$M_1 \to M_2 \to M_3 \to 0$ exact iff $0 \to \text{Hom}_A(M_3, N) \to \text{Hom}_A(M_2, N) \to \text{Hom}_A(M_1, N)$ is exact. Let $M_1 \to M_2 \to M_3 \to 0$ be a sequence of homomorphisms of $A$-modules. Is this sequence exact if and only if the induced sequence of abelian groups$$0 \to \text{Hom}_A(M_3, N) \to \text{Hom}_A(M_2, N) \to \text{Hom}_A(M_1, N)$$is exact for every $A$-module $N$?
 A: Yes, and the proof goes something like this: 
Let $$M_1 \xrightarrow{g} M_2 \xrightarrow{f} M_3 \rightarrow 0$$ be your sequence with the functions written above the arrows. 
First, suppose that the sequence $$0 \rightarrow Hom_A(M_3, N) \xrightarrow{\bar{f}} Hom_A(M_2, N) \xrightarrow{\bar{g}} Hom_A(M_1, N)$$ is exact for all $N$. We know from the conditions for a short sequence to be exact, $\bar{f}$ is injective for all $N$, which in turn implies that $f$ is surjective. 
Additionally, we know that $\bar{g} \circ \bar{f} = 0$ (from the defintion of exact sequences). Well, this then implies that for any $h: M_3 \to N$ we have $f \circ g \circ h = 0$. We then take $N = M_3$, so $f: M_3 \to M_3$ (we can do this because N is arbitrary). Let this map simply be the identity map, and then we have $$f \circ g = 0$$ i.e. $Im(g) \subseteq ker(f)$. 
Now let $N=M_2/Im(g)$ where $Im(g)$ is the image of $M_1$ in $M_2$ and let $\phi: M_2 \to N$ be the projection homomorphism. Then $\phi \in ker(\bar{g})$, and so $\exists$ $\psi: M_3 \to N$ s.t. $\phi = \psi \circ f$. As a result, this tells us that $Im(g) = ker(\phi) \supseteq ker(f)$ and so from the two containments we have $$Im(g) = ker(f)$$ and thus the sequence $$M_1 \xrightarrow{g} M_2 \xrightarrow{f} M_3 \rightarrow 0$$ is exact.
The converse follows along the same lines, it really becomes a matter of "flipping" the above argument. I will leave the details to you.  
