Hi, All: I am trying to show that the connected sum of orientable manifolds M,M' is orientable , i.e., can be given an orientation. I am using the perspective from Simplicial Homology. Consider the perspective of simplicial homology, for orientable manifolds M,M', glued about cycles C,C' respectively. The idea is that we can use the original orientations and then select orientations on C,C', so that they cancel each other out when glued together, and then the remaining orientations on (M-C) and (M'-C') remain the same. Still, I guess I am assumming that manifolds are simplicial complexes; I don't know if we need any additional condition like, e.g., C^1 or higher. Assume WOLG that M,M' are both connected: if an m-manifold M is orientable , this means that the top cycle --call it m'-- can be assigned a coherent orientation, so that m' is a cycle - -that does not bound, since m is the highest dimension--, i.e., the net boundary of m cancels out , e.g., in the simplest case of a loop with boundary a-a=0. This means m', which represents M itself, is a non-trivial cycle, which generates the top homology class. If your coefficient ring is Z, then the top homology will be Z; consider going n-times about the cycle. Now, the key is that the two orientable manifolds can be glued so that, at the circle of gluing, the total boundary cancels out, and the resulting manifold M#M' is still orientable. As a specific example, consider a square a,b,c,d, with arrows going all in the same direction, so that the net boundary is : (b-a)+(c-b)+(d-c)+(a-d)=(b-b)+(a-a)+(c-c)+(d-d)=0 . Now glue a second square a',b',c',d' along a common edge, (say (b,c) with (b',c')), but reverse the orientation of the edge (b',c') in M when gluing, and notice how the simplex resulting from the gluing also has net boundary zero. Now, the key general point is that , at the cycle C where we collapse M with M', we change the orientation of C in either M or M', so that, along the common cycle, where you are doing the gluing, the respective boundaries cancel each other out, and the remaining orientations of M-C and M-C' remain the same, so that M#M' is orientable. Does this Work?