If
$$
0 \to A \to B\to C,$$
is a left exact sequence of $R$-module, then for any $R$-module $M$,
$$
0 \to \operatorname{Hom}_R(M,A)\to \operatorname{Hom}_R(M,B)\to \operatorname{Hom}_R(M,C),
$$
is left exact.
I proved the above, and highlighted what I'm a little unfamiliar with: Let $$
0 \to A\ \xrightarrow{i}\ B\ \xrightarrow{f}\ C,
$$
and
$$
0 \to \operatorname{Hom}(M,A)\ \xrightarrow{\operatorname{Hom}(M,i)}\ \operatorname{Hom}(M,B)\ \xrightarrow{\operatorname{Hom}(M,f)}\ \operatorname{Hom}(M,C).
$$
We need to show that
$$
\ker\left(\operatorname{Hom}(M,f)\right)=\operatorname{Hom}(M,i)(\operatorname{Hom}(M,A)).
$$
Let $i \circ \varphi \in \operatorname{RHS}$. Then $f \circ i \circ \varphi : M \to C$ is $0$ since $f \circ i \circ \varphi(M) \subseteq f( i(A)) = f(\ker(f))=0$.
Conversely, let $\psi \in \operatorname{LHS}$. Then $f \circ \psi = 0$ so that $ f(\psi(M))=0$. Hence $\psi(M) \subseteq \ker(f)=i(A)$. Since the image of $\psi$ is contained in the image of $i$, we may factor $\psi$ as $\psi=i \varphi$ with $\varphi : M \to A$. (Here is my trial, but I'm not fully understanding this: Since $i$ is injective, $i(A)$ is isomorphic with $A$. So $i^{-1}(\psi (M)) \subseteq A$ and if we let $\varphi=i^{-1} \psi$, then $\psi = i \varphi$.)
And I have one more question: The above looks very messy, especially the notation. Is there a better proof/understanding about it?
Best Answer
Well, your proof is okay. Let me suggest a slightly different way of looking at it:
Consider a sequence
$$ 0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C$$ and look at $$ 0\; \xrightarrow{\phantom{j_{\ast}}} \operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\; \operatorname{Hom}{(M,C)},$$ where I write $i_{\ast} = \operatorname{Hom}(M,i)$ and $j_{\ast} = \operatorname{Hom}(M,j)$.
Saying that the first sequence is exact amounts to saying $i = \ker{j}$, that is $ji = 0$ and $i$ has the universal property as depicted in the first diagram below: If $g: M \to B$ is such that $jg = 0$ then there exists a unique $f: M \to A$ such that $if = g$. In other words if $j_\ast g = 0$ then $g = i_{\ast}f$, or yet again $\operatorname{Ker}{j_\ast} \subset \operatorname{Im}{i_{\ast}}$ and $i_{\ast}$ is injective.
On the other hand, the second diagram says: if $g: M \to B$ is of the form $g = if = i_{\ast}f$ then $j_{\ast}g = 0$ (because $j_\ast g = j_{\ast}i_{\ast} f = (ji)_{\ast}f = 0f = 0$). In other words, $\operatorname{Im}{i_{\ast}} \subset \operatorname{Ker}{j_{\ast}}$.
Summing up, we have shown that for all $M$ the sequence
$$ 0\; \xrightarrow{\phantom{j_{\ast}}} \operatorname{Hom}{(M,A)}\; \xrightarrow{i_{\ast}}\operatorname{Hom}{(M,B)} \;\xrightarrow{j_{\ast}}\; \operatorname{Hom}{(M,C)}$$ is exact both at $\operatorname{Hom}{(M,A)}$ ($i_{\ast}$ is injective) and at $\operatorname{Hom}{(M,B)}$ ($\operatorname{Im}{i_{\ast}} = \operatorname{Ker}{j_{\ast}}$)—you seem to have forgotten about the first point here.
Added: As witnessed by the argument above, left exactness of $\operatorname{Hom}$ is essentially the definition of left exactness in the abelian category of $R$-modules. As the comments try to point out, the importance of this fact cannot be overemphasized.
I would like to add two further points:
A functor $F$ is left exact in your definition if and only if $0 \to F(A) \to F(B) \to F(C)$ is left exact for every short exact sequence $0 \to A \to B \to C \to 0$.
Indeed, in a left exact sequence $0 \to A \to B \to C$, we may factor $j: B \to C$ over its image as $B \twoheadrightarrow \operatorname{Im}{j} \rightarrowtail C$ and obtain two exact sequences $$0 \to A \to B \to \operatorname{Im}{j} \to 0 \qquad \text{and} \qquad 0 \to \operatorname{Im}{j} \to C \to \operatorname{Coker}{j} \to 0.$$ Applying $F$ to these two exact sequences, we obtain the left exact sequences $$0 \to F(A) \to F(B) \to F(\operatorname{Im}{j}) \qquad \text{and} \qquad 0 \to F(\operatorname{Im}{j}) \to F(C ) \to F(\operatorname{Coker}{j}).$$ Since the kernel of a map is not changed by postcomposing the map with a monomorphism (check this!), we have $$\operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}))} = \operatorname{Ker}{(F(B) \to F(\operatorname{Im}{j}) \to F(C))},$$ so by functoriality of $F$ we get a left exact sequence $0 \to F(A) \to F(B) \to F(C)$ as desired.
A natural question is: When does $\operatorname{Hom}(M,-)$ send short exact sequences to short exact sequences? In other words, when is $j_\ast = \operatorname{Hom}{(M,j)}$ an epimorphism for all short exact sequences $0\; \xrightarrow{\phantom{ij}} \; A \; \xrightarrow{i\phantom{j}}\; B\; \xrightarrow{j\phantom{i}} \; C \to 0$?
In view of left exactness of $\operatorname{Hom}{(M,-)}$ the question is: Given any morphism $h: M \to C$ and any epimorphism $j: B \to C$, when is $h$ of the form $h = j_\ast g$ for some morphism $g: M \to B$?
As you certainly know, this is precisely the definition of projective modules: $M$ is called projective if and only if $g$ always exists, for all epimorphisms $j: B \twoheadrightarrow C$ and all $h: M \to C$. For emphasis: