TITLE: "The Yoneda embedding reflects exactness" is a direct consequence of Yoneda?
QUESTION [16 upvotes]: Let $A,B,C$ be objects of a category of modules over a ring. It is not hard to see that the Yoneda embedding "reflects exactness" (as Weibel puts it, on p. 28), i.e. if 
$\hom(X,A)\stackrel{f_*}{\to}\hom(X,B) \stackrel{g_*}{\to} \hom(X,C)$ is an exact sequence of abelian groups for every module $X$, then $A\stackrel{f}{\to}B\stackrel{g}{\to}C$ is an exact sequence of modules.
This is done in Weibel, or in this article of Paul Garrett.
There, he states that this is a special case of the Yoneda lemma.
But I'm struggling to see this is the case. Here is my (naive) attempt:
$\ker(g_*)=\ker(\hom(A,g))\simeq \hom(A,\ker(g))$ because the covariant Hom functor is preserves kernels (because it preserves limits, because it has a left adjoint)
I would very much like to have $\operatorname{im}(f_*)\simeq \hom(A,\operatorname{im}(f))\ \ \ \  (*) \ \ $ in the same way; if I had that, the given exactness would get me $\hom(A,\operatorname{im}(f))\simeq \hom(A,\ker(g))$. And then I would be happy, for Yoneda lemma would instantly get me $\operatorname{im}(f)\simeq \ker(g)$.
But I'm not happy because I can't see why $(*)$ holds. It's false that Hom preserves images.
However, I'm baffled because if this is supposed to be an application of Yoneda lemma, then surely it must be because $\hom(A,\operatorname{im}(f))\simeq \hom(A,\ker(g))$... or perhaps not?

REPLY [2 votes]: Yoneda Lemma, In my opinion, is to demonstrate Yoneda embedding functor (which is fully-faithful).
You can prove without any difficulty to show that if an embedding functor is exact at the same time, then "it reflect the exactness" (By homology calculation).
So statement and proof you menstioned in Weibel's textbook just means that exactness of this embedding is not necessary to "reflect the exactness".
Here is the statment:
Assume two abelian categories $\mathscr{A}$, and $\mathscr{B}$, and an addictive fuctor $E$ between them. In additioin, $E$ is fully-faithful, then we have:
Consider $a \xrightarrow{f} b\xrightarrow{g} c$ in category $\mathscr{A}$,
if $E(a) \xrightarrow{E(f)} E(b)\xrightarrow{E(g)} E(c)$ is exact in $\mathscr{B}$, then $a \xrightarrow{f} b\xrightarrow{g} c$ if exact in $\mathscr{A}$.
you take

$E(g)E(f)=E(gf)=0\Rightarrow gf=0$ implies $\mathrm{im}(f)\subseteq \mathrm{ker}(g)$.
consider the inclusion $\iota \in \hom (\ker g, b)\cong\hom(E(\ker g),E(b)) $, by following exact sequence:
$$
0\rightarrow\hom(E(\mathrm{ker}g),E(a)) \rightarrow \hom(E(\mathrm{ker}g),E(b))\rightarrow \hom(E(\mathrm{ker}g),E(c)) .
$$
we can find $\kappa \in \hom (\ker g, a)$ such that $\iota = f\kappa$,
And so, $\ker g = \mathrm{im}\ \iota = \mathrm{im}\ (f\kappa) \subset \mathrm{im}\ f$.