8. Additive Categories

8.1Preadditive Categories

定义 8.1.1. A category $\mathcal{C}$ is called preadditive if ${\mathrm{Hom}}_{\mathcal{C}}(A,B)$ is an additive abelian group for every objects $A,B$ of $\mathcal{C}$, and the composition of morphisms is biadditive, i.e.,$$\left(f_1+f_2\right)\circ g=\left(f_1\circ g\right)+\left(f_2\circ g\right),f\circ \left(g_1+g_2\right)=\left(f\circ g_1\right)+\left(f\circ g_2\right)$$for every suitable morphisms.

We denote by $0_{XY}:X\to Y$ the zero element of the abelian group ${\mathrm{Hom}}_{\mathcal{C}}(X,Y)$.

引理 8.1.2. Let $f:W\to X$ and $g:Y\to Z$ be morphisms of a preadditive category $\mathcal{C}$. Then $0_{XY}\circ f=0_{WY}$ and $g\circ 0_{XY}=0_{XZ}$.

证明. In order to show that an element of an abelian group is the zero element, it suffices to establish that its sum with itself is itself. Using the biadditivity of the composition, we have:$$\begin{array}{rl}& \left(0_{XY}\circ f\right)+\left(0_{XY}\circ f\right)=\left(0_{XY}+0_{XY}\right)\circ f=0_{XY}\circ f,\\ & \left(g\circ 0_{XY}\right)+\left(g\circ 0_{XY}\right)=g\circ \left(0_{XY}+0_{XY}\right)=g\circ 0_{XY},\end{array}$$which show the conclusion.

$\square$

命题 8.1.3. The following statements are equivalent for an object $Y$ of a preadditive category $\mathcal{C}$:

(1)	$Y$ is a zero object.
(2)	$Y$ is initial.
(3)	$Y$ is terminal.
(4)	${\mathrm{Hom}}_{\mathcal{C}}(Y,Y)$ is the zero group.
(5)	$0_{YY}=1_Y$.

证明. (1)$\Rightarrow$(2)$\Rightarrow$(4)$\Rightarrow$(5) and (1)$\Rightarrow$(3)$\Rightarrow$(4)$\Rightarrow$(5) These follow from the definition of everything involved: a zero object is by definition an initial and terminal object, initial and terminal objects by definition have a single endomorphism, and any two morphisms of the zero group are equal.

(5)$\Rightarrow$(1) Let $X$ and $Z$ be any objects of $\mathcal{C}$. For every arrows $f:X\to Y$ and $g:Y\to Z$, we have $f=1_Y\circ f=0_{YY}\circ f=0_{XY}$ and $g=g\circ 1_Y=g\circ 0_{YY}=0_{YZ}$ by Lemma 8.1.2. These imply that the morphism sets ${\mathrm{Hom}}_{\mathcal{C}}(X,Y)$ and ${\mathrm{Hom}}_{\mathcal{C}}(Y,Z)$ consist of precisely one morphism each. Because $X$ and $Z$ were arbitrary, we may conclude that $Y$ is a zero object.

$\square$

例 8.1.4. (1) As a first but not very typical example of a preadditive category, we consider any unitary ring $R$. We may thus consider the category ${\mathcal{C}}_R$ induced by the multiplicative monoid of the ring - in other words, the category with a single object $R$, and the ring elements as morphisms with composition given by ring multiplication. Endow the only morphism set ${\mathrm{Hom}}_{{\mathcal{C}}_R}(R,R)=R$ with the additive group structure of the ring. That is, addition of morphisms is done by adding them as ring elements. Note now that this indeed defines a preadditive category: the only thing to verify is that the composition is biadditive, but this is exactly the requirement that multiplication distributes over addition in the ring, which is one of the ring axioms. One may also note the converse, that any preadditive category with a single object gives rise to a ring of morphisms with addition given by morphism addition and multiplication given by composition. The converse generalizes to categories with more than one object: in any preadditive category $\mathcal{C}$, every endomorphism set is a unitary ring with the ring addition given by morphism addition and the ring multiplication given by morphism composition. Two takeaways from this are that unitary rings may be thought of as precisely preadditive categories with a single object and that preadditive categories generalize unitary rings in the same way that arbitrary categories generalize monoids.

(2) We may by Proposition 8.1.3 conclude that there is no preadditive structure that Set can be endowed with to make a preadditive category, because it has initial objects that are not terminal and viceversa. Alternatively, note that there are empty morphism sets in $\mathbf{Set}$, namely ${\mathrm{Hom}}_{\mathbf{Set}}(X,\varnothing )$ for any non-empty set $X$, while the morphism sets are non-empty in preadditive categories, because they are groups.

(3) $\mathbf{Mon}$, $\mathbf{Ring}$, $\mathbf{Pos}$ and $\mathbf{Top}$ are not preadditive, because they do not have a zero object. One may also show that $\mathbf{Grp}$ is not preadditive.

命题 8.1.5. Let $f:X\to Y$ be a morphism in a preadditive category. Then $f$ is a monomorphism if and only if $0:0\to X$ is a kernel of $f$.

Dually, $f$ is an epimorphism if and only if $0:Y\to 0$ is a cokernel of $f$.

证明. Assume first that $f$ is a monomorphism. Clearly, $f\circ 0=0$. Next let $l:L\to X$ be a morphism such that $f\circ l=0$. Then $f\circ l=f\circ 0$, which implies that $l=0$, because $f$ is a monomorphism.

Then there is a unique morphism $g:L\to 0$, namely $g=0$, such that $0\circ g=l=0$. Hence $0:0\to X$ is a kernel of $f$.

Conversely, assume that $0:0\to X$ is a kernel of $f$. Let $l_1,l_2:L\to X$ be morphisms such that $f\circ l_1=f\circ l_2$. Then $f\circ \left(l_1-l_2\right)=0$.

Since $0:0\to X$ is a kernel of $f$, there is a unique morphism $g:L\to 0$ such that $0\circ g=l_1-l_2$. Hence $l_1=l_2$, which shows that $f$ is a monomorphism.

The statement on epimorphisms follows by duality.

$\square$

命题 8.1.6. In a preadditive category, the Hom functors may be viewed as functors into the category of all abelian groups rather than just sets. In other words, for any preadditive category $\mathcal{C}$ and fixed object $A$ of $\mathcal{C}$, we have ${\mathrm{Hom}}_{\mathcal{C}}(A,-):\mathcal{C}\to \mathbf{Ab}$ and ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Ab}$.

证明. The morphism sets in a preadditive category are abelian groups, so the mapping of the objects is sound. It remains to be shown that morphisms in $\mathcal{C}$ are mapped to morphisms in $\mathbf{Ab}$. For any morphism $f:X\to Y$, consider the function ${\mathrm{Hom}}_{\mathcal{C}}(A,f):{\mathrm{Hom}}_{\mathcal{C}}(A,X)\to {\mathrm{Hom}}_{\mathcal{C}}(A,Y)$ and verify that it is additive. For any morphisms $g,h\in {\mathrm{Hom}}_{\mathcal{C}}(A,X)$ we have:$$\left[{\mathrm{Hom}}_{\mathcal{C}}(A,f)\right](g+h)=f\circ (g+h)=(f\circ g)+(f\circ h)=\left[{\mathrm{Hom}}_{\mathcal{C}}(A,f)\right](g)+\left[{\mathrm{Hom}}_{\mathcal{C}}(A,f)\right](h).$$Similarly, we get for the contravariant Hom functor a function ${\mathrm{Hom}}_{\mathcal{C}}(f,A):{\mathrm{Hom}}_{\mathcal{C}}(Y,A)\to {\mathrm{Hom}}_{\mathcal{C}}(X,A)$, and for any morphisms $g,h\in {\mathrm{Hom}}_{\mathcal{C}}(Y,A)$ we find that:$$\left[{\mathrm{Hom}}_{\mathcal{C}}(f,A)\right](g+h)=(g+h)\circ f=(g\circ f)+(h\circ f)=\left[{\mathrm{Hom}}_{\mathcal{C}}(f,A)\right](g)+\left[{\mathrm{Hom}}_{\mathcal{C}}(f,A)\right](h).$$Hence every morphism in $\mathcal{C}$ is mapped to a suitable morphism in $\mathbf{Ab}$.

$\square$

8.2Additive Categories

定义 8.2.1. An additive category is a preadditive category with a zero object in which every pair of objects has a product.

注 8.2.2. The requirement that there exists a zero object may equivalently be viewed as a requirement that there are empty products. By this observation, we may equivalently define an additive category as a preadditive category in which every finite collection of objects has a product. Later on, we shall see that “product” may be replaced by “coproduct” and even “biproduct” in the definition of an additive category, which will explain the apparent asymmetry of the definition.

例 8.2.3. Let us show that $\mathbf{Ab}$ is an additive category.

The zero group $\{0\}$ is a zero object of $\mathbf{Ab}$. Any two objects of $\mathbf{Ab}$ have a product, namely the direct product of the groups with the canonical projections.

Claim 8.2.4. ${\mathrm{Hom}}_{\mathbf{Ab}}(M,N)$ is an abelian group for any abelian groups $M$ and $N$.

All the necessary properties for the morphisms are inherited from the abelian group structure on $N$. The addition is well-defined in the sense that a sum of two morphisms is a morphism:$$(f+g)(m+n)=f(m+n)+g(m+n)=f(m)+f(n)+g(m)+g(n)=f(m+n)+g(m+n)$$for any $m,n\in M$. The remaining properties are verified in a similar fashion using the fact that the zero morphism and valuewise subtraction of a morphism are both morphisms.

Claim 8.2.5. Composition of morphisms is biadditive in $\mathbf{Ab}$.

We need to prove that$$\left(f_1+f_2\right)\circ \left(g_1+g_2\right)=\left(f_1\circ g_1\right)+\left(f_1\circ g_2\right)+\left(f_2\circ g_1\right)+\left(f_2\circ g_2\right)$$holds for any suitable morphisms $f_1,f_2,g_1$ and $g_2$ in $\mathbf{Ab}$. For any $m$ in the domain of $g_1$ and $g_2$, we have:$$\begin{array}{rl}\left[\left(f_1+f_2\right)\circ \left(g_1+g_2\right)\right](m)& =\left(f_1+f_2\right)\left[\left(g_1+g_2\right)(m)\right]=\left(f_1+f_2\right)\left[g_1(m)+g_2(m)\right]\\ & =f_1\left[g_1(m)+g_2(m)\right]+f_2\left[g_1(m)+g_2(m)\right]\\ & =f_1\left[g_1(m)\right]+f_1\left[g_2(m)\right]+f_2\left[g_1(m)\right]+f_2\left[g_2(m)\right]\\ & =\left(f_1\circ g_1\right)(m)+\left(f_1\circ g_2\right)(m)+\left(f_2\circ g_1\right)(m)+\left(f_2\circ g_2\right)(m).\end{array}$$Now the proposition follows from the claims and the existence of zero objects and products.

例 8.2.6. (1) Additive categories: $\mathbf{Vect}(K)$, the category $\mathcal{T}$ of torsion abelian groups.

(2) Any unitary ring $R$ viewed as a category with only one object is a preadditive category, which is not additive, because it does not have a zero object.

(3) Non-additive categories: $\mathbf{Set}$, $\mathbf{Mon}$, $\mathbf{Ring}$, $\mathbf{Pos}$, $\mathbf{Top}$ (they do not have a zero object), $\mathbf{Grp}$ (finite products and coproducts do not coincide - see Proposition 8.2.7).

命题 8.2.7. Let $X_1,X_2,\dots ,X_n$ be objects of some preadditive category $\mathcal{C}$, which have a product $X=\prod _{j=1}^n{X}_j$ together with ${\pi }_j:X\to X_j$ (for $j=1,\dots ,n$). Set (for $i=1,\dots ,n$) ${\iota }_i={\left({\delta }_{ij}\right)}_{j=1}^n:X_i\to \prod _{j=1}^n{X}_j$, where ${\delta }_{ij}:X_i\to X_j$ is the morphism$${\delta }_{ij}=\{\begin{array}{ll}{1}_{X_i}& \text{if}i=j\\ 0_{X_i{X}_j}& \text{if}i\ne j\end{array}$$That is, ${\iota }_i$ is the morphism induced by a family consisting of a single identity morphism and the rest zero elements of the morphism groups. Then $\left(\prod _{j=1}^n{X}_j,{\left({\iota }_j\right)}_{j=1}^n\right)$ is a coproduct of ${\left(X_j\right)}_{j=1}^n$.

Before proving the proposition, we will state and prove a useful result.

命题 8.2.8. ${\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n=1_X$.

证明. Consider the morphism $\left({\pi }_1,\dots ,{\pi }_n\right):X\to X$, i.e., the unique morphism through which ${\pi }_j$ (for $j=1,\dots ,n$) factors as ${\pi }_j={\pi }_j\circ \left({\pi }_1,\dots ,{\pi }_n\right)$. On one hand, it is clearly the identity morphism $1_X$ for it has the required factorization property. On the other hand, we may show that it coincides with ${\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n$, because it too has the required factorization property: use the definition of all the ${\iota }_i$ in terms of the identity morphism and zero elements and the fact that zero elements compose to zero elements to get:$$\begin{array}{rl}{\pi }_j\circ \left({\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n\right)& ={\pi }_j\circ {\iota }_1\circ {\pi }_1+\cdots +{\pi }_j\circ {\iota }_n\circ {\pi }_n\\ & ={\pi }_j\circ {\iota }_1\circ {\pi }_1+\cdots +{\pi }_j\circ {\iota }_j\circ {\pi }_j+\cdots +{\pi }_j\circ {\iota }_n\circ {\pi }_n\\ & =0_{XiX}\circ {\pi }_1+\cdots +1_{Xj}\circ {\pi }_j+\cdots +0_{Xn{X}_j}\circ {\pi }_n\\ & =0_{X{X}_j}+\cdots +{\pi }_j+\cdots +0_{X{X}_j}={\pi }_j.\end{array}$$This shows that ${\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n=1_X$.

$\square$

Now we continue to prove the Proposition 8.2.7.

证明. Let $f_i:X_i\to Z$ (for $i=1,\dots ,n$) be a family of arbitrary morphisms into some object $Z$. We need to find a morphism $f:X\to Z$ through which all $f_i$ factor via the proposed inclusion morphisms ${\iota }_i$, and we need to prove that $f$ is unique with this property. Consider $f=f_1\circ {\pi }_1+\cdots +f_n\circ {\pi }_n$. From the definition of the ${\iota }_i$ in terms of identity and zero elements, we see that $f$ works: for any $j=1,\dots ,n$,$$\begin{array}{rl}f\circ {\iota }_j& =\left(f_1\circ {\pi }_1+\cdots +f_n\circ {\pi }_n\right)\circ {\iota }_j=f_1\circ {\pi }_1\circ {\iota }_j+\cdots +f_n\circ {\pi }_n\circ {\iota }_j\\ & =f_1\circ {\pi }_1\circ {\iota }_j+\cdots +f_j\circ {\pi }_j\circ {\iota }_j+\cdots +f_n\circ {\pi }_n\circ {\iota }_j\\ & =f_1\circ 0_{X_j{X}_1}+\cdots +f_j\circ 1_{X_j}+\cdots +f_n\circ 0_{X_j{X}_n}\\ & =0_{X_{j}Z}+\cdots +f_j+\cdots +0_{X_{j}Z}=f_j.\end{array}$$To prove uniqueness of this $f$, let $f^{\prime }$ be any morphism with the same factorization property as $f$. By Proposition 8.2.8, we may rewrite $f$ and $f^{\prime }$ and use that the $f_i$ factor through them:$$\begin{array}{rl}f& =f\circ 1_X=f\circ \left({\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n\right)=f\circ {\iota }_1\circ {\pi }_1+\cdots +f\circ {\iota }_n\circ {\pi }_n\\ & =f_1\circ {\pi }_1+\cdots +f_n\circ {\pi }_n=f^{\prime }\circ {\iota }_1\circ {\pi }_1+\cdots +f^{\prime }\circ {\iota }_n\circ {\pi }_n\\ & =f^{\prime }\circ \left({\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n\right)=f^{\prime }\circ 1_X=f^{\prime }.\end{array}$$Thus we have shown that $X=\prod _{j=1}^n{X}_j$ together with the ${\iota }_j$ (for $j=1,\dots ,n$ ) is a coproduct of the objects $X_1,\dots ,X_n$.

$\square$

注 8.2.9. As a consequence of the proposition, one may equivalently define additive categories to also have coproducts for any finite collection of objects. Thus, the dual of an additive category is an additive category (where the morphism addition in the dual category is understood to be the very same as in the original category).

8.3Biproducts

Using Proposition 8.2.7 and its dual, we may conclude that coproducts and products coincide as objects in an additive category and may hence consider the objects equipped with both projection and inclusion morphisms. This notion of a simultaneous product and coproduct turns out to be useful in the study of additive categories and is captured in the following definition.

定义 8.3.1. Let $\mathcal{C}$ be a preadditive category and let $X_1,\dots ,X_n$ be a finite collection of objects in $\mathcal{C}$. A biproduct of $X_1,\dots ,X_n$ is an object $X$ together with morphisms ${\pi }_1,\dots ,{\pi }_n$ (called projection morphisms) and ${\iota }_1,\dots ,{\iota }_n$ (called inclusion morphisms) satisfying for every $i,j\in \{1,\dots ,n\}:{\pi }_j\circ {\iota }_i={\delta }_{ij}$ and ${\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n=1_X$.

注 8.3.2. The notion of a biproduct is essentially self-dual in the sense that if $X$ together with ${\pi }_1,\dots ,{\pi }_n$ and ${\iota }_1,\dots ,{\iota }_n$ is a biproduct of $X_1,\dots ,X_n$ in some preadditive category $\mathcal{C}$, then $X$ together with ${\iota }_1,\dots ,{\iota }_n$ and ${\pi }_1,\dots ,{\pi }_n$ is a biproduct of the same objects $X_1,\dots ,X_n$ in ${\mathcal{C}}^{\mathrm{op}}$ (with the same morphism addition).

注 8.3.3. In terms of biproducts, Proposition 8.2.8 in the setting of Proposition 8.2.7 may be interpreted as saying that any product (or, by duality, coproduct) in a preadditive category may be expanded to a biproduct.

注 8.3.4. One may note that the definition of the biproduct of a family of objects conveniently refers only to these objects and the biproduct object (the definition is “internal”), whereas the definition of the product and coproduct in terms of their universal properties refers to virtually every object of the category (the definition is “external”).

By the very definition of the biproduct and in accordance with the previous remark, it is not obvious that every biproduct may be viewed as a simultaneous product and coproduct; the biproduct would seem to be more general than that. The following proposition shows, however, that biproducts indeed are simultaneous products and coproducts (with the obvious morphisms). Hence, finite products, finite coproducts, and biproducts coincide in preadditive categories.

命题 8.3.5. Let $X$ together with ${\pi }_1,\dots ,{\pi }_n$ and ${\iota }_1,\dots ,{\iota }_n$ be a biproduct of some objects $X_1,\dots ,X_n$ in a preadditive category. Then:

(1)	$\left(X,\left({\pi }_1,\dots ,{\pi }_n\right)\right)$ is a product of $X_1,\dots ,X_n$;
(2)	$\left(X,\left({\iota }_1,\dots ,{\iota }_n\right)\right)$ is a coproduct of $X_1,\dots ,X_n$.

证明. In order to prove that $\left(X,\left({\iota }_1,\dots ,{\iota }_n\right)\right)$ is a coproduct, we may proceed exactly as in the proof of Proposition 8.2.7, where the identities ${\pi }_j\circ {\iota }_i={\delta }_{ij}$ and ${\iota }_1\circ {\pi }_1+\cdots +{\iota }_n\circ {\pi }_n=1_X$ were sufficient to prove the universal property of the coproduct (the former implied existence and the latter uniqueness of the factoring morphism). It follows by the above and that biproducts are “essentially” self-dual (Remark 8.3.2) that $(X,\left({\pi }_1,\dots ,{\pi }_n\right))$ is a coproduct in the opposite category and hence a product in the original category.

$\square$

The following corollary to Propositions 8.2.7 and 8.3.5 summarizes the relationship between products, coproducts, and biproducts in a preadditive category.

推论 8.3.6. For a collection $X_1,\dots ,X_n$ of objects in a preadditive category, $X_1,\dots ,X_n$ have a product if and only if they have a coproduct if and only if they have a biproduct.

例 8.3.7. Recall that the direct sum and the direct product of finitely many abelian groups $M_1,\dots ,M_n$ coincide. Interpreted categorically, we see that this is no coincidence: we know $\mathbf{Ab}$ to be additive Example 8.2.3) and that the direct sum and direct product together with their inclusion and projection morphisms are a coproduct and, respectively, a product. By the uniqueness of products and coproducts and the above propositions, we may conclude that the direct sum and the direct product would at the very least have to be isomorphic. In fact, the direct sum and the direct product equipped with their canonical morphisms is a biproduct, which we may verify with ease. For any $i\in \{1,\dots ,n\}$, consider the morphism ${\pi }_i{\iota }_i:M_i\to M_i$ and show that it is the identity morphism by noting that it fixes any element $m_i\in M_i$:$$\left({\pi }_i{\iota }_i\right)\left(m_i\right)={\pi }_i\left(\left(0,\dots ,0,m_i,0,\dots ,0\right)\right)=m_i.$$Also, ${\pi }_j{\iota }_i=0_{M_i{M}_j}$ for every $j\ne i$. Next, consider$${\iota }_1{\pi }_1+\cdots +{\iota }_n{\pi }_n:\underset{i=1}{\overset{n}{⨁}}{M}_i\to \underset{i=1}{\overset{n}{⨁}}{M}_i$$and show that it is the identity morphism by noting that it too fixes any element, $\left(m_1,\dots ,m_n\right)$ say:$$\begin{array}{rl}\left({\iota }_1{\pi }_1+\cdots +{\iota }_n{\pi }_n\right)\left(\left(m_1,\dots ,m_n\right)\right)& =\left({\iota }_1{\pi }_1\right)\left(\left(m_1,\dots ,m_n\right)\right)+\cdots +\left({\iota }_n{\pi }_n\right)\left(\left(m_1,\dots ,m_n\right)\right)\\ & ={\iota }_1\left(m_1\right)+\cdots +{\iota }_n\left(m_n\right)\\ & =\left(m_1,0,\dots ,0\right)+\cdots +\left(0,\dots ,0,m_n\right)\\ & =\left(m_1,\dots ,m_n\right).\end{array}$$

命题 8.3.8. If $\left(X,{\left({\pi }_i\right)}_{i=1}^n,{\left({\iota }_i\right)}_{i=1}^n\right)$ and $\left(X^{\prime },{\left({\pi }_i^{\prime }\right)}_{i=1}^n,{\left({\iota }_i^{\prime }\right)}_{i=1}^n\right)$ are biproducts of the same family of objects $X_1,\dots ,X_n$, then there is a unique isomorphism $\phi :X\to X^{\prime }$ that respects the projection and inclusion morphisms, i.e., with ${\pi }_i={\pi }_i^{\prime }\circ \phi$ and ${\iota }_i^{\prime }=\phi \circ {\iota }_i$ for every $i$.

证明. Proof. For uniqueness, we may view the biproducts as products (or coproducts) via Proposition 8.3.5 and use the uniqueness of products (or coproducts) up to isomorphism to conclude that there is precisely one isomorphism that respects the projection morphisms (or inclusion morphisms), and hence that there is at most one isomorphism that respects both projection and inclusion morphisms. We may also show the uniqueness more directly as follows:$$\phi =1_{X^{\prime }}\circ \phi =\left(\sum _{i=1}^n{\iota }_i^{\prime }\circ {\pi }_i^{\prime }\right)\circ \phi =\sum _{i=1}^n\left({\iota }_i^{\prime }\circ {\pi }_i^{\prime }\circ \phi \right)=\sum _{i=1}^n\left({\iota }_i^{\prime }\circ {\pi }_i\right).$$Alternatively, we have:$$\phi =\phi \circ 1_X=\phi \circ \left(\sum _{i=1}^n{\iota }_i\circ {\pi }_i\right)=\sum _{i=1}^n\left(\phi \circ {\iota }_i\circ {\pi }_i\right)=\sum _{i=1}^n\left({\iota }_i^{\prime }\circ {\pi }_i\right).$$

For existence, we may invoke Proposition 8.3.5 twice (once for the product and once for the coproduct). This yields two isomorphisms $\phi ,{\phi }^{\prime }:X\to X^{\prime }$ with $\phi$ respecting the projection morphisms and ${\phi }^{\prime }$ respecting the inclusion morphisms. By the chain of equalities above, $\phi =$ $\sum _{i=1}^n\left({\iota }_i^{\prime }\circ {\pi }_i\right)={\phi }^{\prime }$. Hence the isomorphisms are the same and they respect both the projection morphisms and the inclusion morphisms. Another more explicit approach for the existence is to show that $\phi =\sum _{i=1}^n\left({\iota }_i^{\prime }\circ {\pi }_i\right)$ respects the projection and inclusion morphisms and to show that the morphism $\psi =\sum _{i=1}^n\left({\iota }_i\circ {\pi }_i^{\prime }\right)$ is a two-sided inverse of $\phi$. In any case, we have shown existence and uniqueness of an isomorphism between the biproduct objects that respects the projection and inclusion morphisms, which proves the proposition.

$\square$

注 8.3.9. Much like for products and coproducts, the uniqueness of the biproduct object up to isomorphism guaranteed by Proposition 8.3.8 justifies the notation $\underset{i=1}{\overset{n}{⨁}}{X}_i$ for the object of any biproduct of $X_1,\dots ,X_n$. Somewhat sloppily, we will often use $\underset{i=1}{\overset{n}{⨁}}{X}_i$ to refer to the entire biproduct (including the projection and inclusion morphisms).

8.4Matrix Notation

The idea of viewing morphisms as matrices, with addition and composition of morphisms as matrix addition and multiplication, is fruitful not only in linear algebra (i.e., in the category $\mathbf{Vect}(K)$), but in any additive category.

命题 8.4.1. Let $X=\underset{j=1}{\overset{n}{⨁}}{X}_j$ and $Y=\underset{i=1}{\overset{m}{⨁}}{Y}_i$ be biproducts in an additive category $\mathcal{A}$. A morphism $\phi :X\to Y$ induces a family of morphisms $\left\{{\phi }_{ij}\mid i=1,\dots ,m;j=1,\dots ,n\right\}$ with ${\phi }_{ij}:X_j\to Y_i$ via ${\phi }_{ij}={\pi }_i\circ \phi \circ {\iota }_j$. Conversely, any family of morphisms $\{{\phi }_{ij}\mid i=1,\dots ,m;j=$ $1,\dots ,n\}$ with ${\phi }_{ij}:X_j\to Y_i$ induces a unique map $\phi :X\to Y$ via the universal properties of the biproducts viewed as a coproduct and a product. The setting is depicted in the following diagram:

Given ${\phi }_{ij}$, one gets ${\phi }_i$ by universal mapping property of the coproduct $\underset{j=1}{\overset{n}{⨁}}{X}_j$, and then $\phi$ by universal mapping property of the product $\underset{i=1}{\overset{m}{⨁}}{Y}_i$. For any morphism $\phi$ as above, arrange its corresponding morphisms into a matrix $[\phi ]$ as follows:$$[\phi ]:=\left[\begin{array}{ccc}{\phi }_{11}& \cdots & {\phi }_{1n}\\ ⋮& \ddots & ⋮\\ {\phi }_{m1}& \cdots & {\phi }_{mn}\end{array}\right]$$

(1)

The correspondence is an isomorphism between the abelian groups ${\mathrm{Hom}}_{\mathcal{A}}(X,Y)$ and $\underset{j=1}{\overset{n}{⨁}}\underset{i=1}{\overset{m}{⨁}}{\mathrm{Hom}}_{\mathcal{A}}\left(X_j,Y_i\right)$. In particular, the matrix of a sum is obtained by matrix addition: $$\left[\phi +{\phi }^{\prime }\right]=[\phi ]+\left[{\phi }^{\prime }\right],$$where the addition of matrix entries in the right-hand side is just the morphism addition in the additive category.

(2)

Let $W=\underset{k=1}{\overset{p}{⨁}}{W}_k$ be a third biproduct. If $\psi :W\to X$ and $\phi :X\to Y$ are morphisms, then $$(\phi \circ \psi {)}_{ik}=\sum _{j=1}^n{\phi }_{ij}\circ {\psi }_{jk}.$$In other words, the matrix of the composition $\phi \circ \psi$ is obtained by matrix multiplication $$[\phi \circ \psi ]=[\phi ]\cdot [\psi ],$$where multiplication of the entries is done by composition and addition is the addition on morphisms in the additive category.

Pullbacks and pushouts exhibit a connection with kernels and cokernels in additive categories.

命题 8.4.2. Consider in an additive category a square and the following diagonal morphisms that it gives rise to:

Then the square is:

(1)	a pullback if and only if $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$ is a kernel of $[\phi -\psi ]$.
(2)	a pushout if and only if $[\phi -\psi ]$ is a cokernel of $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$.

证明. We only prove the first statement, the other one following by duality.

Assume first that the square is a pullback. Then $[\phi -\psi ]\cdot \left[\begin{array}{c}\alpha \\ \beta \end{array}\right]=\phi \circ \alpha -\psi \circ \beta =0$. Next let $\left[\begin{array}{c}\gamma \\ \delta \end{array}\right]:X^{\prime }\to A\oplus B$ be such that $[\phi -\psi ]\cdot \left[\begin{array}{c}\gamma \\ \delta \end{array}\right]=0$, that is, $\phi \circ \gamma =\psi \circ \delta$. By the universal mapping property of the pullback, there is a unique morphism $\chi :X^{\prime }\to X$ such that $\alpha \circ \chi =\gamma$ and $\beta \circ \chi =\delta$, that is, $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]\cdot [\chi ]=\left[\begin{array}{c}\alpha \circ \chi \\ \beta \circ \chi \end{array}\right]=\left[\begin{array}{c}\gamma \\ \delta \end{array}\right]$. Hence $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$ is a kernel of $[\phi -\psi ]$.

Conversely, assume that $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]$ is a kernel of $[\phi -\psi ]$. Then $\phi \circ \alpha -\psi \circ \beta =[\phi -\psi ]\cdot \left[\begin{array}{c}\alpha \\ \beta \end{array}\right]=0$, hence $\phi \circ \alpha =\psi \circ \beta$. Next let $\gamma :X^{\prime }\to A$ and $\delta :X^{\prime }\to B$ be morphisms such that $\phi \circ \gamma =\psi \circ \delta$. Then $[\phi -\psi ]\cdot \left[\begin{array}{c}\gamma \\ \delta \end{array}\right]=\phi \circ \gamma -\psi \circ \delta =0$. By the universal mapping property of the kernel, there is a unique morphism $\chi :X^{\prime }\to X$ such that $\left[\begin{array}{c}\alpha \\ \beta \end{array}\right]\cdot [\chi ]=\left[\begin{array}{c}\gamma \\ \delta \end{array}\right]$, that is, $\alpha \circ \chi =\gamma$ and $\beta \circ \chi =\delta$. Hence the square is a pullback.

$\square$

注 8.4.3. The placement of the minus sign in front of $\psi$ in the matrices for the diagonal morphisms is arbitrary. As noted in the end of the proof, the minus could equally well be placed in front of $\beta$, and one readily shows in similar fashion that placing it in front of either $\alpha$ or $\phi$ instead yields an equivalent statement.

8.5Additive Functors

定义 8.5.1. Let $\mathcal{C}$ and $\mathcal{D}$ be preadditive categories. Then a functor $F:\mathcal{C}\to \mathcal{D}$ is called additive if $F(f+g)=F(f)+F(g)$ for any suitable morphisms $f$ and $g$.

注 8.5.2. (1) If $F:\mathcal{C}\to \mathcal{D}$ is an additive functor, then $F\left(0_{XY}\right)=0_{XY}$ for every objects $X$, $Y$ of $\mathcal{C}$, and $F(-f)=-F(f)$ for every morphism $f$ in $\mathcal{C}$.

(2) The composition of two additive functors is again an additive functor.

命题 8.5.3. If $F:\mathcal{C}\to \mathcal{D}$ is an additive functor between preadditive categories and $X$ is a zero object of $\mathcal{C}$, then $F(X)$ is a zero object of $\mathcal{D}$.

证明. Using the characterization of zero objects from Proposition 8.1.3, we have $0_{XX}=1_X$. Applying $F$, we have $0_{F(X)F(X)}=1_{F(X)}$, which is precisely to say that $F(X)$ is a zero object (using the characterization again).

$\square$

命题 8.5.4. The following are equivalent for a functor $F:\mathcal{A}\to \mathcal{B}$ between additive categories:

(1)	$F$ is additive.
(2)	$F$ respects binary products. In other words, if $\left(X\prod Y,\left({\pi }_X,{\pi }_Y\right)\right)$ is a product of $X,Y\in \mathcal{A}$, then $\left(F\left(X\prod Y\right),\left(F\left({\pi }_X\right),F\left({\pi }_Y\right)\right)\right)$ is a product of $F(X),F(Y)\in \mathcal{B}$.
(3)	$F$ respects binary coproducts. In other words, if $\left(X\coprod Y,\left({\iota }_X,{\iota }_Y\right)\right)$ is a coproduct of $X,Y\in \mathcal{A}$, then $\left(F(X\coprod Y),\left(F\left({\iota }_X\right),F\left({\iota }_Y\right)\right)\right)$ is a coproduct of $F(X),F(Y)\in \mathcal{B}$.
(4)	$F$ respects binary biproducts. In other words, if $\left(X\oplus Y,\left({\pi }_X,{\pi }_Y\right),\left({\iota }_X,{\iota }_Y\right)\right)$ is a biproduct of $X,Y\in \mathcal{A}$, then $\left(F(X\oplus Y),\left(F\left({\pi }_X\right),F\left({\pi }_Y\right)\right),\left(F\left({\iota }_X\right),F\left({\iota }_Y\right)\right)\right)$ is a biproduct of $F(X),F(Y)\in$ $\mathcal{B}$.

证明. (1)$\Rightarrow$(4) This follows readily from the fact that every additive functor respects the ingredients of the biproduct definition: identity morphisms, zero elements of the morphism groups, and morphism composition and addition. Let us verify this by assuming that $F$ is additive, applying it to a biproduct $\left(X\oplus Y,\left({\pi }_X,{\pi }_Y\right),\left({\iota }_X,{\iota }_Y\right)\right)$ in $\mathcal{A}$, and seeing that a biproduct in $\mathcal{B}$ is indeed obtained. We have:$$\begin{array}{c}F\left({\pi }_X\right)\circ F\left({\iota }_X\right)=F\left({\pi }_X\circ {\iota }_X\right)=F\left(1_X\right)=1_{F(X)},\\ F\left({\pi }_Y\right)\circ F\left({\iota }_X\right)=F\left({\pi }_Y\circ {\iota }_X\right)=F\left(0_{XY}\right)=0_{F(X)F(Y)}.\end{array}$$Finally, the sum of compositions in the opposite order is the identity on the biproduct object:$$F\left({\iota }_X\right)F\left({\pi }_X\right)+F\left({\iota }_Y\right)F\left({\pi }_Y\right)=F\left({\iota }_X{\pi }_X\right)+F\left({\iota }_Y{\pi }_Y\right)=F\left({\iota }_X{\pi }_X+{\iota }_Y{\pi }_Y\right)=F\left(1_{X\oplus Y}\right)=1_{F(X\oplus Y)}.$$Thus we have shown that an additive functor respects binary biproducts.

(4)$\Rightarrow$(2) and (4)$\Rightarrow$(3) These are immediate by Remark 8.3.3 and Proposition 8.3.8 stating that products and coproducts may be expanded to biproducts and that a biproduct may be viewed as a product and coproduct. Given a (binary) product or coproduct in $\mathcal{A}$, expand it to a biproduct, use the assumption that $F$ respects biproducts to obtain a biproduct in $\mathcal{B}$, and view this biproduct as a product or coproduct in $\mathcal{B}$ (which will be of the required form, i.e., with the appropriate projection or inclusion morphisms) to conclude that $F$ respects products and coproducts.

(2)$\Rightarrow$(1) and (3)$\Rightarrow$(1) Let $f,g:A\to A^{\prime }$ be morphisms in $\mathcal{A}$. Then $f+g$ is the composition of the following morphisms:$$A\stackrel{\left[\begin{array}{c}1\\ 1\end{array}\right]}{\to }A\oplus A\stackrel{\left[\begin{array}{cc}f& 0\\ 0& g\end{array}\right]}{\to }{A}^{\prime }\oplus A^{\prime }\stackrel{\left[\begin{array}{cc}1& 1\end{array}\right]}{\to }{A}^{\prime }$$

Since $F$ preserves binary (co)products, it follows that:$$\begin{array}{rl}F(f+g)& =F\left(\left[\begin{array}{cc}1& 1\end{array}\right]\cdot \left[\begin{array}{cc}f& 0\\ 0& g\end{array}\right]\cdot \left[\begin{array}{c}1\\ 1\end{array}\right]\right)=[F(1)F(1)]\cdot \left[\begin{array}{cc}F(f)& 0\\ 0& F(g)\end{array}\right]\cdot \left[\begin{array}{c}F(1)\\ F(1)\end{array}\right]\\ & =\left[\begin{array}{cc}1& 1\end{array}\right]\cdot \left[\begin{array}{cc}F(f)& 0\\ 0& F(g)\end{array}\right]\cdot \left[\begin{array}{c}1\\ 1\end{array}\right]=F(f)+f(g).\end{array}$$Hence $F$ is additive.

$\square$

The Hom functors for an additive category, which by Proposition 8.1.6 may be viewed as functors into the additive category $\mathbf{Ab}$, are an important example of additive functors.

命题 8.5.5. Let $\mathcal{C}$ be a preadditive category and $A$ a fixed object of $\mathcal{C}$. Then the Hom functors ${\mathrm{Hom}}_{\mathcal{C}}(A,-):\mathcal{C}\to \mathbf{Ab}$ and ${\mathrm{Hom}}_{\mathcal{C}}(-,A):\mathcal{C}\to \mathbf{Ab}$ are additive.

证明. Note first that the proposition makes sense, seeing as $\mathbf{Ab}$ is additive by Example 8.2.3. The proposition is a rather immediate consequence of the definition of the Hom functors in terms of composition and the biadditivity of the composition in an additive category.

$\square$

例 8.5.6. By Propositions 8.5.4 and 8.5.5, the morphism set for two biproducts $X_1\oplus X_2$ and $Y_1\oplus Y_2$ of an additive category $\mathcal{A}$ decomposes as$$\begin{array}{rl}& {\mathrm{Hom}}_{\mathcal{A}}\left(X_1\oplus X_2,Y_1\oplus Y_2\right)\\ =& {\mathrm{Hom}}_{\mathcal{A}}\left(X_1\oplus X_2,-\right)\left(Y_1\oplus Y_2\right)\\ \cong & {\mathrm{Hom}}_{\mathcal{A}}\left(X_1\oplus X_2,-\right)\left(Y_1\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(X_1\oplus X_2,-\right)\left(Y_2\right)\\ =& {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_1\right)\left(X_1\oplus X_2\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_2\right)\left(X_1\oplus X_2\right)\\ \cong & {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_1\right)\left(X_1\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_1\right)\left(X_2\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_2\right)\left(X_1\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(-,Y_2\right)\left(X_2\right)\\ =& {\mathrm{Hom}}_{\mathcal{A}}\left(X_1,Y_1\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(X_1,Y_2\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(X_2,Y_1\right)\oplus {\mathrm{Hom}}_{\mathcal{A}}\left(X_2,Y_2\right).\end{array}$$