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.
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.