surface by slightly curved triangles. I'm going to assume that the triangles are "small", in the sense that if you drew the normal lines to all points of the surface within the triangle, they'd all point almost in the same direction (e.g., the angle between any two would be no more than, say, 30 degrees). If that's not the case, you can subdivide repeatedly until it is. 
Suppose that you have such a triangulated surface, and the triangles of the surface are oriented  by listing the triangle vertices in a particular order, like $(A, B, C)$ (the letters indicate vertices). Then you can compute $v = (B - A) \times (C - A)$ (that's the cross product of vectors) to get a vector that's more or less "normal" to the triangle. If you look at any normal line at a point of the triangle, you can choose one of two directions; $v$ will be close to one of these and far from the other. Pick the one that $v$ is close to. So given an ordering of the vertices of a triangle on a smooth surface, I've shown how to pick a normal vector to the surface at each point of that triangle. Applying this to each triangle of your triangulation, you get a normal vector at every point of your surface. 
There's one question that remains, though: do these choices of normal vectors "agree" with each other at adjacent triangles, or might they "flip" as you cross an edge? Let's look at two adjacent small triangles, 
and let's assume (by further subdivision if necessary) that all normal lines to BOTH triangles lies within 30 degrees of each other. 
TO make things easier, let's suppose that the two triangles are ABC and BCD, and put down a coordinate plane $P$ whose origin is at B, whose $y = 1$ point is at $C$, and for which $A$ is in the left half-plane ($x < 0$). Then because the normal vectors for ABC and BCD are within 30 degrees of each other, it must be the case that the orthogonal projection $D'$  of $D$ onto the plane $P$ lies in the right half-plane. 
We can extend the coordinates on $P$ to coordinates on all of space, by drawing the $z$-axis in a right-handed way. When we do, we see that the vector $v = (B - A) \times (C - A)$ points in the positive $z$ direction. 
Recall that $ABC$ was oriented as $(A, B, C)$, so its boundary consists of oriented edges $(B, A), (C, B), (A, C)$. Since ABD shares the edge $AB$ with this triangle, the orientation of $ABD$ must be $(A, D, B)$, so that its boundary is $(A, B), (D, A), (B, D)$, and the edges $(A,B)$ (from $ABC$) and $(B,A)$ (form $ABD$) cancel. 
The normal vector for $ABD$ is then computed as $w = (D-A) \times (B - A) = - (B-A) \times (D - A)$. 
Now $(D - A) \approx (D' - A)$, by the assumption of "normal lines not varying too much", so all we need to check is that $-(B-A) \times (D' - A)$ points in the right direction (i.e., along the positive-z axis in our coordinate system). But that's straightforward if you just draw a picture. 
