Additive localization and projector decompositions

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by the AI that wrote it (GPT-6.1 Sol, Ultra); no independent review of this lesson has occurred. Public domain (CC0).

Addition of morphisms lets us describe finite decompositions by matrices. It also survives an ordinary category localization when incoming fractions exist. These two facts explain why additive methods remain available after separating direct summands or making denominators invertible.

Retain the full zero-map and both biproduct universal-property proofs in Recovering addition, Section 1. For projectors retain the full splitting criterion; Section 2 below proves the compatible comparison of two splittings and the finite decompositions. For localization retain the full descent and opposite-category interface in One-sided fractions, Section 5, together with the canonical construction and its remaining checks in Section 3 below. The full finite-product transport proof turns these products into addition on localized morphisms.

Basic references are the Stacks Project's treatments of additive categories and localization, and Mathlib's preadditive localization construction. We give the needed arguments below.

1. Finite sums in familiar categories

A category is preadditive when each Hom set is an abelian group and composition is additive in both variables. It is additive when it is preadditive and has a zero object and binary biproducts. Thus all finite biproducts exist. A functor between preadditive categories is additive when its maps on Hom sets are group homomorphisms.

Proposition 1.1. For every unital ring \(R\), each of the following full categories is additive: all left \(R\)-modules, finitely generated left \(R\)-modules, and finitely presented left \(R\)-modules.

Proof. The sum of two module homomorphisms is defined pointwise. It is again \(R\)-linear, and composition distributes over these sums. The zero module is a zero object. For modules \(M,N\), the usual inclusions and projections of \(M\oplus N\) satisfy

\[ \begin{gathered} p_a i_b=\delta_{ab}1,\qquad a,b\in\{1,2\},\\ i_1p_1+i_2p_2=1_{M\oplus N}. \end{gathered} \tag{1.1} \]

Here \(\delta_{ab}1\) means the appropriate identity when \(a=b\) and the zero morphism otherwise. These identities give both universal properties. A finite union of generating sets generates the sum of finitely generated modules. If \(M,N\) have finite presentations, take their direct sum:

\[ \begin{gathered} R^{a+c}\longrightarrow R^{b+d}\\ \longrightarrow M\oplus N\longrightarrow0. \end{gathered} \tag{1.2} \]

The first map is the direct sum of the two presentation maps, and its image is the kernel of the second map coordinate by coordinate. This is a finite presentation over any unital ring. No closure under kernels of arbitrary maps is needed. \(\square\)

Proposition 1.2. Complex Banach spaces and bounded linear maps form an additive category.

Proof. Bounded linear maps have pointwise sums and negatives. The operator-norm inequality \(\|f+g\|\leq\|f\|+\|g\|\) shows that their sum is bounded, and composition is bilinear. The zero space is a zero object. Give \(X\times Y\) the norm

\[ \|(x,y)\|_{\max}=\max(\|x\|,\|y\|). \tag{1.3} \]

A Cauchy sequence has Cauchy coordinates, whose limits converge in this norm, so the space is complete. The coordinate inclusions and projections are bounded. For bounded \(a:W\to X\), \(b:W\to Y\), the map \(w\mapsto(a(w),b(w))\) is bounded and is the unique map with those coordinates. For bounded \(u:X\to W\), \(v:Y\to W\), the map \((x,y)\mapsto u(x)+v(y)\) satisfies

\[ \begin{aligned} &\|u(x)+v(y)\|\\ &\leq(\|u\|+\|v\|)\|(x,y)\|_{\max}. \end{aligned} \tag{1.4} \]

It is the unique map with restrictions \(u,v\). This proves both biproduct properties. \(\square\)

Proposition 1.3. If \(I\) is small and \(\mathsf C\) is additive, the functor category \(\operatorname{Fct}(I,\mathsf C)\) is additive.

Proof. Add natural transformations componentwise. If \(a,b:F\to G\) are natural and \(t:i\to j\), bilinearity gives

\[ G(t)(a_i+b_i)=(a_j+b_j)F(t). \tag{1.5} \]

Negatives and zero transformations are natural by the same calculation. Composition is bilinear componentwise. The constant zero functor is a zero object. Define \((F\oplus G)(i)=F(i)\oplus G(i)\), with each arrow acting diagonally on the two coordinates. The diagonal maps preserve identity and composition by the biproduct equations. The pointwise inclusions and projections are natural and satisfy (1.1). A map into the sum is uniquely determined by its two projections, and a map out of it by its two restrictions; the maps obtained pointwise are natural because their components are. Thus this is a biproduct in the functor category. \(\square\)

2. Projectors and complementary summands

An endomorphism \(e:X\to X\) is idempotent if \(e^2=e\). A splitting consists of

\[ \begin{gathered} X\xrightarrow{r}Y\xrightarrow{i}X,\\ ri=1_Y,\qquad ir=e. \end{gathered} \tag{2.1} \]

A category is idempotent complete when every idempotent admits a splitting. This does not assert that arbitrary morphisms have images. For a split idempotent, however, its summand has a useful image-like universal property.

Lemma 2.1. In an additive category, the inclusion \(i\) in (2.1) is a kernel of \(1_X-e\), and \(r\) is a cokernel of \(1_X-e\).

Proof. From (2.1), \(ei=i\), \(re=r\), and \((1-e)i=r(1-e)=0\). If \(h:T\to X\) satisfies \((1-e)h=0\), then

\[ h=eh=i(rh). \tag{2.2} \]

This factorization is unique: \(iu=h\) implies \(u=riu=rh\). If \(k:X\to T\) satisfies \(k(1-e)=0\), then \(k=ke=(ki)r\). The factorization through \(r\) is unique because \(vr=k\) implies \(v=vri=ki\). These are exactly the two universal properties. \(\square\)

Accordingly we may write \(Y\cong\operatorname{Im}e\) for the splitting object. Any two splittings of the same idempotent are uniquely isomorphic in a way compatible with both their inclusions and retractions: for \((i,r)\) and \((i',r')\), the maps are \(r'i\) and \(ri'\), and their composites are identities by \(ir=i'r'=e\).

Theorem 2.2. In an idempotent-complete additive category, every idempotent \(e:X\to X\) gives a decomposition

\[ \begin{gathered} X\cong Y\oplus Z,\\ e\longleftrightarrow \begin{pmatrix}1_Y&0\\0&0\end{pmatrix}. \end{gathered} \tag{2.3} \]

Proof. The complement \(1-e\) is idempotent since \((1-e)^2=1-2e+e^2=1-e\). Choose splittings \(e=ir\) and \(1-e=jt\), with \(ri=1_Y\) and \(tj=1_Z\). We have

\[ \begin{gathered} rj=re(1-e)j=0,\\ ti=t(1-e)ei=0,\\ ir+jt=1_X. \end{gathered} \tag{2.4} \]

Here \(re=r\), \((1-e)j=j\), \(t(1-e)=t\), and \(ei=i\) follow from the splitting equations. Assemble \((r,t):X\to Y\oplus Z\) and \((i,j):Y\oplus Z\to X\). Equations (2.4) and the two identity retractions show that their composites are identities. In these coordinates, \(e\) fixes the \(Y\) summand and kills the \(Z\) summand, giving (2.3). \(\square\)

Theorem 2.3. Let \(I\) be a finite set and \(e_i:X\to X\) endomorphisms in an idempotent-complete additive category. Suppose

\[ \begin{gathered} \sum_{i\in I}e_i=1_X,\\ e_i e_j=0\quad(i\ne j). \end{gathered} \tag{2.5} \]

Then each \(e_i\) is idempotent and

\[ X\cong\bigoplus_{i\in I}X_i, \qquad X_i\cong\operatorname{Im}e_i. \tag{2.6} \]

Under this isomorphism \(e_i\) is the corresponding coordinate projector.

Proof. First, idempotence follows from the stated hypotheses rather than needing a separate assumption:

\[ e_i=e_i1_X=\sum_{j\in I}e_i e_j=e_i^2. \tag{2.7} \]

Split \(e_i=a_i b_i\), with \(b_i a_i=1_{X_i}\). Then \(b_i e_i=b_i\), \(e_i a_i=a_i\), and for distinct indices

\[ b_i a_j=b_i e_i e_j a_j=0. \tag{2.8} \]

The map \(b:X\to\bigoplus_iX_i\) has components \(b_i\); the map \(a:\bigoplus_iX_i\to X\) has restrictions \(a_i\). The matrix of \(ba\) is the identity matrix by (2.8) and \(b_i a_i=1\). Its entries determine the morphism by the two biproduct universal properties. In the other order, \(ab=\sum_i a_i b_i=\sum_i e_i=1_X\). Hence \(a,b\) are inverse, and \(e_i=a_i b_i\) is the coordinate projector.

If \(I\) is empty, (2.5) states \(1_X=0\). For any \(f:X\to T\), \(f=f1_X=f0=0\), and for any \(g:T\to X\), \(g=1_Xg=0g=0\). Thus \(X\) is a zero object, which is the empty biproduct. \(\square\)

These decompositions express algebraic complementarity. They involve neither an inner product nor a choice of perpendicular subspaces.

3. Incoming fractions preserve finite products

Let \(S\) be a multiplicative system with a calculus of incoming fractions in \(\mathsf C\), and let

\[ Q:\mathsf C\longrightarrow\mathsf L=\mathsf C[S^{-1}] \tag{3.1} \]

be the ordinary localization with the same objects. Every localized arrow is represented as

\[ \begin{gathered} Qf\,(Qs)^{-1}:QX\longrightarrow QY,\\ s:Z\to X\text{ in }S,\quad f:Z\to Y. \end{gathered} \tag{3.2} \]

For a fixed \(X\), the denominator category \(I_X\) has objects \(s:Z\to X\) in \(S\) and arrows given by commuting triangles over \(X\). It is cofiltered. A comparison arrow between denominators need not itself belong to \(S\). The opposite category \(J_X=I_X^{\rm op}\) is filtered, and the fraction formula is

\[ \begin{gathered} \operatorname{Hom}_{\mathsf L}(QX,QY)\\ \cong\mathop{\operatorname{colim}}_{s\in J_X} \operatorname{Hom}_{\mathsf C}(Z_s,Y). \end{gathered} \tag{3.3} \]

The transition maps precompose with the comparison arrows. Retain the complete common-denominator and equality proofs and the filtered outgoing denominator proof from the AI-integrated Stacks baseline. Apply them to \(\mathsf C^{\mathrm{op}}\): an outgoing denominator there is an incoming denominator here, and its comparison arrows reverse to precisely the arrows of \(J_X\). The roof equivalence is consequently the same common-refinement equivalence in (3.2), giving (3.3). In particular a comparison need not lie in \(S\); only its composite denominator must do so. The full opposite-localization and natural-transformation comparison is retained from the linked fractions lesson. If necessary, choose a larger universe containing the denominator category and its Hom sets; no smallness of \(\mathsf C\) in its original Hom-set universe is assumed here.

Fraction construction checks. Retain the equivalence-relation proof, independence of the Ore square, refinement of the preceding roof, and associativity in the canonical outgoing construction. Here are its remaining refinement and identity checks, before taking opposites. An outgoing roof \(p_i:X\to Y\) is a pair \((f_i,s_i)\), where

\[ \begin{gathered} f_i:X\to Y_i,\qquad s_i:Y\to Y_i\text{ in }S. \end{gathered} \]

Suppose \(a:Y_1\to Y_2\) refines the first roof to the second, so \(f_2=af_1\) and \(s_2=as_1\). The arrow \(a\) can be arbitrary. Let the preceding roof \(q:W\to X\) be \((g,t)\), with \(g:W\to X'\) and \(t:X\to X'\) in \(S\). An Ore square supplies

\[ \begin{gathered} b:X'\to Z,\qquad c:Y_1\to Z\text{ in }S,\\ bt=cf_1. \end{gathered} \]

Thus \(p_1q\) is represented by \((bg,cs_1)\). Apply the Ore axiom once more, to \(c\) and \(a\), obtaining

\[ \begin{gathered} d:Z\to Z',\qquad e:Y_2\to Z'\text{ in }S,\\ dc=ea. \end{gathered} \]

Now \((db)t=ef_2\), so \((db,e)\) is an Ore square for \(p_2q\). Its composite roof is \((dbg,es_2)\). Since \(es_2=dcs_1\) belongs to \(S\), this roof is the refinement by \(d\) of \((bg,cs_1)\). Hence \(p_1q=p_2q\). An arbitrary equivalence is witnessed by a common refinement, so this calculation proves the omitted invariance for all equivalent following roofs. The retained proof handles refinement of the preceding roof; composition is therefore well defined in both arguments.

For \(p=(f:X\to Y',s:Y\to Y')\), composition after the identity of \(X\) uses the Ore square with arrows \(f\) and \(1_{Y'}\), and composition before the identity of \(Y\) uses the square with arrows \(1_{Y'}\) and \(s\). Both return the pair \((f,s)\). Ordinary arrows compose using identity denominators. For \(s:Y\to Y'\) in \(S\), the roof \((1_{Y'},s)\) is inverse to \(Q(s)\): one composite is an identity roof and the other is \((s,s)\), the refinement by \(s\) of the identity of \(Y\). This supplies both inverse equations. Together with the retained associativity proof, these checks give the ordinary localization functor used above. The full descent proof in the fractions lesson verifies functoriality and every natural transformation. Applying its opposite-category interface gives exactly the incoming construction, including its identities and composition. No addition is used in these fraction checks.

Lemma 3.1. If \(\mathsf C\) has finite products, \(Q\) preserves them.

Proof. Let \(P=A\times B\), with projections \(p_A,p_B\). For any \(X\), the localized product cone induces

\[ \begin{gathered} \operatorname{Hom}_{\mathsf L}(QX,QP)\\ \longrightarrow\operatorname{Hom}_{\mathsf L}(QX,QA)\\ \times\operatorname{Hom}_{\mathsf L}(QX,QB). \end{gathered} \tag{3.4} \]

To prove surjectivity, represent the two target arrows at denominators \(s_1,s_2\). Filteredness of \(J_X\) gives a denominator receiving arrows from both, so both representatives may be taken at a single \(s:Z\to X\). Write their numerators as \(f:Z\to A\), \(g:Z\to B\). The fraction with numerator \(\langle f,g\rangle:Z\to P\) has the required two projections.

For injectivity, first represent the two candidate arrows at a common denominator. Equality of their \(A\) coordinates holds after a further refinement, and equality of their \(B\) coordinates holds after another. Both equalities can be made to hold at one stage. Explicitly, filteredness gives a common target for the two refinement stages. The two resulting arrows from the original stage to that target can differ; the parallel-arrow condition for a filtered category equalizes them after one more transition. At that final stage both coordinate equalities hold along the same transition. The product universal property in \(\mathsf C\) then makes the two numerators equal. Formula (3.3) makes their localized arrows equal. Thus (3.4) is bijective, proving the product property for \(QP\).

For a terminal object \(1\), every \(\operatorname{Hom}_{\mathsf C}(Z_s,1)\) is a singleton. A filtered category is nonempty and any two of its objects have a common target, so their colimit is a singleton. Formula (3.3) shows that \(Q1\) is terminal. Terminal objects and binary products give all finite products by iteration. \(\square\)

The Stacks Project proves the stronger finite-limit preservation statement in Lemma 4.27.17. The proof above uses only finite products and the incoming fraction formula.

4. Addition after localization

Theorem 4.1. If \(\mathsf C\) is additive and \(S\) admits incoming fractions, \(\mathsf L=\mathsf C[S^{-1}]\) has a unique preadditive structure for which \(Q\) is additive. With this structure \(\mathsf L\) is additive, and \(Q\) preserves finite biproducts.

Proof. Every object \(Y\) of \(\mathsf C\) carries a commutative internal group structure: addition is the codiagonal \(\nabla_Y:Y\oplus Y\to Y\), the identity is \(0\to Y\), and inverse is \(-1_Y\). On every \(\operatorname{Hom}_{\mathsf C}(T,Y)\) this is the original group law, since

\[ f+g=\nabla_Y\langle f,g\rangle. \tag{4.1} \]

Every morphism \(Y\to Y'\) preserves these group structures because postcomposition is additive. By Lemma 3.1 and the finite-product transport theorem, each \(QY\) inherits a commutative internal group, and every \(Qf\) is a morphism of these groups.

The inverse of an invertible group-object morphism is also a group-object morphism. Indeed, the induced map on every representable Hom group is both a group homomorphism and a bijection, whose inverse is a group homomorphism. Thus \((Qs)^{-1}\) preserves the transported groups. Formula (3.2) now shows that every arrow of \(\mathsf L\) does so.

Give each \(\operatorname{Hom}_{\mathsf L}(U,V)\) the abelian group associated with the internal group on \(V\). Precomposition is a group homomorphism by the definition of a group object. Postcomposition is a group homomorphism because every localized arrow preserves these groups. This proves bilinearity of composition, hence preadditivity. Product transport sends pairings to pairings. Applying \(Q\) to (4.1) consequently gives

\[ Q(f+g)=Qf+Qg. \tag{4.2} \]

It also sends the zero and inverse operations to their counterparts, so \(Q\) is additive.

The object \(Q0\) is terminal by Lemma 3.1. Its endomorphism group has one element, so \(1_{Q0}=0\). Bilinearity then gives \(u=u1_{Q0}=u0=0\) for every map out of \(Q0\); there is at least a zero morphism to every object. Hence \(Q0\) is initial as well. For a source biproduct, applying the additive functor \(Q\) to (1.1) gives the same biproduct identities in \(\mathsf L\). The images therefore form a biproduct. This proves additivity and finite-biproduct preservation.

Finally, any two fractions \(a,b:QX\to QY\) have a common incoming denominator. Write

\[ \begin{gathered} a=Qf\,(Qs)^{-1},\\ b=Qg\,(Qs)^{-1},\\ a+b=Q(f+g)\,(Qs)^{-1}. \end{gathered} \tag{4.3} \]

The last equality is forced by bilinearity and the additivity of \(Q\) in any candidate preadditive structure. Every pair of arrows has this form, so addition is unique. Its zero and inverses are then uniquely determined. Existence has already been proved by transporting internal groups, so no separate independence-of-denominator assertion is being assumed in this uniqueness argument. \(\square\)

For example, making multiplication by a prime invertible on finite free abelian groups produces finite matrices with entries in \(\mathbb Z[1/p]\), as Exercise 2 proves. The sum in (4.3) becomes ordinary matrix addition after choosing a common power of \(p\).

Corollary 4.2. Suppose \(F:\mathsf C\to\mathsf D\) is additive, \(\mathsf D\) is preadditive, and \(F\) sends every arrow of \(S\) to an isomorphism. Its ordinary localized extension \(\overline F:\mathsf L\to\mathsf D\) is additive.

Proof. For the strict extension with \(\overline FQ=F\), use a common denominator in (4.3):

\[ \begin{aligned} \overline F(a+b) &=F(f+g)(Fs)^{-1}\\ &=(Ff+Fg)(Fs)^{-1}\\ &=\overline F(a)+\overline F(b). \end{aligned} \tag{4.4} \]

The same calculation applies to zero and negatives, or they follow from additivity of sums. Any other extension naturally isomorphic to this one is additive: conjugation by its component isomorphisms is a group homomorphism on Hom sets, by bilinearity in \(\mathsf D\). \(\square\)

Mathlib's preadditive construction starts with outgoing fractions and needs only a preadditive source. Passing to opposite categories converts that direction to the incoming one: reverse arrows, keep their Hom-group operations, and reverse composition. The localization comparison from the prerequisite on fractions identifies the two underlying ordinary categories. Finite biproducts enter separately when the source is additive, as in Theorem 4.1.

5. Exercises with complete solutions

Exercise 1 (introductory: two norms on a finite sum). For complex Banach spaces \(X,Y\), put both \(\|(x,y)\|_{\max}\) and \(\|(x,y)\|_1=\|x\|+\|y\|\) on \(X\times Y\). Prove that both give the same biproduct up to bounded isomorphism. Compute the norm of \(w\mapsto(a(w),b(w))\) for the maximum norm, and of \((x,y)\mapsto u(x)+v(y)\) for the sum norm.

Solution. We have

\[ \begin{gathered} \|(x,y)\|_{\max}\leq\|(x,y)\|_1,\\ \|(x,y)\|_1\leq2\|(x,y)\|_{\max}. \end{gathered} \tag{5.1} \]

Thus a sequence is Cauchy for one norm exactly when it is Cauchy for the other, and convergence is equivalent. The maximum-norm space is complete by Proposition 1.2, so the sum-norm space is complete too. The identity from the maximum-norm space to the sum-norm space has norm at most \(2\); its inverse has norm at most \(1\). These are bounded inverse linear maps.

For \(A(w)=(a(w),b(w))\),

\[ \|A\|_{\max}=\max(\|a\|,\|b\|). \tag{5.2} \]

The upper bound follows from the pointwise maximum. Each coordinate projection has norm at most \(1\), giving both lower bounds \(\|a\|\leq\|A\|\), \(\|b\|\leq\|A\|\). For \(U(x,y)=u(x)+v(y)\), the sum norm gives

\[ \|U\|_1=\max(\|u\|,\|v\|). \tag{5.3} \]

Indeed, put \(M=\max(\|u\|,\|v\|)\). The triangle inequality gives

\[ \|U(x,y)\|\leq M(\|x\|+\|y\|). \tag{5.3a} \]

Restriction to each coordinate axis gives the lower bounds. This also covers zero spaces by the convention that the zero operator has norm \(0\). The coordinate inclusions and projections are bounded for both norms and satisfy the same identities (1.1), proving the two biproduct properties. The norms change quantitative estimates, while the bounded isomorphism identifies their categorical biproducts.

Exercise 2 (intermediate: matrices with one prime inverted). Let \(p\) be prime. Take the category whose objects are \(\mathbb Z^n\), \(n\geq0\), and whose maps are integer matrices. Let \(S\) consist of \(p^a1_{\mathbb Z^n}\), \(a\geq0\). Check the incoming fraction axioms and identify its localization with the category of finite matrices over \(\mathbb Z[1/p]\).

Solution. Identities belong to \(S\), and composition adds exponents. For \(f:\mathbb Z^m\to\mathbb Z^n\) and \(s=p^a1_{\mathbb Z^n}\), choose \(t=p^a1_{\mathbb Z^m}\) and \(h=f\). They give the incoming square

\[ s h=f t. \tag{5.4} \]

If \(sf=sg\), then multiplying every entry of \(f-g\) by \(p^a\) gives zero. The additive group of integers is torsion-free, so \(f=g\); the cancellation axiom holds with the identity refinement. The zero-rank case has a unique matrix and satisfies the same equations. Hence incoming fractions exist.

An incoming denominator into \(\mathbb Z^m\) is necessarily \(p^a1_{\mathbb Z^m}\), so a fraction is represented by a matrix \(A\) with value \(p^{-a}A\). Every finite matrix over \(\mathbb Z[1/p]\) has such a representation: choose one exponent at least as large as the finitely many entry denominators. If

\[ \begin{gathered} p^{-a}A=p^{-b}B,\\ \text{then}\quad p^bA=p^aB. \end{gathered} \tag{5.5} \]

Refine the first denominator by \(p^b1\) and the second by \(p^a1\). Their resulting denominators are both \(p^{a+b}1\), and their numerators agree by (5.5). Thus they are equal as localized arrows. Conversely, an equality of localized arrows remains an equality of their matrices because the integer-matrix functor into \(\mathbb Z[1/p]\)-matrices inverts \(S\). This proves a bijection on every Hom set.

For composable representatives, centrality of scalar multiplication gives

\[ (p^{-b}B)(p^{-a}A)=p^{-(a+b)}BA. \tag{5.6} \]

This is both matrix composition and localized composition. Identities correspond to identity matrices, so these Hom bijections give an isomorphism of categories with the same rank objects. Formula (4.3) gives their ordinary matrix addition. The rank sum \(m+n\), with block inclusions and projections, remains a biproduct. The target category is equivalently the category of finite free \(\mathbb Z[1/p]\)-modules with chosen standard bases.

Exercise 3 (intermediate: a non-diagonal projector over the integers). On \(X=\mathbb Z^2\), define \(e(x,y)=(x+y,0)\). Split \(e\) and \(1-e\), and exhibit their direct-sum coordinates without extending scalars.

Solution. Direct calculation gives \(e^2(x,y)=e(x+y,0)=(x+y,0)\), so \(e\) is idempotent. Set

\[ \begin{array}{ll} i(a)=(a,0),&r(x,y)=x+y,\\ j(b)=(-b,b),&t(x,y)=y. \end{array} \tag{5.7} \]

Then \(ri=1_{\mathbb Z}\), \(ir=e\), \(tj=1_{\mathbb Z}\), and

\[ jt(x,y)=(-y,y)=(1-e)(x,y). \tag{5.8} \]

Also \(rj=0\) and \(ti=0\). The inverse coordinate maps are

\[ \begin{aligned} \Phi(x,y)&=(x+y,y),\\ \Psi(a,b)&=(a-b,b). \end{aligned} \tag{5.9} \]

Their two composites are identities by substitution, and all maps have integer coefficients. In these coordinates \(\Phi e\Psi(a,b)=(a,0)\). Thus the two summands are the subgroups generated by \((1,0)\) and \((-1,1)\); every \((x,y)\) has the unique decomposition \((x+y)(1,0)+y(-1,1)\). This is an algebraic direct sum over \(\mathbb Z\).

Exercise 4 (advanced: why finite projector sums matter). Let \(k\) be a field, \(V=\prod_{n\geq0}k\), and \(e_n\) the projector retaining only coordinate \(n\). Show that the \(e_n\) split and are pairwise orthogonal, but their retractions do not define a map \(V\to\bigoplus_{n\geq0}k\). Identify the obstruction to applying Theorem 2.3.

Solution. Let \(i_n:k\to V\) insert a scalar in coordinate \(n\) and zero elsewhere, and let \(r_n:V\to k\) select that coordinate. Then

\[ \begin{gathered} r_n i_n=1_k,\qquad i_n r_n=e_n,\\ e_n e_m=0\quad(n\ne m). \end{gathered} \tag{5.10} \]

These equations prove splitting and orthogonality. The family \((r_n)\) defines the identity map into the product by its universal property. A map \(R:V\to\bigoplus_n k\) with all coordinates \(r_n\), however, would send the vector \(v=(1,1,1,\ldots)\) to a vector with every coordinate equal to \(1\). No such vector belongs to the direct sum, whose elements have finite support. Hence \(R\) cannot exist.

The inclusions do define a map \(A:\bigoplus_nk\to V\), because each input has finite support. Its image consists precisely of the finite-support vectors, so \(v\) is outside its image and \(A\) is not an isomorphism. For every finite subset \(F\subset\mathbb N\), \(\sum_{n\in F}e_n\) keeps just the coordinates in \(F\); it is not \(1_V\). An abelian Hom group provides finite addition, with no infinite summation operation specified. In particular \(\sum_{n\geq0}e_n=1_V\) is not an equation in that group. Pointwise coordinate recovery in a product cannot replace the finite-sum hypothesis of Theorem 2.3.

References