Let $X,X',X''$ be $A$-modules and denote by $\mbox{Hom}_A(X',X)$ the set of $A$-homomorphisms of $X'$ into $X$.
Proposition 2.1 in the Lang's Algebra text states the following:
A sequence $$ X' \xrightarrow{\lambda} X \xrightarrow{\mu} X'' \to 0 $$ is exact iff the sequence $$ \mbox{Hom}_A(X',Y) \xleftarrow{\lambda^*} \mbox{Hom}_A(X,Y) \xleftarrow{\mu^*} \mbox{Hom}_A(X'',Y) \leftarrow 0 $$ is exact for all $Y$. where $\lambda^*(g) = g \circ \lambda$ and $\mu^*$ is defined simliarly.
I cannot prove the 'if' ($\Leftarrow$) part; quite stuck. Any help or suggestion?
What I've done so far;
I proved $\mbox{Im} \lambda \subseteq \mbox{Ker} \mu$ by drawing commute diagrams.
To prove exactness at $X''$ (surjectivity of $\mu$), take $Y = \mathrm{coker}(\mu)$. The canonical map $q \colon X'' \to Y$ is mapped to $0$ by $\mu^{*}$, hence was already $0$. Since $q$ is surjective, $Y = 0$.
You can deduce $\mu \circ \lambda = 0$ by taking $Y = X''$, and looking at what happens with $\mathrm{id}_{Y}$.
Finally, look at $Y = \mathrm{coker}(\lambda)$. The canonical map $q \colon X \to Y$ is mapped to $0$ by $\lambda^{*}$, hence comes from a map $f$ in $\mathrm{Hom}(X'', Y)$. In other words, $f \circ \mu = q$. In particular, ther kernel of $\mu$ is contained in the kernel of $q$ (which is the image of $\lambda$).