Finding chains with a given boundary I am attempting the following exercise: Given a delta complex (below) on a torus with an open disk taken out, find a chain of 2-simplices whose boundary is the boundary of the hole. (All the 2-simplices are oriented anti-clockwise)

My attempt: to get the boundaries of the disk (the inner square) we need a chain $T_2 + T_4 + T_6 + T_8 $, but the boundary of this includes the diagonals, so include the $T_1, T_3, T_5, T_7$ simplices. So now we have the chain $T_1 + T_2 + \dots + T_8$, whose boundary is definitly that of the disk, but also doesn't seem a very 'interesting' answer, as all I'm saying is that $\delta(T - D^\circ) = \delta D$.
I can't see how to efficiently find a chain with a given boundary (or what it would be in this case)? Any pointers on how to proceed with this type of problem? Thanks.
 A: Your answer is perfectly fine, and is about as "interesting" as it will get.  Indeed, the statement that $\delta(T - D^\circ) = \delta D$ (at least, up to sign) is exactly the sort of intuition you want to have here.  That intuitive geometric statement tells you that you should be looking for a chain which geometrically represents the entire space $T-D^\circ$, roughly, with the obvious choice then being just the sum of all the triangles.
In fact, your derivation actually shows this is the only chain with the desired boundary, since such a chain must have exactly one copy of each of $T_2,T_4,T_6,$ and $T_8$, and then must have exactly one of each other simplex as well to cancel out the diagonals.  Alternatively, any two chains with the desired boundary must differ by a 2-cycle.  Assuming you already know that $H_2(T-D^\circ)$ is trivial, every 2-cycle is a boundary, but there are no nontrivial boundaries, since there are no 3-simplices.  So there can only be one 2-chain with any given boundary.
