Claim: The inclusion $R\text{-mod}\hookrightarrow R\text{-Mod}$ preserves kernels.
Once this is known, it follows that a kernel in $R\text{-mod}$, if it exists, must be isomorphic to the corresponding kernel in $R\text{-Mod}$; in particular, the latter is finitely generated.
Specializing this observation to projections $\pi_I: R\to R/I$ for an ideal $I$ in $R$, it follows that $\ker(\pi_I)$ exists in $R\text{-mod}$ if and only if $I$ is finitely generated. Letting $I$ vary, we see that $R\text{-mod}$ is abelian only if all ideals of $R$ are finitely generated.
Proof of claim: Suppose $f: M\to N$ is a morphism of finitely generated $R$-modules and $k: K\to M$ is a kernel of $f$ in $R\text{-mod}$. Further, let $g: T\to M$ be another $R$-module homomorphism with $fg=0$ and $T$ arbitrary. Then, by assumption, for any finitely generated submodule $\iota: S\subseteq T$ the composite $g\iota$ factors uniquely through $k$ via some $t_S: S\to K$. Since any module is the union of its finitely generated submodules, it follows that a factorization of $g$ through $k$ is unique, if it exists. In turn, applying this uniqueness it moreover follows that for any other $\iota^{\prime}: S^{\prime}\subseteq T$, the factorizations $S\to K$ and $S^{\prime}\to K$ of $\iota$ resp. $\iota^{\prime}$ agree on $S\cap S^{\prime}$. Therefore, all $t_S$ glue to a factorization $t: T\to K$ of $g$ through $k$, proving that $k$ is a kernel in $R\text{-Mod}$.
Addendum (independent of the rest): If you like it more technically, you can package the same argument as follows: Consider any category ${\mathscr C}$ (generalizing $R\text{-Mod}$), any diagram $D: I\to {\mathscr C}$ over some index category $I$ (generalizing $\bullet\rightrightarrows\bullet$), and any cone $c: D\to X$ over it, that is, you have $X\in{\mathscr C}$ and for any $i\in I$ you have a morphism $c_i: X\to D(i)$ such that $$X\xrightarrow{c_i} D(i)\xrightarrow{D(\alpha)} D(j) = X\xrightarrow{c_j} D(j)$$ for any arrow $\alpha$ in $I$. In other words, $X\to D$ is a candidate for a limit-cone for $I$, and you might ask:
Question: Which objects of ${\mathscr C}$ indeed 'see' $X$ as the limit of $D$?
Formally, this means that for some $Y\in{\mathscr C}$ you can check whether the natural morphism in $\textsf{Set}$,
$${\mathscr C}(Y,X)\to {\lim}_I{\mathscr C}(Y,D(i))$$
is an isomorphism. Call the respective subcategory ${\mathscr C}_D$ for lack of a better name. Now you have two facts:
- Since inverse limits commute, ${\mathscr C}_D$ is closed under colimits in ${\mathscr C}$.
- If $X$ and the codomain of $I$ are contained in some full subcategory ${\mathscr D}$ in which $X\to D$ is indeed an inverse limit, then ${\mathscr D}\subseteq{\mathscr C}_D$.
Combining both, it follows that ${\mathscr C}_D={\mathscr C}$ if there's a full subcategory ${\mathscr D}\subset{\mathscr C}$ containing $X$ and the codomain of $D$, for which $X\to D$ is a limit, and such that any object of ${\mathscr C}$ is a colimit of a diagram in ${\mathscr D}$.
This applies to $R\text{-mod}\subset R\text{-Mod}$ and shows that the latter embedding preserves all limits, in particular kernels.
By step 4 we have that $A \to \operatorname{Im}{f}$ and $\operatorname{Coim}{f} \to B$ are epi and mono, respectively. Therefore $\bar{f}$ is both epi and mono and we're done by step 3.
Best Answer
the natural map comes from the following construction:
consider: $A \xrightarrow{f} B$ , denote by $K \hookrightarrow A$ and $B \twoheadrightarrow C$ the kernel and cokernel of $f$. Furthermore denote by $K_C \hookrightarrow B$ the kernel of the cokernel of f (the image) and by $A \twoheadrightarrow C_K $ the cokernel of the kernel (the coimage).
Then we know that $K \hookrightarrow A \xrightarrow{f} B$ is $0$ hence this factors over $C_K$ as $C_K \xrightarrow{\alpha} B$, now since $A \twoheadrightarrow C_K$ is an epic, we can deduce, since $A \xrightarrow{f} B \twoheadrightarrow C$ is $0$ that $\alpha$ factors over $K_C$ as $\beta: C_K \to K_C$ and this is your desired map. Remark, kernels are in general monics and cokernels epics, but not generally the other way around, hence this comes into play for showing that this is an iso.