Proof

The first condition implies that if $f$ is a monomorphism then $f \cong \ker(\coker(f))$ (in the category of objects over $B$) so $f$ is a kernel. Dually if $f$ is an epimorphism it follows that $f \cong coker(ker(f))$. So (1) implies (2).

The converse can be found in, among other places, Chapter VIII of (MacLane).

Proof

We give a streamlined version of the argument in Freyd & Scedrov 1990 §1.597-§1.598, which proceeds via the notion of exact categories (where the analog of Step 3 (iii) is immediate). The construction of the subtraction morphism $\sigma_B$ (Step 4) and of the addition from it (Step 6) is that of loc. cit. §1.597 (beware of misprints in the proof on p. 93 there: “$x \colon C \to \mathrm{Cok}(\theta)$” should be “$z \colon \cdots$”, “$y$ is monic” should be “$y$ is epic”, and “forces $y$ to be an isomorphism” should be “forces $\theta$ to be an isomorphism”).

We write $0$ also for the zero morphisms $A \to 0 \to B$; these form a two-sided ideal under composition. For binary products we write $p_1, p_2$ for the projections, $\langle f, g \rangle$ for the pairing of morphisms, $\Delta_B \coloneqq \langle \mathrm{id}_B, \mathrm{id}_B \rangle \colon B \to B \times B$ for the diagonal morphism, and

Step 0 (reduction to binary products). All hypotheses except the second are invariant under passage to the opposite category, while the second turns binary products into binary coproducts and vice versa. The conclusion is invariant as well: $Ab$-enrichments of $C$ are the same as those of $C^{op}$, and $C$ is abelian iff $C^{op}$ is (see below). Hence it suffices to consider the case that $C$ has binary products, which we assume from now on.

Step 1 (normal monos and epis).

(i) If $m$ is the kernel of some morphism $g$, then $m$ is the kernel of its own cokernel $c \coloneqq \mathrm{coker}(m)$: Since $g \circ m = 0$, we have $g = g' \circ c$ for some $g'$, so that $c \circ z = 0$ implies $g \circ z = 0$, whence $z$ factors uniquely through $m$. Dually, every epimorphism, being a cokernel by assumption, is the cokernel of its kernel.

(ii) Every split monomorphism is a monomorphism, hence by assumption a kernel, hence by (i) the kernel of its cokernel. Dually for split epimorphisms. This applies in particular to $\Delta_B$ and $\iota_1$ (split by $p_1$), to $\iota_2$ (split by $p_2$), and to $p_1$, $p_2$ (split by $\iota_1$, $\iota_2$, respectively).

Step 2. The morphism $\iota_1 \colon A \to A \times B$ is a kernel of $p_2$, and $p_2$ is a cokernel of $\iota_1$: First, $p_2 \circ \iota_1 = 0$, and if $p_2 \circ z = 0$ then $z = \langle p_1 \circ z, 0 \rangle = \iota_1 \circ p_1 \circ z$, uniquely so since $\iota_1$ is mono. Hence $\iota_1 = \mathrm{ker}(p_2)$, and so $p_2 = \mathrm{coker}(\iota_1)$ by Step 1. Symmetrically, $\iota_2 = \mathrm{ker}(p_1)$ and $p_1 = \mathrm{coker}(\iota_2)$.

Step 3 (equalizers, and a criterion for isomorphisms).

(i) $C$ has equalizers: For $f, g \colon A \to B$ let $e \coloneqq \mathrm{ker}\big( \mathrm{coker}(\Delta_B) \circ \langle f, g \rangle \big)$. Since $\Delta_B = \mathrm{ker}\big(\mathrm{coker}(\Delta_B)\big)$ by Step 1, a morphism $z$ into $A$ factors (necessarily uniquely) through $e$ iff $\langle f \circ z, g \circ z \rangle$ factors through $\Delta_B$, hence iff $f \circ z = g \circ z$. So $e$ is an equalizer of $f$ and $g$.

(ii) If $f \colon A \to B$ has vanishing kernel, in that $f \circ z = 0$ implies $z = 0$, then $f$ is a monomorphism: Let $k \colon P \to A \times A$ be the equalizer of $f \circ p_1$ and $f \circ p_2$ (the kernel pair of $f$), and write $q_i \coloneqq p_i \circ k$. Since $f \circ p_1 \circ \Delta_A = f \circ p_2 \circ \Delta_A$, there is $s \colon A \to P$ with $k \circ s = \Delta_A$, hence with $q_1 \circ s = \mathrm{id}_A = q_2 \circ s$. Now if $q_1 \circ z = 0$, then $f \circ q_2 \circ z = f \circ q_1 \circ z = 0$, hence $q_2 \circ z = 0$, hence $k \circ z = \langle q_1 \circ z, q_2 \circ z \rangle = 0$, hence $z = 0$, since $k$ is mono. Thus the kernel of the split epimorphism $q_1$ is $0 \to P$, so that, by Step 1, $q_1$ is a cokernel of $0 \to P$, hence an isomorphism, with inverse $s$. Therefore $q_2 = q_2 \circ s \circ q_1 = q_1$. Finally, if $f \circ x = f \circ y$ then $\langle x, y \rangle = k \circ u$ for some $u$, and so $x = q_1 \circ u = q_2 \circ u = y$.

(iii) If $f \colon A \to B$ has vanishing kernel and vanishing cokernel (in that also $t \circ f = 0$ implies $t = 0$), then $f$ is an isomorphism: By (ii), $f$ is mono, hence by Step 1 a kernel of $\mathrm{coker}(f)$. But $\mathrm{coker}(f) = 0$, and the kernel of a zero morphism out of $B$ is $\mathrm{id}_B$. Since kernels are unique up to unique isomorphism, $f$ is an isomorphism.

Step 4 (the subtraction morphism). For any object $B$, let $q \colon B \times B \to Q$ be a cokernel of $\Delta_B$, and consider

• $r$ has vanishing kernel: If $r \circ z = 0$, then $\iota_1 \circ z$ factors through $\Delta_B = \mathrm{ker}(q)$ (by Step 1), say as $\iota_1 \circ z = \Delta_B \circ w$. Applying $p_2$ to this equation gives $0 = w$, and then applying $p_1$ gives $z = w = 0$.

• $r$ has vanishing cokernel: If $t \circ r = 0$, then $t \circ q$ vanishes on $\iota_1$ and hence factors through $p_2 = \mathrm{coker}(\iota_1)$ (by Step 2), say as $t \circ q = s \circ p_2$. Then $s = s \circ p_2 \circ \Delta_B = t \circ q \circ \Delta_B = 0$, so that $t \circ q = 0$, and hence $t = 0$, since $q$ is epi.

By Step 3 (iii), $r$ is an isomorphism, and we set

This is a cokernel of $\Delta_B$ which satisfies

(For $C =$ Ab this is the map $(b_1, b_2) \mapsto b_1 - b_2$.) Moreover, $\sigma$ is natural: For $f \colon A \to B$ we have $\sigma_B \circ (f \times f) = f \circ \sigma_A$. Namely, $\sigma_B \circ (f \times f) \circ \Delta_A = \sigma_B \circ \Delta_B \circ f = 0$, so that $\sigma_B \circ (f \times f) = f' \circ \sigma_A$ for some $f'$, and then

Step 5 (properties of subtraction). For $x, y \colon A \to B$ define

(the notation is justified by Step 6). Then, for all composable morphisms $x, y, z, w$:

1. $x - 0 = x$, since $\langle x, 0 \rangle = \iota_1 \circ x$;

2. $x - x = 0$, since $\langle x, x \rangle = \Delta_B \circ x$;

3. $z \circ (x - y) = z \circ x - z \circ y$, by naturality of $\sigma$ (Step 4) and since $(z \times z) \circ \langle x, y \rangle = \langle z \circ x, z \circ y \rangle$;

4. $(x - y) \circ w = x \circ w - y \circ w$, since $\langle x, y \rangle \circ w = \langle x \circ w, y \circ w \rangle$.

From this follows the middle-two interchange law: For $u, v, x, y \colon A \to B$ we have

To see this, observe first that for $a, b \colon A \to D$ and $c, d \colon D \to B$, items 3 and 4 give

and then specialize to $D = B \times B$, $a = \langle u, v \rangle$, $b = \langle x, y \rangle$, $c = p_1$, $d = p_2$.

Step 6 (the abelian group structure). Define

Then $x + 0 = x - 0 = x$, by items 2 and 1 of Step 5, and

by items 1, 2 and the interchange law. Applying the interchange law three times gives

Setting here $u = y = 0$ shows that $+$ is commutative, and setting $v = 0$ shows that it is associative (this is the Eckmann-Hilton argument). Moreover, $(0 - x) + x = (0 - x) - (0 - x) = 0$, so that every $x$ has the inverse $-x \coloneqq 0 - x$ (and $x + (-y) = x - y$, by the formula for $0 + x$ above). Hence each hom-set $C(A,B)$ is an abelian group, whose neutral element is the zero morphism. Finally, composition is bilinear, since by items 3 and 4 of Step 5

and similarly $(x + y) \circ w = x \circ w + y \circ w$. This is an $Ab$-enrichment of $C$.

Step 7 (abelianness). With respect to this enrichment, $\iota_1 \circ p_1 + \iota_2 \circ p_2 = \mathrm{id}_{A \times B}$, as seen by composing with $p_1$ and with $p_2$. Together with $p_1 \circ \iota_1 = \mathrm{id}_A$, $p_2 \circ \iota_2 = \mathrm{id}_B$ and $p_2 \circ \iota_1 = 0$, $p_1 \circ \iota_2 = 0$, this means that binary products are biproducts, so that $C$ is an additive category, hence (having kernels and cokernels) a pre-abelian category in which, by assumption, every monomorphism is a kernel and every epimorphism is a cokernel. Therefore $C$ is abelian, by the second condition in the definition above.

Step 8 (uniqueness). Let $+'$ be any $Ab$-enrichment of $C$. Its neutral elements are the zero morphisms, because $C(A,0)$ is a singleton, so that $A \to 0$ is neutral, and composition is bilinear. Consider

By bilinearity of composition we have, for $x, y \colon A \to B$,

as well as $\nabla' \circ \iota_1 = \mathrm{id}_B +' 0 = \mathrm{id}_B$ and similarly $\nabla' \circ \iota_2 = \mathrm{id}_B$. Hence, using Step 7,

and therefore $x +' y = x + y$.

Proof

This result can be found as Theorem 7.34 on page 150 of Peter Freyd’s book Abelian Categories. His terminology is a bit outdated, in that he calls an abelian category “fully abelian” if admits a full and faithful exact functor to a category of $R$-modules. See also the Wikipedia article for the idea of the proof.

Proof

This theorem, minus the explicit description of $R$, can be found as Exercise F on page 103 of Peter Freyd’s book Abelian Categories. The first part of this theorem can also be found as Prop. 2.1.7. of Victor Ginzburg’s Lectures on noncommutative geometry. Conversely, it is easy to see that $R$ is a compact projective generator of $R Mod$.