Substitute the coordinate formulas. At the face \(\sigma _{x,y}\), \[\begin {aligned} (d_1d_0f)(\sigma _{x,y}) &=\bigl (f(x+1,y)-f(x,y)\bigr )\\ &\quad +\bigl (f(x+1,y+1)-f(x+1,y)\bigr )\\ &\quad -\bigl (f(x+1,y+1)-f(x,y+1)\bigr )\\ &\quad -\bigl (f(x,y+1)-f(x,y)\bigr ). \end {aligned}\] The 8 terms cancel in pairs, so \(d_1d_0f=0\). The pairing separates chains, and therefore the boundary of every face boundary is zero as well.

If \(d_0f=0\), then \(f(x+1,y)=f(x,y)\) and \(f(x,y+1)=f(x,y)\) for every vertex. Following the horizontal steps from \((0,0)\) and then the vertical steps gives \(f(x,y)=f(0,0)\). The kernel consists of the constant functions, so \(\dim \ker d_0=1\) and the rank of \(d_0\) is \(V-1\).

On a single cycle of length \(N\ge 1\), a real function \(g\) on the residues modulo \(N\) is the coboundary of a vertex function if and only if the sum of its values is zero. The forward difference of any vertex function sums to zero. Conversely, if that sum vanishes, set \(f(0)=0\) and let \(f(k)\) be the sum of \(g(0),\ldots ,g(k-1)\) for \(1\le k\le N-1\), with the empty sum equal to zero when \(N=1\). For \(0\le k\le N-2\) the forward difference equals \(g(k)\). The remaining difference is \(f(0)-f(N-1)\). It equals the negative of the sum of \(g(0),\ldots ,g(N-2)\), which the vanishing-sum hypothesis rewrites as \(g(N-1)\). When \(N=1\) both that difference and \(g(0)\) are zero. Thus \(g\) is a coboundary. The functions with vanishing difference are the constants, a one-dimensional space, so the coboundary on the cycle has rank \(N-1\).

Let \(\alpha \) satisfy \(d_1\alpha =0\). Write \(P^h(y)=\sum _x\alpha ^h(x,y)\) and \(P^v(x)=\sum _y\alpha ^v(x,y)\). Sum the face equation over \(x\). The vertical terms cancel because \(x+1\) runs through every residue, and the result is \(P^h(y)=P^h(y+1)\). Sum the face equation over \(y\). The horizontal terms cancel, and the result is \(P^v(x)=P^v(x+1)\). Thus both quantities are constants; call them \(P^h\) and \(P^v\). If \(\alpha =d_0f\), each is a sum of a telescoping difference, so \((P^h,P^v)=(0,0)\).

Conversely, suppose \(d_1\alpha =0\) and \((P^h,P^v)=(0,0)\). For each fixed \(y\), the horizontal row sums to zero, so the cycle criterion supplies a function \(f(\cdot ,y)\) with \(f(0,y)=0\) and horizontal difference \(\alpha ^h\). Define \(\delta (x,y)=\alpha ^v(x,y)-\bigl (f(x,y+1)-f(x,y)\bigr )\). Its horizontal difference is \[\alpha ^v(x+1,y)-\alpha ^v(x,y) -\bigl (\alpha ^h(x,y+1)-\alpha ^h(x,y)\bigr ).\] The face equation rearranges to the vanishing of that expression. Hence \(\delta (x,y)\) depends only on \(y\). The normalization \(f(0,y)=0\) gives the common value \(k(y)=\alpha ^v(0,y)\). The sum of \(k\) is \(P^v=0\), so the cycle criterion supplies \(c\) with \(c(y+1)-c(y)=k(y)\). The function \(f(x,y)+c(y)\) then has horizontal difference \(\alpha ^h\) and vertical difference \(\alpha ^v\). Every cocycle with vanishing periods is therefore a coboundary.

The identity \(d_1d_0=0\) puts the image of \(d_0\) inside \(\ker d_1\). Coboundaries have vanishing periods, and every cocycle with vanishing periods is a coboundary, so that image is exactly the set of cocycles whose periods vanish. The edge field with every horizontal value \(1\) and every vertical value \(0\) has coboundary zero on each face, because the face sees \(+1\) on its lower edge and \(-1\) on its upper edge. Its periods are \((N_1,0)\). The field with the two roles exchanged is likewise a cocycle, with periods \((0,N_2)\). For real \(a\) and \(b\), the combination with horizontal value \(a/N_1\) and vertical value \(b/N_2\) is a cocycle of periods \((a,b)\). The period map on \(\ker d_1\) is therefore a surjective linear map to \(\mathbb R^2\), with kernel equal to the image of \(d_0\). Rank nullity gives \[\dim \ker d_1=(V-1)+2=V+1,\] and the quotient dimension is \[b_1=(V+1)-(V-1)=2.\] The rank of \(d_1\) is \(2V-(V+1)=V-1\).

Summing a coboundary over every face cancels: each horizontal value appears once with each sign as \(y\) runs through a full set of residues, and each vertical value does the same as \(x\) does. The sum of the face values is therefore a linear functional vanishing on the image of \(d_1\). The constant function \(1\) has sum \(V\ne 0\), so the functional is nonzero, its kernel has dimension \(V-1\), and that constant is not a coboundary. The image already has dimension \(V-1\), so the image equals the kernel of the sum. Therefore \(b_2=1\).

On a cochain of one degree the two summands of \((d+d^\dagger )\circ (d+d^\dagger )\) are \(d(d^\dagger \gamma )\) and \(d^\dagger (d\gamma )\), because \(d\circ d\) and \(d^\dagger \circ d^\dagger \) vanish. Thus the square is \(\Delta \). Self-adjointness gives \(\langle \Delta \Phi ,\Phi \rangle =\langle (d+d^\dagger )\Phi ,(d+d^\dagger )\Phi \rangle \) for a sum \(\Phi \) of cochains of all three degrees. Expanding the degree-wise formula gives \[\langle \Delta \gamma ,\gamma \rangle =\langle d\gamma ,d\gamma \rangle +\langle d^\dagger \gamma ,d^\dagger \gamma \rangle .\] Both summands are nonnegative, so \(\Delta \gamma =0\) if and only if \(d\gamma =0\) and \(d^\dagger \gamma =0\). The same identity shows that \((d+d^\dagger )\Phi =0\) if and only if every degree of \(\Phi \) is harmonic.

On vertex functions the incoming adjoint is zero, so \(\Delta f=d^\dagger (df)\). The counting adjoint on edge fields is this \(d^\dagger \), and for the constant weight \(1\) the weighted graph Laplacian is \(d_R^\dagger d_R\). The two operators agree.

Fix a degree, and write \(d_-\) for the coboundary into that degree. Its image is contained in the kernel of the outgoing coboundary, because \(d\circ d=0\). The rank of \(d_-^\dagger \) equals the rank of \(d_-\), so the kernel of the adjoint has dimension equal to the dimension of the cochain space minus that rank. Added to the dimension of the image, the total is the whole space. If \(d_-\alpha \) lies in the kernel of \(d_-^\dagger \), then \(\langle d_-\alpha ,d_-\alpha \rangle =\langle \alpha ,d_-^\dagger d_-\alpha \rangle =0\), so \(d_-\alpha =0\). The image therefore meets that kernel only at zero, and the cochain space is their direct sum. A closed cochain splits uniquely as \(d_-\alpha +\rho \) with \(d_-^\dagger \rho =0\). The outgoing coboundary kills \(d_-\alpha \) and kills the sum, so it kills \(\rho \). Thus \(\rho \) is harmonic. The splitting of the closed cochains is direct, and the dimension of the harmonic space is the Betti number of that degree.

On vertices the kernel of \(d\) is the constants. The Betti count gives that kernel dimension \(1\), so the rank of \(d\) is one less than the number of vertices. The image of \(d^\dagger \) from edge fields is orthogonal to that kernel, because \(\langle d^\dagger \alpha ,f\rangle =\langle \alpha ,df\rangle \). The rank of this adjoint equals the rank of \(d\), so the image is the orthogonal complement of the constants. The pairing of a vertex function against the constant \(1\) is the sum of its values, and therefore the complement is the set of vertex functions whose values sum to zero.