$I$-adic completion of an exact sequence of finite $A$-modules is again exact I'm reading Hideyuki Matsumuras Commutative Ring Theory, and I'm having a trouble understanding a statement he makes in theorem 8.7.
For some context, the following two theorems have previously been proven:

1) $A$ noetherian, $M$ finite $A$-module, $N\subset M$ submodule and $I$ an ideal of $A$. Then the $I$-adic topology of $N$ coincides with the topology induced by the $I$-adic topology of $M$ on the subspace $N$.

And 

2) In the $I$-adic topology the sequence $0\rightarrow \hat{N} \rightarrow \hat{M} \rightarrow \hat{M/N}\rightarrow 0$ is exact.

Now I'm reading, that from these two theorems, we can conclude that 

Let $A$ be a noetherian ring and $I$ an ideal. Then the $I$-adic completion of an exact sequence of finite $A$-modules is again exact.

But I don't get why that is the case? I don't understand how we can generalize the fact that if $0\rightarrow {N} \rightarrow {M} \rightarrow {M/N}\rightarrow 0$ is an exact sequence, then taking I-adic completion we get that $0\rightarrow \hat{N} \rightarrow \hat{M} \rightarrow \hat{M/N}\rightarrow 0$ beign exact, to given an arbitrary sequence $0\rightarrow {A} \rightarrow {B} \rightarrow {C}\rightarrow 0$ which is exact, then taking I-adic completion Them $0\rightarrow \hat{A} \rightarrow \hat{B} \rightarrow \hat{C}\rightarrow 0$ is exact?
Now I know that the $I$-adic topology on $M$ can be defined by $\{I^nM\}_{n\in \mathbb{N}}$, the one on $M/N$ by $\{M/(N+I^nM)\}_{n\in \mathbb{N}}$ and the one on $N$ by $\{N/(N\cap I^nM)\}_{n\in \mathbb{N}}$, so I was thinking we somehow could use that to generalize, but I haven't come Any futher.
 A: If $F$ is an additive functor from finite $A$-modules to abelian groups, then the conditions that 
$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form 
$$
0\to N\to M\to P\to0,
$$
$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form 
$$
N\to M\to P,
$$ 
$\bullet\ F$ preserves the exactness of any exact sequence of finite $A$-modules,
are equivalent.
See Exercise 8.17 in this text.
EDIT 1. Here is a version which tries to take into account the OP's comment.
Let $Q$ be an $A$-module, and, for any $A$-module $X$ set 
$$
FX:=Q\otimes_AX.
$$ 
Note that any morphism 
$$f:X\to Y
$$ 
yields an obvious morphism 
$$
Ff:FX\to FY.
$$ 
In particular, applying $F$ to an exact sequence of finite $A$-modules gives rise to a complex of $A$-modules (that is, a sequence of morphisms such that the composition of any two successive morphisms is $0$). (This complex will not be exact in general.)
Then the conditions that 
$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form 
$$
0\to N\to M\to P\to0,
$$
$\bullet\ F$ preserves the exactness of exact sequence of finite $A$-modules of the form 
$$
N\to M\to P,
$$ 
$\bullet\ F$ preserves the exactness of any exact sequence of finite $A$-modules,
are equivalent.
The case you are interested in is $Q=\widehat A$.
EDIT 2. In fact, if $F$ is as in Edit 1, and if $F$ preserves the exactness of $0\to J\to A$ for all finitely generated ideals $J$, then $F$ preserves the exactness of arbitrary exact sequences of arbitrary $A$-modules.
EDIT 3. I think there is a nano-gap in Matsumura's proof of Theorem 8.7: 
Write $FX$ for $\widehat X$. Let $X,Y,Z,U,V,W$ be finite $A$-modules. It is proved in Theorem 8.1 that $$0\to X\to Y\to Z\to0$$ exact implies $$0\to FX\to FY\to FZ\to0$$ exact. Let's denote this property by $(\star)$. But the property used in Theorem 8.7 is that $$X\to Y\to Z\to0$$ exact implies $$FX\to FY\to FZ\to0$$ exact.
It's better (I think) to prove that $(\star)$ implies that $F$ preserves the exactness of any exact sequence of finite $A$-modules. 
Assuming $(\star)$, let $U\to V\to W$ be exact. It suffices to show that $FU\to FV\to FW$ is exact.
Setting $$U':=\text{Ker}(U\to V),\ V':=\text{Im}(U\to V)=\text{Ker}(V\to W),\ W':=\text{Im}(V\to W),$$ we get the exact sequences
$$
\begin{matrix}
FU&\to&FV'&\to&0\\ \\ 
0&\to&FV'&\to&FV&\to&FW'&\to&0\\ \\ 
&&&&0&\to&FW'&\to&FW.
\end{matrix}
$$ 
It follows that $FU\to FV\to FW$ is exact, as required.
