Let $(A,\mathfrak{m})=k[[x,y]]$ with $char(k)=0$ and $K=Quot(A)$. Set $X=Spec(A)$, $U=Spec(A)\backslash \lbrace \mathfrak{m} \rbrace$ the pointed spectrum. Furthermore given an $A$-algebra $B$, which can be embedded in $C=M_n(A)$, where $B$ is free $A$-module of rank $n^2$. One can see the algebra as a sheaf on $X$ resp. $U$ and $B^{\times}$ denotes its group of units.

Now we get an exact sequence $0\rightarrow B^{\times} \rightarrow C^{\times}=Gl_n(A) \rightarrow F \rightarrow 0$ and $F$ is supported on $Y=\lbrace x=0 \rbrace$, where it can be identified with some flag space.

Now in the article i'm reading, there are 3 facts i don't quite understand (now it is just one):

a) $H^0(X,C^{\times})=H^0(U,C^{\times})$ because $C$ is a free $A$-module

b) $H^0(X,C^{\times}) \rightarrow H^0(X,F)$ is surjective, because $X$ is local (see vytas comment)

c) $H^1(U,C^{\times})=0$ because $C$ and $A$ are Morita equivalent and this holds for $A$ (here one has to use "reflexive" Morita equivalence, then this follows from the fact that every reflexive A-module is free)

My question is: a) why does this follow from the freeness. If it was just $C$ i would believe this, because sections extend uniquely to codimension 2 points, which works here because $dim(A)=2$. But why should this be true for the group/sheaf of units? More generally this should be true if we replace $A$ by a free $A$-algbera $D$, i.e $H^0(X,Gl_n(D))=H^0(U,Gl_n(D))$. I tried using the Cech complex for $U$, as suggested in the comments, but it didn't help.

The article/text i'm referring to is $\href{http://www.math.lsa.umich.edu/courses/711/ordpages60-85.ps}{this}$. It is called "Stable orders and the Riemann-Roch Theorem". It is Lemma 3.1.9 on page 62 ( Page 3 in this document ).