Exact sequence of $A$-modules I was trying to demonstrate the Proposition 2.9 of Atiyah and MacDonald's Introduction to Commutative Algebra. But I couldn't do the following:

Let $M$, $M'$, and $M''$ be $A$-modules, $v$ and $u$ be homomorphisms, and
  $$0 \longrightarrow\operatorname{Hom}(M'',N) \stackrel{\bar{v}}\longrightarrow\operatorname{Hom}(M,N) \stackrel{\bar{u}}\longrightarrow\operatorname{Hom}(M',N)$$
  be a sequence exact for all $A$-module $N$. Then the sequence
  $$M' \stackrel{u}\longrightarrow M \stackrel{v}\longrightarrow M'' \longrightarrow 0$$
  is exact. Here, $\bar{v}(f)=f\circ v$ and $\bar{u}(g)=g\circ u$.

I suppose there is a particular A-module N that holds this, but I couldn't find it.
Somebody can help me? Thanks.
 A: You will find several solutions in the comments and in some books, but actually the statement is, from a "higher" perspective, almost the definition of an exact sequence. Let me explain this.
By definition, $M' \xrightarrow{u} M \xrightarrow{v} M'' \to 0$ is exact iff $v$ is surjective and $\mathrm{im}(u)=\mathrm{ker}(v)$. This means equivalently that $v$ induces an isomorphism $\mathrm{coker}(u)=M/\mathrm{im}(u) \cong M''$. Now, cokernels have a universal property, and it is a general fact that objects are determined by their universal property (Yoneda Lemma). Hence, $M' \xrightarrow{u} M \xrightarrow{v} M'' \to 0$ is exact iff $v$ is a cokernel of $u$ which means that $vu=0$ and for all morphisms $M \to T$ which vanish when composed with $u$ there is a unique factorization over $v$. But this exactly means that $0 \to \hom(M'',T) \to \hom(M',T) \to \hom(M,T)$ is exact for all $T$.
Similarly, a sequence $0 \to M'' \to M'  \to M$ is exact iff $M'' \to M'$ is a kernel of $M' \to M$ iff $0 \to \hom(T,M'') \to \hom(T,M') \to \hom(T,M)$ is exact for all $T$.
(When you want to prove this directly, you will just repeat the proof of the Yoneda Lemma in a special case; this happens all the time because category theoretic principles are almost ignored in teaching...)
