## 12.3 Preadditive and additive categories

Here is the definition of a preadditive category.

Definition 12.3.1. A category $\mathcal{A}$ is called *preadditive* if each morphism set $\mathop{Mor}\nolimits _\mathcal {A}(x, y)$ is endowed with the structure of an abelian group such that the compositions

\[ \mathop{Mor}\nolimits (x, y) \times \mathop{Mor}\nolimits (y, z) \longrightarrow \mathop{Mor}\nolimits (x, z) \]

are bilinear. A functor $F : \mathcal{A} \to \mathcal{B}$ of preadditive categories is called *additive* if and only if $F : \mathop{Mor}\nolimits (x, y) \to \mathop{Mor}\nolimits (F(x), F(y))$ is a homomorphism of abelian groups for all $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$.

In particular for every $x, y$ there exists at least one morphism $x \to y$, namely the zero map.

Lemma 12.3.2. Let $\mathcal{A}$ be a preadditive category. Let $x$ be an object of $\mathcal{A}$. The following are equivalent

$x$ is an initial object,

$x$ is a final object, and

$\text{id}_ x = 0$ in $\mathop{Mor}\nolimits _\mathcal {A}(x, x)$.

Furthermore, if such an object $0$ exists, then a morphism $\alpha : x \to y$ factors through $0$ if and only if $\alpha = 0$.

**Proof.**
First assume that $x$ is either (1) initial or (2) final. In both cases, it follows that $\mathop{Mor}\nolimits (x,x)$ is a trivial abelian group containing $\text{id}_ x$, thus $\text{id}_ x = 0$ in $\mathop{Mor}\nolimits (x, x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_ x = 0$ in $\mathop{Mor}\nolimits (x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$ and let $f \in \mathop{Mor}\nolimits (x ,y)$. Denote $C : \mathop{Mor}\nolimits (x,x) \times \mathop{Mor}\nolimits (x,y) \to \mathop{Mor}\nolimits (x,y)$ the composition map. Then $f = C(0, f)$ and since $C$ is bilinear we have $C(0, f) = 0$. Thus $f = 0$. Hence $x$ is initial in $\mathcal{A}$. A similar argument for $f \in \mathop{Mor}\nolimits (y, x)$ can be used to show that $x$ is also final. Thus (3) implies both (1) and (2).
$\square$

Definition 12.3.3. In a preadditive category $\mathcal{A}$ we call *zero object*, and we denote it $0$ any final and initial object as in Lemma 12.3.2 above.

Lemma 12.3.4. Let $\mathcal{A}$ be a preadditive category. Let $x, y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$. If the product $x \times y$ exists, then so does the coproduct $x \amalg y$. If the coproduct $x \amalg y$ exists, then so does the product $x \times y$. In this case also $x \amalg y \cong x \times y$.

**Proof.**
Suppose that $z = x \times y$ with projections $p : z \to x$ and $q : z \to y$. Denote $i : x \to z$ the morphism corresponding to $(1, 0)$. Denote $j : y \to z$ the morphism corresponding to $(0, 1)$. Thus we have the commutative diagram

\[ \xymatrix{ x \ar[rr]^1 \ar[rd]^ i & & x \\ & z \ar[ru]^ p \ar[rd]^ q & \\ y \ar[rr]^1 \ar[ru]^ j & & y } \]

where the diagonal compositions are zero. It follows that $i \circ p + j \circ q : z \to z$ is the identity since it is a morphism which upon composing with $p$ gives $p$ and upon composing with $q$ gives $q$. Suppose given morphisms $a : x \to w$ and $b : y \to w$. Then we can form the map $a \circ p + b \circ q : z \to w$. In this way we get a bijection $\mathop{Mor}\nolimits (z, w) = \mathop{Mor}\nolimits (x, w) \times \mathop{Mor}\nolimits (y, w)$ which show that $z = x \amalg y$.

We leave it to the reader to construct the morphisms $p, q$ given a coproduct $x \amalg y$ instead of a product.
$\square$

Definition 12.3.5. Given a pair of objects $x, y$ in a preadditive category $\mathcal{A}$, the *direct sum* $x \oplus y$ of $x$ and $y$ is the direct product $x \times y$ endowed with the morphisms $i, j, p, q$ as in Lemma 12.3.4 above.

Lemma 12.3.7. Let $\mathcal{A}$, $\mathcal{B}$ be preadditive categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an additive functor. Then $F$ transforms direct sums to direct sums and zero to zero.

**Proof.**
Suppose $F$ is additive. A direct sum $z$ of $x$ and $y$ is characterized by having morphisms $i : x \to z$, $j : y \to z$, $p : z \to x$ and $q : z \to y$ such that $p \circ i = \text{id}_ x$, $q \circ j = \text{id}_ y$, $p \circ j = 0$, $q \circ i = 0$ and $i \circ p + j \circ q = \text{id}_ z$, according to Remark 12.3.6. Clearly $F(x), F(y), F(z)$ and the morphisms $F(i), F(j), F(p), F(q)$ satisfy exactly the same relations (by additivity) and we see that $F(z)$ is a direct sum of $F(x)$ and $F(y)$. Hence, $F$ transforms direct sums to direct sums.

To see that $F$ transforms zero to zero, use the characterization (3) of the zero object in Lemma 12.3.2.
$\square$

Definition 12.3.8. A category $\mathcal{A}$ is called *additive* if it is preadditive and finite products exist, in other words it has a zero object and direct sums.

Namely the empty product is a finite product and if it exists, then it is a final object.

Definition 12.3.9. Let $\mathcal{A}$ be a preadditive category. Let $f : x \to y$ be a morphism.

A *kernel* of $f$ is a morphism $i : z \to x$ such that (a) $f \circ i = 0$ and (b) for any $i' : z' \to x$ such that $f \circ i' = 0$ there exists a unique morphism $g : z' \to z$ such that $i' = i \circ g$.

If the kernel of $f$ exists, then we denote this $\mathop{\mathrm{Ker}}(f) \to x$.

A *cokernel* of $f$ is a morphism $p : y \to z$ such that (a) $p \circ f = 0$ and (b) for any $p' : y \to z'$ such that $p' \circ f = 0$ there exists a unique morphism $g : z \to z'$ such that $p' = g \circ p$.

If a cokernel of $f$ exists we denote this $y \to \mathop{\mathrm{Coker}}(f)$.

If a kernel of $f$ exists, then a *coimage of $f$* is a cokernel for the morphism $\mathop{\mathrm{Ker}}(f) \to x$.

If a kernel and coimage exist then we denote this $x \to \mathop{\mathrm{Coim}}(f)$.

If a cokernel of $f$ exists, then the *image of $f$* is a kernel of the morphism $y \to \mathop{\mathrm{Coker}}(f)$.

If a cokernel and image of $f$ exist then we denote this $\mathop{\mathrm{Im}}(f) \to y$.

In the above definition, we have spoken of “the kernel” and “the cokernel”, tacitly using their uniqueness up to unique isomorphism. This follows from the Yoneda lemma (Categories, Section 4.3) because the kernel of $f : x \to y$ represents the functor sending an object $z$ to the set $\mathop{\mathrm{Ker}}(\mathop{Mor}\nolimits _\mathcal {A}(z, x) \to \mathop{Mor}\nolimits _\mathcal {A}(z, y))$. The case of cokernels is dual.

We first relate the direct sum to kernels as follows.

Lemma 12.3.10. Let $\mathcal{C}$ be a preadditive category. Let $x \oplus y$ with morphisms $i, j, p, q$ as in Lemma 12.3.4 be a direct sum in $\mathcal{C}$. Then $i : x \to x \oplus y$ is a kernel of $q : x \oplus y \rightarrow y$. Dually, $p$ is a cokernel for $j$.

**Proof.**
Let $f : z' \to x \oplus y$ be a morphism such that $q \circ f = 0$. We have to show that there exists a unique morphism $g : z' \to x$ such that $f = i \circ g$. Since $i \circ p + j \circ q$ is the identity on $x \oplus y$ we see that

\[ f = (i \circ p + j \circ q) \circ f = i \circ p \circ f \]

and hence $g = p \circ f$ works. Uniqueness holds because $p \circ i$ is the identity on $x$. The proof of the second statement is dual.
$\square$

Lemma 12.3.11. Let $\mathcal{C}$ be a preadditive category. Let $f : x \to y$ be a morphism in $\mathcal{C}$.

If a kernel of $f$ exists, then this kernel is a monomorphism.

If a cokernel of $f$ exists, then this cokernel is an epimorphism.

If a kernel and coimage of $f$ exist, then the coimage is an epimorphism.

If a cokernel and image of $f$ exist, then the image is a monomorphism.

**Proof.**
Part (1) follows easily from the uniqueness required in the definition of a kernel. The proof of (2) is dual. Part (3) follows from (2), since the coimage is a cokernel. Similarly, (4) follows from (1).
$\square$

Lemma 12.3.12. Let $f : x \to y$ be a morphism in a preadditive category such that the kernel, cokernel, image and coimage all exist. Then $f$ can be factored uniquely as $x \to \mathop{\mathrm{Coim}}(f) \to \mathop{\mathrm{Im}}(f) \to y$.

**Proof.**
There is a canonical morphism $\mathop{\mathrm{Coim}}(f) \to y$ because $\mathop{\mathrm{Ker}}(f) \to x \to y$ is zero. The composition $\mathop{\mathrm{Coim}}(f) \to y \to \mathop{\mathrm{Coker}}(f)$ is zero, because it is the unique morphism which gives rise to the morphism $x \to y \to \mathop{\mathrm{Coker}}(f)$ which is zero (the uniqueness follows from Lemma 12.3.11 (3)). Hence $\mathop{\mathrm{Coim}}(f) \to y$ factors uniquely through $\mathop{\mathrm{Im}}(f) \to y$, which gives us the desired map.
$\square$

Example 12.3.13. Let $k$ be a field. Consider the category of filtered vector spaces over $k$. (See Definition 12.19.1.) Consider the filtered vector spaces $(V, F)$ and $(W, F)$ with $V = W = k$ and

\[ F^ iV = \left\{ \begin{matrix} V
& \text{if}
& i < 0
\\ 0
& \text{if}
& i \geq 0
\end{matrix} \right. \text{ and } F^ iW = \left\{ \begin{matrix} W
& \text{if}
& i \leq 0
\\ 0
& \text{if}
& i > 0
\end{matrix} \right. \]

The map $f : V \to W$ corresponding to $\text{id}_ k$ on the underlying vector spaces has trivial kernel and cokernel but is not an isomorphism. Note also that $\mathop{\mathrm{Coim}}(f) = V$ and $\mathop{\mathrm{Im}}(f) = W$. This means that the category of filtered vector spaces over $k$ is not abelian.

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