Hi, everyone: I am trying to understand the geometric interpretation of two simplicial cycles being homologous to each other. Let C_k(X) be the k-th chain group in the simplicial complex X, and let c_k be a chain in C_k(X) The algebraic definition is clear: two k-cycles x=c_k and y=c_k' are homologous, i.e., x~y , iff (def.) x-y is a boundary, i.e., if there is a cycle c_(k+1) in C_(k+1)(X) with del(c_(k+1))= c_k-c_k' . Still: how about geometrically: is there a nice geometric way of telling that two cycles are homologous.?. I am having trouble translating the subtraction of cycles into a geometric situation; it would seem like we could translate the expression of c_k-c_k' is a boundary into saying that the curves c_k and -c_k' (i.e., c_k with reversed orientation) are cobordant, in that there is a surface embedded in X-- the ambient complex--that is bounded by c_k and -c_k' . Is this correct.? Thanks.