Two definitions of an abelian category

Question

In the definition of an abelian category, we require

1. it has a zero object,

2. it has all binary biproducts,

3. it has all kernels and cokernels,

4. all monomorphisms and epimorphisms are normal.

Alternatively, one can start from a preadditive cateogry, and require that every finite set of objects has a biproduct, and satisfies 3 and 4.

However, this seems strange! Abelianess is a property of a category, while a preadditive category is a category together with a choice of enrichment over the monoidal category $\mathrm{Ab}$ of abelian groups. How does one reconcile the first choiceless definition, with the second "choicefull" definition?

Answer 1

Let $\mathcal{A}$ be a preadditive category, i.e. an $\mathbf{Ab}$-enriched category. If $\mathcal{A}$ admits a zero object and binary biproducts, i.e. if $\mathcal{A}$ is additive, then the addition of morphisms can be retrieved from the categorical structure of $\mathcal{A}$:

Suppose first that $\mathcal{A}$ has a zero object $0$. Then for every two objects $X, Y \in \operatorname{Ob}(\mathcal{A})$ the zero morphism $0_{X,Y} \colon X \to Y$ is given by the unique composition $X \to 0 \to Y$. Hence the zero morphism does not depend on the choice of enrichment.

Suppose now that $\mathcal{A}$ also has binary biproducts and is hence an additive category. We have for every object $X$ a diagonal morphism $\Delta_X \colon X \to X \oplus X$ and a codiagonal morphism $\nabla_X \colon X \oplus X \to X$. The diagonal morphism $\Delta_X$ is given via the universal property of the product as $$ p_1 \circ \Delta_X = \mathrm{id}_X \,, \quad p_2 \circ \Delta_X = \mathrm{id}_X \,, $$ and the codiagonal morphism $\nabla_X$ is given via the universal property of the coproduct as $$ \nabla_X \circ i_1 = \mathrm{id}_X \,, \quad \nabla_X \circ i_2 = \mathrm{id}_X \,. $$ If you’re familir with matrix notation in additive categories then this means that $$ \Delta_X = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \,, \quad \nabla_X = \begin{bmatrix} 1 & 1 \end{bmatrix} \,. $$

Let now $f, g \colon X \to Y$ be two parallel morphisms in $\mathcal{A}$. We denote the structure morphisms of the biproducts $X \oplus X$ and $Y \oplus Y$ by \begin{gather*} i_1, i_2 \colon X \to X \oplus X \,, \quad p_1, p_2 \colon X \oplus X \to X \,, \\ j_1, j_2 \colon Y \to Y \oplus Y \,, \quad q_1, q_2 \colon Y \oplus Y \to Y \,. \end{gather*} There exists by these universal properties of the product and coproduct a unique morphism $f \oplus g \colon X \oplus X \to Y \oplus Y$ with \begin{gather*} q_1 \circ (f \oplus g) \circ i_1 = f \,, \quad q_1 \circ (f \oplus g) \circ i_2 = 0 \,, \\ q_2 \circ (f \oplus g) \circ i_1 = 0 \,, \quad q_2 \circ (f \oplus g) \circ i_2 = g \,. \end{gather*} This means in matrix notation that $$ f \oplus g = \begin{bmatrix} f & 0 \\ 0 & g \end{bmatrix} \,. $$ We now find that $$ \nabla_Y \circ (f \oplus g) \circ \Delta_X = f + g \,. $$ Indeed, we have $\mathrm{id}_{X \oplus X} = i_1 \circ p_1 + i_2 \circ p_2$ and $\mathrm{id}_{Y \oplus Y} = j_1 \circ q_1 + j_2 \circ q_2$ and thus \begin{align*} {}& \nabla_Y \circ (f \oplus g) \circ \Delta_X \\ ={}& \nabla_Y \circ \mathrm{id}_{Y \oplus Y} \circ (f \oplus g) \circ \mathrm{id}_{X \oplus X} \circ \Delta_X \\ ={}& \nabla_Y \circ (j_1 \circ q_1 + j_2 \circ q_2) \circ (f \oplus g) \circ (i_1 \circ p_1 + i_2 \circ p_2) \circ \Delta_X \\ ={}& \nabla_Y \circ j_1 \circ q_1 \circ (f \oplus g) \circ i_1 \circ p_1 \circ \Delta_X \\ {}& + \nabla_Y \circ j_2 \circ q_2 \circ (f \oplus g) \circ i_1 \circ p_1 \circ \Delta_X \\ {}& + \nabla_Y \circ j_1 \circ q_1 \circ (f \oplus g) \circ i_2 \circ p_2 \circ \Delta_X \\ {}& + \nabla_Y \circ j_2 \circ q_2 \circ (f \oplus g) \circ i_2 \circ p_2 \circ \Delta_X \\ ={}& \mathrm{id}_Y \circ f \circ \mathrm{id}_X + \mathrm{id}_Y \circ 0 \circ \mathrm{id}_X + \mathrm{id}_Y \circ 0 \circ \mathrm{id}_X + \mathrm{id}_Y \circ g \circ \mathrm{id}_X \\ ={}& f + g \,. \end{align*} This means in matrix notation that $$ \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} f & 0 \\ 0 & g \end{bmatrix} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = f + g \,. $$

We see from this that the preadditive structure of $\mathcal{A}$ can be described by using the zero object and the binary biproducts of $\mathcal{A}$, which do not depend on the (choice of) enrichement.

(If more generally $\mathcal{A}$ is a category that admits a zero objects and binary biproducts then one can use the above construction to define for any two objects $X, Y \in \operatorname{Ob}(\mathcal{A})$ on $\mathcal{A}(X,Y)$ the structure of an abelian monoid. That $\mathcal{A}$ is additive then means that these monoid structures are actually groups structures.)

Answer 2

Here is the key fact:

Theorem: Let $C$ be a category with finite biproducts. Then $C$ admits a unique enrichment in commutative monoids.

Proof sketch: By Yoneda's lemma, an enrichment in commutative monoids is the same thing as giving a natural commutative monoid object structure on each object of $C$ (where "natural" means that every morphism becomes a homomorphism of commutative monoid objects). Now in any category with finite coproducts, considered as a symmetric monoidal category with respect to coproducts, every object admits a unique natural commutative monoid object structure and every morphism is a homomorphism with respect to these unique monoid structures. Specifically, the addition map $\mu:X\coprod X\to X$ is the codiagonal map and the unit map is the unique map $0\to X$. (To see uniqueness, note that the identity axioms force the compositions $X\cong X\coprod 0\to X\coprod X\stackrel{\mu}\to X$ and $X\cong 0\coprod X\to X\coprod X\stackrel{\mu}\to X$ to be the identity, but this says precisely that $\mu$ is the codiagonal map.)

Now since our category $C$ has finite biproducts, a commutative monoid structure on an object $X$ in the usual sense (i.e., with respect to products, so the addition map would be a map $X\times X\to X$) is the same as a commutative monoid structure with respect to coproducts. Thus every object of $C$ admits a unique natural commutative monoid structure.

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Now note that this gives just an enrichment in commutative monoids, not abelian groups. But once you have an enrichment in commutative monoids, the existence of additive inverses is a property, not an additional structure. It is a nontrivial theorem that the additional axioms satisfied by an abelian category actually force the commutative monoid structure on each Hom-set to have inverses and be an abelian group.

As a final note, if you found the proof above overly abstract, here is the concrete construction of the addition of morphisms in a category with finite biproducts. Given morphisms $f,g:X\to Y$, their sum can be defined as the composition $$X\to X\oplus X\stackrel{f\oplus g}\to Y\oplus Y\to Y$$ where the first map is the diagonal and the last map is the codiagonal. It can be checked directly if somewhat laboriously that this really does give an enrichment in commutative monoids and it is the unique such enrichment. For instance, to check that it is commutative, you can use commutativity of $\oplus$ and consider $Y\oplus Y$ as a biproduct of $Y$ with the order of the two inclusions and projections swapped, which will end up swapping the roles of $f$ and $g$ in the definition above. Similarly, associativity ultimately comes from associativity of $\oplus$ and the fact that the zero morphisms are identities with respect to addition comes from the natural isomorphisms $X\cong X\oplus 0$. For uniqueness, see the matrix calculation in Jendrik Stelzner's answer, which shows that any enrichment in commutative monoids must be given by this formula.