Hartshorne, Chapter 3, Exercise 3.7(a):
$A$ is Noetherian, $\mathfrak{a}$ is an ideal, $U=\operatorname{Spec} A\setminus V(\mathfrak{a})$. For any $A$-module $M$, $\Gamma(U,\tilde{M})\cong \varinjlim\operatorname{Hom}(\mathfrak{a}^n,M)$.
Suppose $a=(f_1,...,f_m)$, then $U=D(f_1)\cup\dots\cup D(f_m)$. By above isomorphism and sheaf property, there is an exact sequence: $$0\rightarrow \varinjlim\operatorname{Hom}(\mathfrak{a}^n,M)\rightarrow \coprod_i\varinjlim\operatorname{Hom}((f_i)^n,M)\rightarrow \coprod_{i,j}\varinjlim\operatorname{Hom}((f_if_j)^n,M).$$
Can this exact sequence be induced by the exact sequence of the form: $$0\rightarrow \operatorname{Hom}(\mathfrak{a}^n,M)\rightarrow \coprod_i\operatorname{Hom}((f_i)^n,M)\rightarrow \coprod_{i,j}\operatorname{Hom}((f_if_j)^n,M),$$ where differential is like sheaf property.
For $n=1$, $0\rightarrow \operatorname{Hom}(\mathfrak{a},M)\xrightarrow\alpha \coprod_i\operatorname{Hom}((f_i),M)\xrightarrow\beta \coprod_{i,j}\operatorname{Hom}((f_if_j),M)$, where $\alpha(g)=(\operatorname{res}(g))_i$ and $\beta((g_i)_i)=(\operatorname{res}(g_i)-\operatorname{res}(g_j))_{ij}$. It is clear that $\alpha$ is monic. If $\beta((g_i)_i)=0$, can the morphism can be extended to $\mathfrak{a}$?
Thank you in advance.
The answer is no.
Let our ring be $A=\Bbb{Z}/90\Bbb{Z}$, and let $M=A$. Let $\mathfrak{a} = (6,15) = (3)$. Then $(6)(15)=(0)$, so regardless of $g_1$ and $g_2$, they will agree on $(f_if_j)$. Defining $g_1(6)=12$, $g_2(15)=15$, we see that $g_1(30) = 60$, but $g_2(30)=30$. Thus $(g_1,g_2)$ can't be extended to a morphism on $(3)$.
We'd need $g_i$ and $g_j$ to agree on $(f_i)\cap (f_j)$ rather than $(f_if_j)$.