Extending maps, splittings and exact functors

Written by GPT-6.1 Sol (OpenAI), October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text public domain (CC0); referenced Stacks proofs retain GFDL-1.2 and linked course proofs retain their stated component terms. The ordinary arrow-projectivity proof is supplied in full below.

An injective object lets a map defined on a subobject extend to the containing object. Applying that property to its own identity makes every short exact sequence beginning with it split. This explains both the finite closure properties of injectives and why a quotient of an injective object can fail to be injective: the kernel must supply the splitting.

Work in locally small abelian categories, with all diagrams small in a fixed universe. We use choice for vector-space bases. The complete prerequisites are Testing exactness with doubled objects, Sections 1 and 3, for Hom and finite exactness tests; Splittings, support and sequential exactness, Section 1, for compatible splitting maps; and Locally nilpotent operators and coinduced duals, Section 1, for the full linear-dual adjunction and its field-valued injectivity proof. These proofs retain their original scope.

1. Retain the exactness tests

Let \(F:\mathcal A\to\mathcal B\) be additive. The complete proofs in Testing exactness, Sections 1 and 3, and Stacks, Lemma 12.7.2, give the following equivalent descriptions. Left exactness is preservation of kernels, hence of finite limits. Right exactness is preservation of cokernels, hence of finite colimits. The respective sequence tests are

\[ \begin{gathered} \text{left: }0\to A\to B\to C,\\ 0\to FA\to FB\to FC,\\ \text{right: }A\to B\to C\to0,\\ FA\to FB\to FC\to0. \end{gathered} \tag{1.1} \]

Within each pair, exactness of the first sequence is assumed and exactness of its image in the next line is required. Testing only short exact sequences also suffices, using the left or right three terms of their images. Full exactness is equivalent to preserving every three-term sequence exact at its middle object, including sequences with a zero endpoint; the image factorization and converse endpoint tests are proved in Section 1 of that prerequisite.

For a contravariant additive functor \(G:\mathcal A\to\mathcal B\), apply these statements to \(\mathcal A^{\mathrm{op}}\). Thus left exactness takes \(A\to B\to C\to0\) to \(0\to GC\to GB\to GA\). Right exactness takes \(0\to A\to B\to C\) to \(GC\to GB\to GA\to0\). In particular, \(\operatorname{Hom}(-,T)\) and \(\operatorname{Hom}(T,-)\) are left exact, by the full Hom universal-property proofs in the prerequisite. For an adjunction between abelian categories, the left adjoint is right exact and the right adjoint is left exact. Retain the complete preservation-of-colimits proof of Stacks, Lemma 4.24.5 and its finite-limit dual; Lemma 4.24.6 gives precisely this interface. No exactness of both adjoints is implied.

Preserving individual monomorphisms and epimorphisms is weaker. For a field \(k\), put \(A=k[x]\) and \(F(M)=xM\), with morphisms restricted to these images. Splittings, Section 2, Theorem 2.1 proves the stronger criterion for every commutative ring and scalar: this functor preserves monomorphisms and epimorphisms, and is left or right exact precisely when the scalar generates an idempotent ideal. In the present case it sends the short exact sequence with submodule \(xA\) to

\[ 0\to x^2A\to xA\to0\to0. \tag{1.2} \]

The inclusion \(x^2A\subset xA\) is proper: no polynomial multiplied by \(x^2\) equals \(x\). Consequently neither three-term exactness test holds at \(xA\). This is the precise specialization of the retained theorem.

2. A splitting survives every additive functor

Fix a short exact sequence

\[ 0\to A\xrightarrow f E\xrightarrow g Q\to0. \tag{2.1} \]

It is split if it has any of these equivalent structures: a section of \(g\); a retraction of \(f\); maps \(r:E\to A\) and \(s:Q\to E\) with \(fr+sg=1_E\); mutually inverse maps \([f,s]:A\oplus Q\to E\) and \([r,g]:E\to A\oplus Q\); surjectivity of \(\operatorname{Hom}(T,E)\to\operatorname{Hom}(T,Q)\) for every \(T\); or surjectivity of \(\operatorname{Hom}(E,T)\to\operatorname{Hom}(A,T)\) for every \(T\).

Retain the full equivalence proof of Splittings, Theorem 1.1 and its following paragraph. Here is the exact interface with the six structures just listed. If \(fr+sg=1_E\), composing with \(g\) and canceling the epimorphism \(g\) gives \(gs=1_Q\); composing with \(f\) and canceling the monomorphism \(f\) gives \(rf=1_A\). Thus the sum identity alone supplies the compatible maps of that theorem. The two surjectivity conditions give a section by testing \(T=Q\) and a retraction by testing \(T=A\), respectively. Conversely a section or retraction gives the indicated lifts by composition. Middle exactness and endpoint injectivity of the Hom sequences are already the universal Hom tests.

The compatible maps satisfy

\[ \begin{gathered} rf=1_A,\qquad gs=1_Q,\\ rs=0,\qquad fr+sg=1_E. \end{gathered} \tag{2.2} \]

An additive functor preserves these equations and the specified biproduct maps. Its image of (2.1) is therefore a split biproduct row, hence short exact. This proves preservation of split short exact sequences even for a functor that fails both general exactness tests. Exercise 2 describes the possible compatible maps rather than treating a splitting as unique.

3. Extending maps is the injective test

An object \(I\) is injective if the contravariant functor \(\operatorname{Hom}(-,I)\) is exact. The category has enough injectives if every object admits a monomorphism into some injective object. Dually, \(P\) is projective if \(\operatorname{Hom}(P,-)\) is exact, and there are enough projectives if every object is an epimorphic image of a projective object. Equivalently these full subcategories are, respectively, cogenerating and generating. These definitions make no claim that an arbitrary abelian category has either supply.

Proposition 3.1. An object \(I\) is injective if and only if every map \(a:A\to I\) extends along every monomorphism \(f:A\hookrightarrow E\).

Proof. Complete \(f\) to (2.1) by its cokernel. The contravariant Hom functor is left exact, so its image is already exact at its first two positions. The only missing condition for a short exact sequence is surjectivity of the restriction map

\[ \begin{gathered} \operatorname{Hom}(E,I)\longrightarrow\operatorname{Hom}(A,I),\\ b\longmapsto bf. \end{gathered} \tag{3.1} \]

That is precisely extension of each \(a\). The short-exact test in Section 1 proves both directions. \(\square\)

Passing to the opposite category gives the projective test: every map \(P\to Q\) lifts through every epimorphism \(E\twoheadrightarrow Q\). It is the same proof with the arrows reversed, using covariant Hom left exactness and surjectivity at the last position.

Corollary 3.2. A short exact sequence beginning with an injective object splits. A short exact sequence ending with a projective object splits.

Proof. If \(A=I\) is injective, extend \(1_I\) along \(f\) to obtain a retraction. If \(Q=P\) is projective, lift \(1_P\) through \(g\) to obtain a section. Section 2 identifies either structure with a splitting. \(\square\)

Every additive functor sends these sequences to split short exact sequences. It need not preserve other short exact sequences, nor send the injective or projective object itself to an injective or projective object. Exercise 3 identifies an adjunction hypothesis that does preserve them.

4. Finite sums, summands and quotient conditions

Proposition 4.1. A finite biproduct is injective if and only if every summand is injective. Every retract of an injective object is injective.

Proof. For a binary biproduct \(I_1\oplus I_2\), a map from \(A\) is its pair of component maps. If both targets are injective, extend the components along \(A\hookrightarrow E\) and combine the extensions. Conversely, suppose \(I_1\oplus I_2\) is injective. Compose a map \(A\to I_1\) with the first inclusion, extend to \(E\), then compose with the first projection. This extends the original map to \(I_1\). The second summand is identical with the indices interchanged. The zero object is injective because its Hom groups vanish; induction proves the finite statement, including the empty biproduct.

More generally, if \(j:I\to J\) and \(p:J\to I\) satisfy \(pj=1_I\), extend \(ja\) to \(b:E\to J\) and take \(pb\). Its restriction is \(pja=a\). This proves the retract assertion without requiring a chosen complementary object. \(\square\)

The proof in the opposite category gives the corresponding finite and retract assertions for projectives.

Corollary 4.2. If

\[ 0\to I'\to I\to Q\to0 \tag{4.1} \]

is short exact and both \(I'\) and \(I\) are injective, then \(Q\) is injective.

Proof. Injectivity of \(I'\) splits the sequence by Corollary 3.2. Hence \(Q\) is a retract of \(I\), so Proposition 4.1 applies. \(\square\)

The premise on \(I'\) matters. Exercise 1 gives an injective module with a quotient that is not injective. This argument also makes no assertion about infinite coproducts of injectives: a map into such a coproduct does not have a finite family of components to extend by this proof.

5. Semisimple categories and choice

An abelian category is semisimple if every short exact sequence splits. Section 2 then shows that every additive functor from it to an abelian category is exact. Indeed it preserves all short exact sequences, so the retained short-exact test applies. Equivalently every object is injective, or every object is projective: splitting every row makes both Hom functors exact; conversely injectivity of the first term, or projectivity of the last, splits each row by Corollary 3.2.

For \(k\) a field, every vector space is injective by the basis-extension argument retained in Locally nilpotent operators, Section 1: take \(R=k\), when evaluation identifies \(D_k(V)\) with \(V\). Hence every short exact sequence of vector spaces splits, and the category is semisimple. This includes infinite-dimensional spaces in the chosen universe. The resulting splittings are choices; they are not asserted to be natural in all short exact sequences.

The category of abelian groups is not semisimple. For any integer \(n\ge2\), consider

\[ 0\to\mathbb Z\xrightarrow{n}\mathbb Z\to\mathbb Z/n\to0. \tag{5.1} \]

A section would send the class of one to an integer killed by \(n\), hence to zero. Its reduction modulo \(n\) could not be the class of one. Thus this sequence has no section and does not split. The special case \(n=2\) already distinguishes an abelian category from a semisimple one.

6. Four graded exercises with full solutions

Exercise 1 (warm-up: an injective module with a bad quotient). Let \(k\) be a field of any characteristic and \(R=k[\varepsilon]/(\varepsilon^2)\). Show that the regular module \(R\) is both injective and projective, while \(K=R/(\varepsilon)\) is neither. Identify which hypothesis of Corollary 4.2 fails for \(0\to\varepsilon R\to R\to K\to0\).

Solution. Let \(\lambda:R\to k\) extract the coefficient of \(\varepsilon\). Define

\[ \begin{gathered} \Phi:R\longrightarrow D_R(k),\\ \Phi(b)(a)=\lambda(ab). \end{gathered} \tag{6.1} \]

The retained dual action is \((c\xi)(a)=\xi(ac)\), so \(\Phi(cb)(a)=\lambda(acb)=(c\Phi(b))(a)\). Thus \(\Phi\) is \(R\)-linear. If \(b=u+v\varepsilon\), its values on \(1,\varepsilon\) are \(v,u\). These two values give every linear functional uniquely, so \(\Phi\) is an isomorphism in every characteristic. The complete field-valued injectivity proof for \(D_R(k)\) in the prerequisite proves that \(R\) is injective. The complete arbitrary-ring finite-support lifting proof in Ordinary modules, Section 3 proves projectivity of the regular left module \(R\).

The map \(\varepsilon R\to K\) taking \(\varepsilon\) to the class of one is \(R\)-linear, since \(\varepsilon\) kills both source and target. An extension \(R\to K\) would have to send \(\varepsilon\) to \(\varepsilon\) times its value at one, which is zero. Therefore \(K\) is not injective. A section of \(R\twoheadrightarrow K\) would send the class of one to \(u+v\varepsilon\) with \(u=1\), while \(R\)-linearity would require \(\varepsilon(u+v\varepsilon)=0\), giving \(u=0\). No section exists, so \(K\) is not projective. Finally \(\varepsilon R\simeq K\) by \(\varepsilon\mapsto\overline1\), so the kernel is not injective. The middle term alone does not supply Corollary 4.2.

Exercise 2 (intermediate: all compatible splittings). In (2.1), assume \(A\) is injective. Choose compatible maps \(r,s\) as in (2.2). Parametrize all compatible pairs \(r',s'\), and describe the resulting idempotents on \(E\).

Solution. For every \(a:Q\to A\), put

\[ \begin{gathered} r_a=r+ag,\\ s_a=s-fa. \end{gathered} \tag{6.2} \]

The equations \(gf=0\), \(rf=1\), \(gs=1\), \(rs=0\) give \(r_af=1\), \(gs_a=1\), \(r_as_a=-a+a=0\), and \(fr_a+s_ag=1\). Thus these pairs are compatible. Conversely, \((r'-r)f=0\), so the cokernel property of \(g\) gives a unique \(a\) with \(r'-r=ag\). Since \(g(s'-s)=0\), the kernel property of \(f\) gives a unique \(b:Q\to A\) with \(s'-s=fb\). The equation \(r's'=0\) now reads \(a+b=0\). Hence \(b=-a\), and the displayed pairs are all the compatible pairs, uniquely parametrized by \(\operatorname{Hom}(Q,A)\). This is a torsor description: choosing \(r,s\) chooses an origin.

The endomorphism \(e_a=fr_a\) is idempotent, since \(r_af=1\). It acts as the identity on the specified subobject \(f(A)\), and \(1-e_a=s_ag\). Its image is \(f(A)\), and its kernel is the image of \(s_a\): for a map \(t:T\to E\) killed by \(e_a\), the identity \(1-e_a=s_ag\) gives \(t=s_agt\), uniquely because \(gs_a=1\). Distinct \(a\) give distinct idempotents: \(f(a-a')g=0\) implies \(a=a'\) by monicity of \(f\) and epicity of \(g\). The decomposition is specified but generally not unique.

Exercise 3 (hard: transfer through an adjunction). Let additive functors \(L:\mathcal A\to\mathcal B\) and \(R:\mathcal B\to\mathcal A\) satisfy \(L\dashv R\). Prove that exactness of \(L\) makes \(R\) preserve injectives. If \(\mathcal B\) has enough injectives, prove the converse. State and prove the corresponding projective assertion, including which category needs enough projectives.

Solution. The adjunction is natural in the first variable:

\[ \begin{gathered} \operatorname{Hom}_{\mathcal B}(LX,J)\\ \simeq\operatorname{Hom}_{\mathcal A}(X,RJ). \end{gathered} \tag{6.3} \]

If \(L\) is exact and \(J\) is injective, applying \(L\) and then \(\operatorname{Hom}_{\mathcal B}(-,J)\) takes any short exact sequence to a short exact sequence of groups. Naturality in (6.3) identifies its arrows with the Hom arrows for \(RJ\). Hence \(RJ\) is injective.

Conversely, assume \(R\) preserves injectives and \(\mathcal B\) has enough of them. Let \(i:X\hookrightarrow Y\). Choose a monomorphism \(j:LX\hookrightarrow J\) with \(J\) injective. Its adjoint \(X\to RJ\) extends to \(Y\to RJ\), because \(RJ\) is injective. The adjoint \(h:LY\to J\) then satisfies \(hLi=j\), by naturality. A map whose composite is monic is monic, so \(Li\) is monic. The left adjoint \(L\) is right exact by the retained adjoint theorem. For a short exact sequence, right exactness already gives exactness of its last three terms; the preservation of its first monomorphism adds the remaining condition. Thus \(L\) is exact.

Dually, exactness of \(R\) makes \(L\) preserve projectives: the covariant adjunction \(\operatorname{Hom}_{\mathcal B}(LP,Y)\simeq\operatorname{Hom}_{\mathcal A}(P,RY)\) identifies the two exact Hom sequences. For the converse assume \(\mathcal A\) has enough projectives and \(L\) preserves them. For an epimorphism \(q:Y \twoheadrightarrow Z\) in \(\mathcal B\), choose \(p:P \twoheadrightarrow RZ\) with \(P\) projective. Its adjoint \(LP\to Z\) lifts through \(q\), since \(LP\) is projective. The adjoint lift \(t:P\to RY\) satisfies \(Rq\,t=p\). Since this composite is epic, \(Rq\) is epic. The right adjoint \(R\) is left exact, and adding preservation of the final epimorphism to the left three terms makes every short exact sequence exact. Thus \(R\) is exact. The converse injective test needs enough injectives in \(\mathcal B\); the converse projective test needs enough projectives in \(\mathcal A\).

Exercise 4 (advanced: projectivity of a linear arrow). In the abelian category of arrows \(d:V\to W\) of \(k\)-vector spaces, show that an arrow is projective exactly when \(d\) is monic. Give a projective epimorphism onto every arrow. Combine this with the retained injective classification to determine the objects that are both injective and projective.

Solution. We prove the ordinary arrow statements, including arbitrary small coproducts, directly.

Let \(k\) be a field. Work with vector spaces, linear maps and indexing sets in a fixed universe, using choice for bases and simultaneous lifts. Write \(\mathsf{Arr}(k)\) for the category whose objects are linear maps \(d:V\to W\). A morphism

\[ (a,b):(V\xrightarrow d W)\longrightarrow(X\xrightarrow e Y) \]

is a pair of linear maps satisfying \(bd=ea\). The word projective concerns this object of \(\mathsf{Arr}(k)\); the word injective in the criterion below concerns the linear map \(d\).

Kernels and cokernels of \((a,b)\) are the induced arrows \(\ker a\to\ker b\) and \(\operatorname{coker}a\to\operatorname{coker}b\). Their universal properties follow by factoring the two components; commutativity persists under these factors. Zero objects and finite biproducts are componentwise. The componentwise coimage–image isomorphisms form an isomorphism of arrows, so \(\mathsf{Arr}(k)\) is abelian. A sequence is exact precisely when its two component sequences are exact. In particular, an epimorphism has two surjective components. Small coproducts are componentwise direct sums: maps out of either sum are exactly the families of component maps, and the square commutes precisely when every summand square does.

Put

\[ P_1=(k\xrightarrow{1_k}k),\qquad P_0=(0\to k). \]

For any \(D=(V\xrightarrow d W)\), evaluation gives natural linear isomorphisms

\[ \operatorname{Hom}_{\mathsf{Arr}(k)}(P_1,D)\simeq V, \qquad \operatorname{Hom}_{\mathsf{Arr}(k)}(P_0,D)\simeq W. \]

The first sends a commuting pair \((a,b)\) to \(a(1)\). Its inverse sends \(v\) to the pair \(a(\lambda)=\lambda v\), \(b(\lambda)=\lambda d(v)\), which is forced by \(b=da\). The second sends \((0,b)\) to \(b(1)\), with inverse \(b(\lambda)=\lambda w\). Composition with a morphism of target arrows acts by its corresponding component, proving naturality. Component exactness therefore makes both Hom functors exact, so \(P_1\) and \(P_0\) are projective.

Here is the complete lifting proof for arbitrary small coproducts, including mixed coproducts. For small sets \(A,B\), their coproduct has the form

\[ S(A,B)=\left(k^{(A)}\xrightarrow{\iota_1} k^{(A)}\oplus k^{(B)}\right). \]

Take an epimorphism

\[ q=(q_X,q_Y):(X\xrightarrow e Y)\twoheadrightarrow (X'\xrightarrow{e'}Y') \]

and a morphism \(f=(f_X,f_Y):S(A,B)\to(X'\xrightarrow{e'}Y')\). For each \(a\in A\), choose \(x_a\in X\) with \(q_Xx_a=f_X(e_a)\). For each \(b\in B\), choose \(y_b\in Y\) with \(q_Yy_b=f_Y(0,e_b)\). Surjectivity and choice supply these families. Define

\[ \begin{aligned} \ell_X(e_a)&=x_a,\\ \ell_Y(e_a,0)&=e(x_a),\\ \ell_Y(0,e_b)&=y_b. \end{aligned} \]

Extend linearly using finite supports. Then \(\ell_Y\iota_1=e\ell_X\). On each first-block basis vector,

\[ q_Ye(x_a)=e'q_X(x_a)=e'f_X(e_a)=f_Y(e_a,0), \]

and on each second-block basis vector the lift equation is the defining choice of \(y_b\). Thus \(q\ell=f\). This proves projectivity of \(S(A,B)\) by the projective lifting test. If either set is empty, the corresponding formulas impose no choices. If both are empty, the object and its unique lift are zero.

Proposition. The arrow object \(D=(V\xrightarrow d W)\) is projective if and only if the linear map \(d\) is injective. Every arrow is an epimorphic image of a projective arrow.

Proof. For every \(d\), consider

\[ Q=\left(V\xrightarrow{\iota_1}V\oplus W\right), \qquad q=(1_V,t):Q\twoheadrightarrow D, \qquad t(v,w)=d(v)+w. \tag{6.4} \]

It is a morphism because \(t\iota_1=d\), and is epic because both components are onto. If \(D\) is projective, lift \(1_D\) through \(q\), obtaining a section \((1_V,b)\). Its commuting-square equation is \(bd=\iota_1\). Consequently \((\pi_Vb)d=1_V\), so \(d\) is injective.

Conversely, suppose \(d\) is injective. Choose a complement \(C\) with \(W=d(V)\oplus C\). The pair \((1_V,h)\), where \(h(v,c)=d(v)+c\), is an isomorphism

\[ \left(V\xrightarrow{\iota_1}V\oplus C\right) \xrightarrow{\sim}(V\xrightarrow d W). \]

Its square commutes, and both components are invertible. Choose bases \(A\) of \(V\) and \(B\) of \(C\); their basis isomorphisms identify its source with \(S(A,B)\). The lifting proof above makes that source projective. Projectivity transports across an isomorphism: precompose a given map with the isomorphism, lift it, and compose the lift with the inverse. Hence \(D\) is projective.

The same basis construction, with bases of \(V\) and \(W\), identifies \(Q\) with a projective \(S(A,B)\). Thus the displayed \(q\) supplies enough projectives. Zero spaces and empty bases satisfy the same formulas, including the zero arrow. \(\square\)

For injectivity retain the complete Sierpiński arrow proof, Exercise 3, together with the sheaf-to-arrow equivalence, Example 6. On the Sierpiński space with opens \(\varnothing,\{o\},S\), a sheaf is precisely its linear restriction arrow \(F(S)\to F(\{o\})\), with commuting pairs as morphisms. This equivalence matches the two component vector spaces and their exact sequences. The linked complete proof says precisely that such an arrow is injective if and only if it is onto. Combining this with the projective criterion just proved, the objects that are both are exactly the isomorphism arrows. An arrow can have injective and projective component spaces while possessing neither property as a diagram; any linear map that is neither monic nor epic gives such an example.

7. References