Here is the definition of a preadditive category.

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

$\mathop{\mathrm{Mor}}\nolimits (x, y) \times \mathop{\mathrm{Mor}}\nolimits (y, z) \longrightarrow \mathop{\mathrm{Mor}}\nolimits (x, z)$

are bilinear. A functor $F : \mathcal{A} \to \mathcal{B}$ of preadditive categories is called additive if and only if $F : \mathop{\mathrm{Mor}}\nolimits (x, y) \to \mathop{\mathrm{Mor}}\nolimits (F(x), F(y))$ 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

1. $x$ is an initial object,

2. $x$ is a final object, and

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

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 $\mathop{\mathrm{Mor}}\nolimits (x,x)$ is a trivial abelian group containing $\text{id}_ x$, thus $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x, x)$, which shows that each of (1) and (2) implies (3).

Now assume that $\text{id}_ x = 0$ in $\mathop{\mathrm{Mor}}\nolimits (x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$ and let $f \in \mathop{\mathrm{Mor}}\nolimits (x ,y)$. Denote $C : \mathop{\mathrm{Mor}}\nolimits (x,x) \times \mathop{\mathrm{Mor}}\nolimits (x,y) \to \mathop{\mathrm{Mor}}\nolimits (x,y)$ 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 $f \in \mathop{\mathrm{Mor}}\nolimits (y, x)$ 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 $\mathop{\mathrm{Mor}}\nolimits (z, w) = \mathop{\mathrm{Mor}}\nolimits (x, w) \times \mathop{\mathrm{Mor}}\nolimits (y, w)$ 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.

Remark 12.3.6. Note that the proof of Lemma 12.3.4 shows that given $p$ and $q$ the morphisms $i$, $j$ are uniquely determined by the rules $p \circ i = \text{id}_ x$, $q \circ j = \text{id}_ y$, $p \circ j = 0$, $q \circ i = 0$. Moreover, we automatically have $i \circ p + j \circ q = \text{id}_{x \oplus y}$. Similarly, given $i$, $j$ the morphisms $p$ and $q$ are uniquely determined. Finally, given objects $x, y, z$ and 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 $i \circ p + j \circ q = \text{id}_ z$, then $z$ is the direct sum of $x$ and $y$ with the four morphisms equal to $i, j, p, q$.

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.

1. 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 $i' = i \circ g$.

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

3. 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 $p' = g \circ p$.

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

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

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

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

8. 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 $\mathop{\mathrm{Ker}}(\mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, x) \to \mathop{\mathrm{Mor}}\nolimits _\mathcal {A}(z, y))$. 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 $\mathcal{C}$.

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

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

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

4. 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.

Comment #2243 by on

We should modify the statement of 12.3.5(grammar issue - not sure what the statement is).

Comment #3002 by Herman Rohrbach on

Following up on the above two comments:

Given a pair of objects $x,y$ in a pre-additive category $\mathcal{A}$, the \emph{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.

Comment #3003 by Herman Rohrbach on

Proof of lemma 12.3.2:

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

Now assume that $\text{id}_x = 0$ in $Mor(x,x)$. Let $y$ be an arbitrary object of $\mathcal{A}$, $f \in Mor(x,y)$ and $C: Mor(x,x) \times Mor(x,y) \rightarrow Mor(x,y)$ the composition map. Since $C$ is bilinear, it holds that which yields $f = 0$. Hence $x$ is initial in $\mathcal{A}$. A similar argument for $f \in Mor(y,x)$ can be used to show that $x$ is also final. Thus (3) implies both (1) and (2). $\Box$

Comment #3125 by on

@#3002. OK, I made the change in the text of the definition but I do not think it is very different at all. See change.

@#3003. OK, I added your proof of 12.3.2 because it was so short and I even shortened it a bit using that bilinear maps send zero, something to zero. See here. However, maybe this kind of lemma really doesn't need a proof?

Comment #7841 by Zhenhua Wu on

Comment #7842 by Zhenhua Wu on

Comment #8065 by on

Somebody else can write this up (also look at the comment of bouthier on Section 12.4). Please coordinate with me.

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