Let $A=k[[x,y]]$, with say $k=\mathbb{C}$ and $X = \operatorname{Spec}(A)$. Let $B$ denote a free $A$-algebra of rank $e^2$; actually, we have $B=A[u,v]$ with $u^e=x$, $v^e=y$ and $uv = \xi_e vu$, where $\xi_e$ is an $e$-th root of unity. Now we have $M_n(B)$ and the subalgebra $D$, where entries under the diagonal are in $uB$ and otherwise in $B$.

This gives us an exact sequence of noncommutative groups, where $i: \lbrace x=0 \rbrace \hookrightarrow X$: $$ 1\rightarrow D^{\times} \rightarrow Gl_n(B) \rightarrow i_{\ast}F \rightarrow 1 \, .$$ Why is this an exact sequence of sheaves in the Zariski topology on $X$? That is: why can i look at the induced exact sequence in Zariski cohomology?

This is stated on page 3 of http://www.math.lsa.umich.edu/courses/711/ordpages60-85.ps, around the $3\times 3$ diagram. Or do we need the whole diagram to see this?

I mean even in the simple example $1\rightarrow \mu_n \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \rightarrow 1$, where $\mathbb{G}_m \rightarrow \mathbb{G}_m$ is $x \mapsto x^n$, this sequence is not exact when we use the Zariski topology.

If it is easier, one can replace $B$ with $A$ and $uB$ with $xA$. The goal is to see that $H^1(U,D^{\times})=0$, i.e. every fractional reflexive left $D$-ideal is free. Here $U = X\backslash \lbrace \mathfrak{m} \rbrace$. Maybe there is an easier way to see this?