In an additive category, morphisms can be added and finite products are also coproducts. This permits calculations with matrices without treating abstract objects as sets.
3.1. Adding morphisms
Definition 3.1. Preadditive categories
A category is locally small if each \(\operatorname{Hom}(X,Y)\) is a set. It is preadditive if these sets are abelian groups and composition is bilinear:
\[h(f+g)=hf+hg,\qquad (f+g)k=fk+gk.\]The group identity is denoted \(0_{X,Y}\); negatives are denoted \(-f\). Associativity and identities still refer to the category's composition.
Alphabeta Math: Preadditive categories
Proposition 3.2. Zero and subtraction under composition
In a preadditive category, \(h0=0\), \(0k=0\), \(h(-f)=-(hf)\), and \((-f)k=-(fk)\). If a zero object exists, group-zero morphisms agree with the maps factoring through it.
Alphabeta Math: Zero morphisms · Compatibility with a zero object
Proof.
The equation \(h0=h(0+0)=h0+h0\) gives \(h0=0\) by cancellation in the morphism group. The other zero equation follows from \((0+0)k=0k+0k\). Next \(hf+h(-f)=h(f-f)=0\), giving the negative identity; the second follows in the same way from \((f-f)k=0\). A zero object has only one endomorphism, so its identity is its zero morphism. A map \(X\to0\to Y\) is therefore unchanged by inserting this identity and equals zero by the composition identities just proved.
□Example 3.3. Abelian groups and matrices
In \(\mathrm{Ab}\), add homomorphisms pointwise: \((f+g)(a)=f(a)+g(a)\). Since the target group is abelian, \(f+g\) remains a homomorphism; pointwise inverses give negatives, and composition distributes. A second example is the category with objects nonnegative integers and maps \(n\to m\) the \(m\times n\) matrices over a unital ring \(R\). Entrywise addition gives morphism groups, matrix multiplication gives composition, and its finite-sum formula proves bilinearity. Identity matrices and associative multiplication make this a category, including empty matrices when an object is \(0\).
Alphabeta Math: Matrix categories · Matrix-category laws
3.2. Finite biproducts
Definition 3.4. Biproducts and additive categories
A biproduct of \(X_1,\ldots,X_n\) in a preadditive category is an object \(B\) with arrows \(i_j:X_j\to B\), \(p_j:B\to X_j\) satisfying
\[p_j i_k=\begin{cases}\mathrm{id}_{X_j}&j=k,\\0&j\ne k,\end{cases}\qquad \sum_{j=1}^n i_jp_j=\mathrm{id}_B.\]Write \(B=\bigoplus_jX_j\). The empty biproduct is a zero object. An additive category is a preadditive category with a zero object and all finite biproducts.
Alphabeta Math: Biproducts · Additive categories
Proposition 3.5. The two universal properties
A biproduct is both a product through \(p_j\) and a coproduct through \(i_j\). Conversely, any finite product in a preadditive category has canonical biproduct injections; any finite coproduct has canonical biproduct projections.
Alphabeta Math: Products and biproducts · Identity decomposition
Proof.
Given \(t_j:T\to X_j\), set \(u=\sum_j i_jt_j\). Then \(p_ku=t_k\). Any arrow with these composites satisfies \(u=(\sum_j i_jp_j)u=\sum_j i_jt_j\), proving uniqueness. Given \(t_j:X_j\to T\), set \(v=\sum_j t_jp_j\). Then \(v i_k=t_k\), and every arrow with these restrictions satisfies \(v=v\sum_j i_jp_j=\sum_j t_jp_j\).
For a finite product, define \(i_k\) by components identity in position \(k\) and zero elsewhere. Its product property gives these arrows and \(p_j i_k=\delta_{jk}\). The maps \(\sum_k i_kp_k\) and identity have the same composites with every \(p_j\), so they agree. For a finite coproduct, define \(p_j\) by its restrictions identity on summand \(j\) and zero on every other summand. The arrows \(\sum_j i_jp_j\) and identity have the same restrictions to every summand, hence agree. For the empty product its identity is zero, since its endomorphism set has one map. Every map out of it is zero because \(u=u\mathrm{id}=u0=0\); group-zero maps provide existence. Thus it is also initial. The empty coproduct is terminal by the reversed argument.
□Example 3.6. A pair of abelian groups
The direct sum \(A\oplus B\) has injections \(a\mapsto(a,0)\), \(b\mapsto(0,b)\), and coordinate projections. Their composites satisfy the displayed equations because \((a,b)=(a,0)+(0,b)\). Thus \(\mathrm{Ab}\) is additive. The matrix category in Example 3.3 has biproduct \(n\oplus m=n+m\), with block inclusions and projections, and zero object \(0\).
Alphabeta Math: Abelian-group biproducts
3.3. Matrix calculus
Proposition 3.7. A morphism is its matrix
For finite biproducts \(X=\bigoplus_jX_j\), \(Y=\bigoplus_kY_k\), every \(f:X\to Y\) is determined by entries \(f_{kj}=p^Y_k f i^X_j\), and
\[f=\sum_{k,j}i^Y_k f_{kj}p^X_j.\]For \(g:Y\to Z=\bigoplus_\ell Z_\ell\), composition has entries \((gf)_{\ell j}=\sum_k g_{\ell k}f_{kj}\). Addition is entrywise.
Alphabeta Math: Matrix correspondence · Composition
Proof.
Insert the identity decompositions of \(Y\) and \(X\) on either side of \(f\) and distribute to obtain the reconstruction formula. Conversely any family of entries defines an arrow by that formula, and composing with \(p^Y_k\) and \(i^X_j\) recovers exactly the prescribed entry. For composition, insert \(\mathrm{id}_Y=\sum_k i^Y_kp^Y_k\) into \(p^Z_\ell gf i^X_j\); bilinearity gives the claimed sum. For addition, bilinearity gives \(p^Y_k(f+f')i^X_j=f_{kj}+f'_{kj}\).
□For instance, a map \(X\oplus Y\to Z\) is a row \([f\ g]\), and a map \(T\to X\oplus Y\) is a column \([a\ b]^t\); their composite is \(fa+gb\). These are statements about morphisms, regardless of whether the objects have elements.
3.4. Functors and addition
Definition 3.8. Additive, full, and faithful functors
A functor between preadditive categories is additive if each map \(\operatorname{Hom}(X,Y)\to\operatorname{Hom}(F X,F Y)\) is a group homomorphism, equivalently \(F(f+g)=F(f)+F(g)\). A functor is faithful if each such map is injective, full if each is surjective, and fully faithful if each is bijective.
Alphabeta Math: Additive functors · Full and faithful functors
Worked example for Definition 3.8
Counterexample: Additive functors can lose and miss morphisms
Proposition 3.9. Addition and finite biproducts
An additive functor between additive categories preserves finite biproducts. Conversely, a functor between additive categories that preserves zero objects and finite products and coproducts is additive.
Alphabeta Math: Preservation · Converse
Worked example for Proposition 3.9
Counterexample: Additive functors can lose and miss morphisms
Proof.
An additive functor preserves zero morphisms and sums. Applying it to the biproduct equations gives the same equations for \(F(B),F(i_j),F(p_j)\), so Proposition 3.5 applies. It preserves a zero object because \(\mathrm{id}_{F(0)}=F(\mathrm{id}_0)=F(0)=0\). An object whose identity is zero has only zero maps in or out, so is a zero object.
For the converse, zero morphisms factor through zero objects, so their images are zero. Use the binary biproduct \(X\oplus X\) and \(Y\oplus Y\). Define the diagonal \(\Delta_X:X\to X\oplus X\) by both projection composites equal to identity, the codiagonal \(\nabla_Y:Y\oplus Y\to Y\) by both injection restrictions equal to identity, and \(f\oplus g\) by diagonal matrix entries \(f,g\) and off-diagonal zero. Matrix calculus gives
\[f+g=\nabla_Y(f\oplus g)\Delta_X.\]Preservation of the two universal properties identifies the images of the diagonal, codiagonal, and block map with those for \(F(X),F(Y)\): their defining composites are preserved, including the zero entries. Applying \(F\) to the displayed composite gives \(F(f)+F(g)\). Together with preservation of zero this is a group homomorphism; negatives follow from \(F(f)+F(-f)=F(0)=0\).
□There is no infinite identity decomposition \(\mathrm{id}=\sum_i i_ip_i\) here: morphism groups only admit finite sums. Infinite coproducts will be treated by their universal property, with the separate smallness condition of chapter 4.
Worked examples
Each example checks the hypotheses and the calculation for the linked statements. The links in the statements return here.
Worked example 1. Composition with integer matrices
Applies to: Definition 3.1, Proposition 3.2.
Use the category of finite-rank free abelian groups. Hom sets are sets of integer matrices; entrywise addition makes them abelian groups, and finite distributivity of integer multiplication makes composition bilinear. For \(f(n)=2n,g(n)=3n,h(n)=5n\) on \(\mathbb Z\), \(h(f+g)(n)=25n=hf(n)+hg(n)\), \(h(-f)(n)=-10n=-hf(n)\), and \(h0=0\). Also \((-f)h=-10\) and \(0h=0\). The zero matrix factors through the rank-zero group, verifying the categorical and additive meanings of zero agree.
Worked example 2. A biproduct calculation
Applies to: Definition 3.4, Proposition 3.5, Proposition 3.7.
In the same category take \(B=\mathbb Z^2\), \(i_1(n)=(n,0),i_2(n)=(0,n)\), and coordinate projections \(p_1,p_2\). Then \(p_ji_k=\delta_{jk}\) and \(i_1p_1+i_2p_2=I\). A pair \(a,b:T\to\mathbb Z\) factors uniquely by \(i_1a+i_2b\); a pair \(c,d:\mathbb Z\to T\) factors uniquely by \(cp_1+dp_2\). This proves both universal properties. With the rank-zero group these biproducts make the category additive.
Let \(f=\begin{pmatrix}1&2\\3&4\end{pmatrix}\) and \(g=(5\;6)\). The entry \(p_2fi_1\) is multiplication by 3. Reassembling the four \(i_k f_{kj}p_j\) returns \(f\), and direct composition gives \(gf=(23\;34)\), since \(5+18=23\) and \(10+24=34\). This is precisely the biproduct matrix law.
Worked example 3. Forgetting scalars
Applies to: Definition 3.8, Proposition 3.9.
The forgetful functor from complex vector spaces to abelian groups is additive: \(F(f+g)(v)=f(v)+g(v)\). It is faithful because it keeps the underlying functions. It is not full: conjugation \(z\mapsto\bar z\) is additive but not complex-linear, since it sends \(i\) to \(-i\). It preserves the zero object and finite direct sums with their usual inclusions and projections. In contrast the identity functor on complex vector spaces is fully faithful. The reconstruction of addition by \(f+g=\nabla(f\oplus g)\Delta\) is visible here: \(v\mapsto(v,v)\mapsto(f(v),g(v))\mapsto f(v)+g(v)\). Keeping these biproduct maps therefore keeps addition, illustrating both directions of Proposition 3.9.
Further proof details
The forgetful functor’s finite-biproduct preservation uses the actual structure maps. For two vector spaces its underlying group is the pairs \((v,w)\). Given additive incoming maps \(a,b\), their unique product map is \(t\mapsto(a(t),b(t))\). Given additive outgoing maps \(c,d\), their unique coproduct map is \((v,w)\mapsto c(v)+d(w)\), since \((v,w)=(v,0)+(0,w)\). These are the same projections and inclusions as the vector-space biproduct. The zero vector space becomes the group with one element, initial and terminal. Thus preservation here is an explicit universal-property calculation.
Counterexamples and limits of the hypotheses
Each construction below refutes the explicitly stated stronger claim. It does not refute the related theorem with its original hypotheses.
Counterexample 1. Additive functors can lose and miss morphisms
Related statements: Definition 3.8, Proposition 3.9.
Claim refuted. An additive functor must be full or faithful.
Proof.
Let \(F:\mathrm{Ab}\to\mathrm{Ab}\) send \(M\) to \(M/2M\) and a map to its induced quotient map. It is additive because \(F(f+g)([m])=[f(m)+g(m)]=F(f)([m])+F(g)([m])\). On \(\mathbb Z\), the distinct maps multiplication by 2 and zero both induce zero on \(\mathbb Z/2\), so \(F\) is not faithful. It is not full either: every map \(\mathbb Z/2\to\mathbb Z\) is zero, since its generator must map to an integer killed by 2. Yet \(F(\mathbb Z/2)=\mathbb Z/2=F(\mathbb Z)\) admits a nonzero identity map. Thus the induced Hom map is not surjective. Preserving addition and finite biproducts therefore does not imply either Hom-set property.
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