Cohomology with twisted coefficients

I'm studying Cech Cohomology on Differential Forms in Algebraic Topology - Bott, Tu. In order to understand more about this topic I'm doing some exercises.

Exercise 10.7 (Cohomology with twisted coefficients). Let $\mathscr{F}$ be the presheaf on $S^1$ which associates to every open set the group $\mathbb{Z}$. We define the restriction homomorphism on the good cover $\mathfrak{U} = \{U_0, U_1, U_2\}$ (Figure 10.1) by: \begin{align} \rho^0_{01} & = \rho^1_{01} = 1,\\ \rho^1_{12} & = \rho^2_{12} = 1,\\ \rho^2_{02} & = -1,\; \rho^0_{02} = 1, \end{align} where $\rho^i_{ij}$ is the restriction from $U_i$ to $U_i \cap U_j$. Compute $H^*(\mathcal{U}, \mathcal{F})$. (Cf. presheaf on an open cover, p. 142.)

In the exercise above, we have $d^0(a,b,c)=(b-a,-c-a,c-b)$, therefore $\ker d^0=0$ and then $H^0(\mathcal{U,\mathcal{F}})=0$. Can some one help me to find $H^1(\mathcal{U,\mathcal{F}})$?

Note: In this question user171326 says that $H^1(\mathcal{U,\mathcal{F}})=0$ but if my computation of $d^0$ is right, then $\mathsf{Im} \, d_0\neq C^1(\mathcal{U},\mathcal{F})$ ($C^2(\mathcal{U},\mathcal{F})=0$, since $U_{012}=\emptyset$), therefore the result claimed in the question above is wrong.

• I think the link to the question doesn't work. – user90189 Dec 4 '17 at 2:25
• @user90189 yes you are right! Now I'm using my phone and I can't use my pc now. However I will provide to fix the link. – Vincenzo Zaccaro Dec 4 '17 at 9:01

Writing (for clarity) $$C^0(\mathcal{U},\mathcal{F})=\mathcal{F}(U_0)\times \mathcal{F}(U_1)\times \mathcal{F}(U_2)=\mathbb{Z}^3$$ and $$C^1(\mathcal{U},\mathcal{F})=\mathcal{F}(U_0\cap U_1)\times\mathcal{F}(U_1\cap U_2)\times \mathcal{F}(U_0\cap U_2)=\mathbb{Z}^3,$$ the boundary map $d_0$ is, as you wrote, given by $d_0(a,b,c)=(b-a,c-b,-a-c)$, so we have the matrix $$d_0=\begin{bmatrix}-1&0&-1\\1&-1&0\\0&1&-1\end{bmatrix}$$ which has determinant $-2$, so $\mathrm{im}(d_0)$ has index $2$ in $C^1(\mathcal{U},\mathcal{F})$, and thus the cokernel will be $\mathbb{Z}/2\mathbb{Z}$, so we have $H^1(\mathcal{U},\mathcal{F})=\mathbb{Z}/2\mathbb{Z}$.