Consider the category of pairs $(X,Y)$ where $X,Y$ are Abelian groups such that $X\subseteq Y$. I want to show that this category is not Abelian. I have checked that this category is additive, and certainly admits kernels and cokernels, so the problem has to be in the last condition, i.e. there is $\mathbf{coim}(f)\not\cong \mathbf{im}(f)$ for some morphism $f:(A,B)\to (A',B')$. I think there might be some extreme counter-example that I am missing under my nose, but I just can't find that out. Thanks for your time!
Category of Abelian group pairs is not Abelian
abelian-categoriesadditive-categoriescategory-theory
Related Solutions
It simply doesn't follow that $g$ is a morphism in $S$. For example, $A$ might be the category of abelian groups, $S$ might be the full subcategory of finitely generated abelian groups, and the target of $g$ might be an infinitely generated abelian group.
But the proof is very easy to repair: just define $g = \text{coker}(f) \in A$ in the first place. Then in fact we have $g \in S$, so $\text{ker}(g) \in S$ as well. Now, a key feature of full subcategories $S$ is that any limits or colimits, in $A$, of $S$-valued diagrams which happen to land in $S$ must in fact be limits or colimits in $S$: that is, inclusions of full subcategories reflect limits and colimits. Because $f = \text{ker}(g)$ in $A$, it follows that $f = \text{ker}(g)$ in $S$ as well, so $f$ is a kernel. Similarly for epimorphisms.
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.
Best Answer
You're right that there's a simple counterexample.
Consider $f : (0, \mathbb{Z}) \to (\mathbb{Z},\mathbb{Z})$ which is the identity in the second slot. That is:
Can you show that this is a mono and an epi, but is not an iso? This will contradict the fact that every abelian category is balanced (see here, say)
I hope this helps ^_^