Exercise 10.7 in Bott and Tu involves calculating the Cech cohomology, valued in $\mathbb{Z}$, for a particular open cover of the circle. In particular, we have a good cover consisting of $U_0, U_1$, and $U_2$, and restriction morphisms $\rho^i_{ij}:U_i \to U_{ij}$ with $$\rho^2_{02} = -1$$ and $\rho^i_{ij} = 1$ for all other $i,j$.
See Cohomology with twisted coefficients for more details. My question is, how does one compute the boundary map ($d_0$ in the linked questions notation)? I thought you took the alternating sum of the $\rho$s evaluated appropriately, but that gives me
$$(a,b,c)\mapsto (a-b,c-b,a-c),$$ which has determinant $0$ (rather than $-2$)
Here's what I did:
$a \in U_0, b \in U_1, c \in U_2$
$d_0(a,b,c) = (\rho^1_{01}(b) - \rho^0_{01}(a), \rho^2_{02}(c) - \rho^0_{02}(a), \rho^2_{12}(c) - \rho^1_{12}(b)) = (b - a, -c - a, c - b)$
Which gives $$\begin{pmatrix} -1 & 1 & 0 \\ -1 & 0 & -1 \\ 0 & -1 & 1 \end{pmatrix} \begin{pmatrix} a \\ b \\ c \end{pmatrix} = \begin{pmatrix}b - a \\ - c - a\\ c - b \end{pmatrix}$$
This has determinant $-1(0 - 1) - 1(-1 - 0) + 0(1 - 0) = 2$