1st isomorphism theorem. Given a morphism $f : A \rightarrow B$ of abelian groups, the first isomorphism theorem states that the canonical map $A/\ker(f) \rightarrow f(B)$ is an isomorphism. This holds in an abelian category. That is, let $f : A \rightarrow B$ be a morphism in an abelian category. We define
- $\mathrm{im}(f) := \ker(\mathrm{cok}(f))$ and
- $\mathrm{coim}(f) := \mathrm{cok}(\ker(f)) = A/\ker(f).$
More precisely, we have $m : \mathrm{im}(f) \hookrightarrow B$ and $e : A \twoheadrightarrow \mathrm{coim}(f).$ From either universal property, one has a canonical map $\mathrm{coim}(f) \rightarrow \mathrm{im}(f),$ and this is necessarily an isomorphism (e.g., see Mac Lane, p.199, Proposition 1).
Remark. It seems that the 1st isomorphism can be taken as part of axioms of a category being abelian: see Proposition 2.3 of this nLab page.
2nd isomoprhism theorem. Given a subgroups $X, Y$ of an abelian group $A,$ the 2nd isomorphism theorem states that the canonical map $$Y/(X \cap Y) \rightarrow (X + Y)/X$$ is an isomorphism. How do we get this in an abelian category? A priori, this is just the 1st isomorphism theorem applied to the map $Y \twoheadrightarrow \pi_{X}(Y),$ where $\pi_{X} : A \twoheadrightarrow A/X$ is the projection map. We similarly define $\pi_{Y} : A \twoheadrightarrow Y.$ We also denote by $\iota_{X}$ and $\iota_{Y}$ the maps $X \hookrightarrow A$ and $Y \hookrightarrow A,$ respectively.
Taking intersection in an abelian category. Note that
- $X \cap Y = \ker(\iota_{X} \circ \pi_{Y} : X \hookrightarrow A \twoheadrightarrow A/Y),$
so we define $X \cap Y := \ker(\iota_{X} \circ \pi_{Y}).$ Note that we have $$\ker(\iota_{X} \circ \pi_{Y}) \hookrightarrow \ker(\pi_{Y}) = Y$$ and if there is a common subobject $S$ of $X$ and $Y$ (with all the maps into $A$ commute), then, there is a unique map $S \hookrightarrow X \cap Y$ that commutes with the maps from $S$ and $X \cap Y$ into $A$ induced by the universal property. This means that we get a unique isomorphism $$X \cap Y \simeq Y \cap X$$ commuting with maps into $A$ (and hence those into $X$ or $Y$).
Taking sum in an abelian category. Denoting $p_{X} : X \oplus Y \rightarrow X$ and $p_{Y} : X \oplus Y \rightarrow Y$ for the projections that come with the direct product (which comes with the direct sum structure in any abelian category), we define $$X + Y := \mathrm{im}(\iota_{X}p_{X} + \iota_{Y}p_{Y}).$$ By the universal property, we can observe that $X, Y$ are subobjects of $X + Y,$ and their maps to $A$ are all commutative. Now, let $T$ be any subobject of $A$ that take $X, Y$ as subobjects (with all the maps into $A$ commute). Then there is a unique map $$X + Y \simeq \mathrm{coim}(\iota_{X}p_{X} + \iota_{Y}p_{Y}) \rightarrow T,$$ that commutes with maps from $X \oplus Y.$ The induced map is therefore (e.g., by applying Mac Lane, p.199, Proposition 1) commutes with the maps into $A,$ so it is in particular a monomorphism. Hence, it follows that $X + Y$ is the smallest subobject of $A$ that contain $X$ and $Y.$
Proof of 2nd isomorphism theorem. By definition, we may already note that the kernel of the composition $Y \hookrightarrow A \twoheadrightarrow A/X$ is $X \cap Y.$ It remains to show that its image is given by $(X + Y)/X.$ First, it is not difficult to observe that the cokernel of this composition is given by $A/X \twoheadrightarrow A/(X + Y).$ One may then show that the kernel of this map is precisely $(X+Y)/X,$ which finishes the proof.
