Products, disjoint unions and mixed functors

Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Original text: CC0.

A pair of choices and a choice between alternatives lead to different categories. In a product, an arrow changes every coordinate at once. In a disjoint union, an arrow stays inside one alternative. A bifunctor combines two changes, and its two possible orders must agree. This agreement explains both the variance of Hom and the balanced tensor construction.

We use the category and size conventions of Universes and small categories, and the complete functor-category constructions in Natural transformations and composition of functors. All category data and indexing families lie in a sufficiently large ambient universe. A set is \(\mathcal U\)-small when it is bijective to a member of the fixed smaller universe; its actual labels need not belong to that universe. The module examples assume unital rings and unital modules. Their scalar ring \(k\) is commutative and maps into the center of the algebra \(R\), which may be noncommutative. Basic references are [Stacks, Categories], [Riehl] and the complete owned proofs cited below.

1. Independent coordinates

Let \((C_i)_{i\in I}\) be a family of categories indexed by an ambient set. Define its product \(P\) by

\[ \begin{gathered} \operatorname{Ob}P=\prod_{i\in I}\operatorname{Ob}C_i,\\ \operatorname{Hom}_P(X,Y)\\ =\prod_{i\in I}\operatorname{Hom}_{C_i}(X_i,Y_i). \end{gathered} \tag{1.1} \]

An object is an actual tuple of objects. An arrow \(f:X\to Y\) is a tuple of arrows with exactly those endpoints. Set \(1_X=(1_{X_i})_i\), and define composition by

\[ (gf)_i=g_i f_i. \tag{1.2} \]

For composable \(f,g,h\), both bracketings have coordinate \(h_i(g_i f_i)=(h_i g_i)f_i\). Equality at every coordinate is equality of tuples, so composition is associative. The two unit equations follow from \(f_i1_{X_i}=f_i=1_{Y_i}f_i\). These are all category laws. For two indices the notation is \(C_1\times C_2\). The two-factor definition agrees with the exact open product definition; the indexed construction uses the same coordinate operations.

If \(I=\varnothing\), the object product is the singleton containing the empty tuple, and its sole Hom set is also a singleton. Thus the empty product is a one-object, one-arrow category. If some factor has no objects and \(I\ne\varnothing\), no object tuple exists and the product is empty. Neither case requires choosing an object in every factor.

The complete small-family closure theorem in Universes and small categories, Proposition 3.1 applies to (1.1). If \(I\) is \(\mathcal U\)-small and every \(C_i\) is locally \(\mathcal U\)-small, the product is locally \(\mathcal U\)-small. If their object sets are small as well, the product category is small. An arbitrary ambient index set gives an ambient category; it need not give a small Hom set.

Suppose \(F_i:C_i\to D_i\) are functors. Their product sends \(X\) to \((F_iX_i)_i\) and \(f\) to \((F_if_i)_i\). Its identity equation holds at coordinate \(i\) because \(F_i1_{X_i}=1_{F_iX_i}\); its composition equation holds because \(F_i(g_if_i)=F_i(g_i)F_i(f_i)\). For an empty index set it is the unique functor between the two empty products.

The projections \(p_i:P\to C_i\) are functors by the same equations. For functors \(G_i:A\to C_i\), there is exactly one functor \(G:A\to P\) with \(p_iG=G_i\): take the tuple of object values and the tuple of arrow values. The coordinate equations prove its laws and uniqueness. For transformations \(\alpha_i:G_i\Rightarrow H_i\), their component tuples give a transformation \(G\Rightarrow H\); naturality at \(a:x\to y\) is exactly the family of equations \(H_i(a)\alpha_{i,x}=\alpha_{i,y}G_i(a)\). Every transformation into the product has these coordinates, and pointwise identities and compositions remain coordinatewise. Consequently this construction gives an isomorphism of categories

\[ \operatorname{Fun}(A,P) \cong\prod_{i\in I}\operatorname{Fun}(A,C_i). \tag{1.3} \]

It includes the empty index set: each side then has exactly one object and one arrow, even if \(A\) is empty.

2. Alternatives with retained labels

For the same family define \(S=\coprod_{i\in I}C_i\). Its objects are pairs \((i,X)\) with \(X\in C_i\). For tagged objects, H in the following formula denotes their Hom set:

\[ \begin{gathered} H_{ij}(X,Y)\\ =\begin{cases} \operatorname{Hom}_{C_i}(X,Y),&i=j,\\ \varnothing,&i\ne j. \end{cases} \end{gathered} \tag{2.1} \]

An arrow retains its component label. Its identity is the identity of its original object. If two arrows are composable, all three endpoints have the same component label, so their composite is defined in that one \(C_i\). Any composable triple stays there too, and its associativity and unit equations are the equations in \(C_i\). This proves the category laws. The label is essential: equal object names from different factors still denote different objects of \(S\).

For two indices write \(C_1\sqcup C_2\). The empty disjoint union has no objects or arrows. If all factors are empty, so is their union, even for a nonempty index set. Each inclusion \(j_i:C_i\to S\) preserves identities and composition because the operations inside its component are unchanged. It is fully faithful by (2.1).

Local \(\mathcal U\)-smallness of the factors implies local \(\mathcal U\)-smallness of \(S\): each Hom set is either one such Hom set or empty. This conclusion needs no smallness of \(I\). If \(I\) and every factor's object set are \(\mathcal U\)-small, the object set is small by the complete disjoint-union closure theorem cited in Section 1.

A family \(F_i:C_i\to D_i\) gives a functor between the disjoint unions by \((i,X)\mapsto(i,F_iX)\), sending an arrow in component \(i\) to its \(F_i\)-image in that component. Every composable pair lies in one factor; the corresponding functor equation proves the global equation. Identities are preserved there as well. Empty components and an empty family cause no exception.

Functors \(G_i:C_i\to D\) similarly determine exactly one functor \(G:S\to D\): use \(G_i\) on component \(i\). Restrictions recover the original functors. A transformation between two such functors is exactly a family of transformations on their components, since there are no equations crossing components. Components, identities and composites are all recovered under restriction. This proves

\[ \operatorname{Fun}(S,D) \cong\prod_{i\in I}\operatorname{Fun}(C_i,D). \tag{2.2} \]

For an empty family both sides are again a one-object, one-arrow category, for any \(D\), including an empty \(D\).

3. The square that joins two variables

A bifunctor is a functor \(H:C\times D\to E\). Fixing either object gives a functor in the other variable. For \(f:X\to X'\) in \(C\) and \(g:Y\to Y'\) in \(D\), the two factorizations of a product arrow give

\[ \begin{gathered} H(f,g)\\ =H(1_{X'},g)H(f,1_Y)\\ =H(f,1_{Y'})H(1_X,g). \end{gathered} \tag{3.1} \]

Each side has source \(H(X,Y)\) and target \(H(X',Y')\). The product identities and composition law show that the partial assignments preserve identities and composites. Equation (3.1) is the required mixed square.

Conversely, suppose object values \(H(X,Y)\) are given, with partial functors \(A_X:D\to E\) and \(B_Y:C\to E\) taking these values. Assume

\[ A_{X'}(g)B_Y(f)=B_{Y'}(f)A_X(g). \tag{3.2} \]

Here is the precise passage to the complete strict-currying proof in Natural transformations, Exercise 4. Set \(K(X)=A_X\), and let \(K(f)\) have component \(B_Y(f)\) at \(Y\). Equation (3.2) is exactly naturality of this component family. For every \(Y\), the partial functor \(B_Y\) gives \(B_Y(1_X)=1_{H(X,Y)}\) and \(B_Y(f'f)=B_Y(f')B_Y(f)\). Equality of transformations is componentwise, so these equations prove that \(K:C\to\operatorname{Fun}(D,E)\) is a functor.

The cited exercise supplies the complete uncurrying construction, its mixed composition law, its inverse on all arrows, and both empty-input cases. Applying that exact proof to \(K\) gives the unique bifunctor whose mixed arrow is

\[ \begin{aligned} H(f,g)&=A_{X'}(g)B_Y(f)\\ &=B_{Y'}(f)A_X(g). \end{aligned} \tag{3.3} \]

Uniqueness follows also from the product factorization: every mixed arrow is a composite of two separate arrows. Thus separate functoriality together with (3.2) is equivalent to joint functoriality. Separate functoriality alone is insufficient, as the second exercise shows.

4. Hom changes its first direction

For any ambient category \(C\), the literal Hom assignment has values in ambient sets. An arrow \((f^{\mathrm{op}},g):(X,Y)\to(X',Y')\) in \(C^{\mathrm{op}}\times C\) means

\[ f:X'\to X,\qquad g:Y\to Y'. \tag{4.1} \]

Define its action on \(h:X\to Y\) by

\[ H(f^{\mathrm{op}},g)(h)=ghf. \tag{4.2} \]

The result has the required endpoints \(X'\to Y'\). Identity arrows give \(1_Yh1_X=h\). For a second pair \(f':X''\to X'\), \(g':Y'\to Y''\), applying the two maps successively gives

\[ g'(ghf)f'=(g'g)h(ff'). \tag{4.3} \]

This is the map of the composite in \(C^{\mathrm{op}}\times C\), since its first coordinate corresponds to \(ff'\) in \(C\). Associativity in \(C\) proves the equation. Precomposition and postcomposition commute because \(g(hf)=(gh)f\), so the partial actions satisfy the mixed square too. No inverse or cancellation is required.

If \(C\) is locally \(\mathcal U\)-small, every Hom value is small. There is a distinction between small values and literal universe-member values. Let \(\mathcal U\text{-}\mathsf{Set}\) have exactly the members of \(\mathcal U\) as its objects. To make the literal Hom assignment a functor to this category, its actual Hom sets must belong to \(\mathcal U\). Local smallness alone gives a functor with small ambient values; one can transport it to universe-member values as follows.

For each pair \((X,Y)\), choose a member \(\widehat H(X,Y)\in\mathcal U\) and a bijection \(e_{X,Y}:\operatorname{Hom}_C(X,Y)\to\widehat H(X,Y)\). The pairs form an ambient set, so choice can be made in the ambient setting. For the pair of arrows in (4.1), define

\[ \widehat H(f^{\mathrm{op}},g) =e_{X',Y'}H(f^{\mathrm{op}},g)e_{X,Y}^{-1}. \tag{4.4} \]

For an identity, the adjacent inverse bijections cancel and give the identity of \(\widehat H(X,Y)\). For two composable pairs, the middle inverse bijections cancel; (4.3) then gives the composite equation. Thus \(\widehat H\) is a bifunctor to \(\mathcal U\text{-}\mathsf{Set}\). After inclusion into ambient sets, the family \(e_{X,Y}\) is a natural isomorphism from the original Hom functor: (4.4), composed with \(e_{X,Y}\), is exactly its naturality equation. Different encodings have the natural comparison \(e'_{X,Y}e_{X,Y}^{-1}\); the same cancellation proves naturality and compatibility of successive comparisons.

The distinction has a concrete one-object test. In ordinary well-founded set theory, \(\mathcal U\notin\mathcal U\). Take a category with one object and a single arrow whose label is the set \(\mathcal U\); that arrow is the identity and its self-composite is itself. Every category law holds because there is only one arrow. Its Hom set is \(\{\mathcal U\}\), bijective to a singleton in \(\mathcal U\), so the category is locally small. But \(\{\mathcal U\}\notin\mathcal U\): otherwise transitivity would imply \(\mathcal U\in\mathcal U\). Its literal Hom value therefore is not an object of \(\mathcal U\text{-}\mathsf{Set}\), whereas its encoded singleton value is. The small-valued version and the encoded universe-member version keep the Hom example valid with the precise stated size assumptions.

5. Two module examples and a topological contrast

For the central \(k\)-algebra \(R\), retain the complete constructions and proofs in Ring actions and tensor–Hom, Theorems 5.1 and 5.2, together with their exact ordinary-module identification in Hom, tensor and module limits, Section 1. They give

\[ \begin{gathered} \mathsf{Mod}\text{-}R\times R\text{-}\mathsf{Mod} \longrightarrow k\text{-}\mathsf{Mod},\\ (N,M)\longmapsto N\otimes_R M,\\ (R\text{-}\mathsf{Mod})^{\mathrm{op}} \times R\text{-}\mathsf{Mod} \longrightarrow k\text{-}\mathsf{Mod},\\ (M,L)\longmapsto\operatorname{Hom}_R(M,L). \end{gathered} \tag{5.1} \]

The first proof constructs a representing object from a free right-module presentation and a cokernel; the second constructs a representing object from a free left-module presentation and a kernel. Their complete proofs include arbitrary modules, both functor laws, the two-variable compatibility, and independence of presentations. The specialization takes the underlying abelian category to \(k\text{-}\mathsf{Mod}\). Its small products and coproducts are supplied by the complete module constructions retained in the second cited lesson. Thus no finite-generation or field hypothesis is being added.

On module maps \(a:N\to N'\), \(b:M\to M'\), the tensor arrow satisfies \(n\otimes m\mapsto a(n)\otimes b(m)\). On \(a:M'\to M\), \(b:L\to L'\), the Hom arrow is \(h\mapsto bha\). These are the arrows fixed by the representing properties in the cited proofs. For the tensor arrow, balance is preserved because

\[ \begin{aligned} a(nr)\otimes b(m) &=a(n)r\otimes b(m)\\ &=a(n)\otimes rb(m)\\ &=a(n)\otimes b(rm). \end{aligned} \tag{5.2} \]

The Hom arrow is \(R\)-linear since \(bha(rm')=b(h(ra(m')))=rbha(m')\), and it is \(k\)-linear in \(h\) by pointwise scalars. These equations identify the concrete operations with the exact retained bifunctors.

There is also an initial-object description valid for an arbitrary ring \(R\), without a specified ground field. Fix a right module \(N\) and a left module \(M\). Let \(\mathsf B(N,M)\) have objects \((L,c)\), where \(L\) is an abelian group and \(c:N\times M\to L\) is additive in both variables and satisfies \(c(nr,m)=c(n,rm)\). A map \((L,c)\to(L',c')\) is a group homomorphism \(t:L\to L'\) with \(tc=c'\).

An identity satisfies this equation. If \(tc=c'\) and \(sc'=c''\), then \((st)c=sc'=c''\); composition is therefore defined, and associativity and units are those of group homomorphisms. These facts prove that \(\mathsf B(N,M)\) is a category. Choosing one ambient universe containing the needed modules and groups keeps its object and arrow data sets; the same universal assertion holds for every group in any enlargement containing the tensor object.

Every unital ring is a central \(\mathbb Z\)-algebra: \(n\mapsto n1_R\) commutes with every ring element by the distributive laws. Specialize the retained balanced tensor proof to \(k=\mathbb Z\). With \(u(n,m)=n\otimes m\), it gives exactly one homomorphism

\[ t:N\otimes_R M\longrightarrow L, \qquad tu=c \tag{5.3} \]

for every balanced biadditive \(c\). Hence \((N\otimes_R M,u)\) is initial in \(\mathsf B(N,M)\). Existence and uniqueness in (5.3) are the actual complete owned representing proof and its evaluated balanced identification, rather than an assumption that a tensor already has the desired universal property.

Finally, let the topological spaces have underlying sets belonging to \(\mathcal U\). The underlying-set functor to \(\mathcal U\text{-}\mathsf{Set}\) sends a continuous map to the same function. Identities are continuous, and a composite of continuous maps is continuous: its inverse image of an open set is a successive pair of open inverse images. The underlying composite is the composite of the functions, proving the functor laws. Two continuous maps are equal exactly when their underlying functions are equal, so the functor is faithful.

It is not full. Give \(\{0,1\}\) the indiscrete topology in the source and the discrete topology in the target. The underlying identity function has inverse image \(\{0\}\) of the open singleton \(\{0\}\); this inverse image is not open in the indiscrete source. That function is not continuous. It belongs to the target Hom set but not to the image of the source Hom set, so this one Hom map is not surjective.

6. Constants and inclusions

For any category \(I\) and object \(X\in C\), the constant functor \(\Delta_X:I\to C\) sends every object to \(X\) and every arrow to \(1_X\). The complete identity and composition check, including empty \(I\), is in Natural transformations, Exercise 2. An arrow \(a:X\to Y\) gives the transformation with every component equal to \(a\); its naturality equation is \(1_Ya=a1_X\). Pointwise identities and composites make \(\Delta:C\to\operatorname{Fun}(I,C)\) a functor. For empty \(I\), its image is the unique empty diagram, even when \(C\) has many objects.

For a subcategory \(C'\subset C\), the inclusion sends an object or arrow to itself. The definition of subcategory retains identities and composites, so this assignment is a functor. Its maps on Hom sets are literal inclusions and are injective: it is faithful. It is fully faithful precisely when every one of those inclusions is surjective, which is precisely the definition of a full subcategory. The term embedding functor for this inclusion does not add fullness to a general subcategory.

7. Exercises with full solutions

Exercise 1 (beginner: a product and a choice)

Let \(A\) be the walking arrow \(0\to1\). Let \(B\) have objects \(b,c\), identities, and two distinct parallel arrows \(u,v:b\to c\). Count the objects and arrows of \(A\times B\) and \(A\sqcup B\). List the arrows from \((0,b)\) to \((1,c)\) in the product and explain their two factorizations. Compare the empty product with \(A\times\varnothing\).

Solution. The category \(A\) has two objects and three arrows; \(B\) has two objects and four arrows. Each product arrow is exactly a pair, so the product has four objects and twelve arrows. In the union the labels retain both object collections and both arrow collections, giving four objects and seven arrows. Its different components have no arrows between them.

Writing \(s:0\to1\), the requested product Hom set contains exactly \((s,u)\) and \((s,v)\). They are distinct because their second coordinates differ. For either \(w=u\) or \(w=v\), the two factorizations are

\[ \begin{aligned} (s,w)&=(1_1,w)(s,1_b)\\ &=(s,1_c)(1_0,w). \end{aligned} \tag{7.1} \]

Their intermediate objects are respectively \((1,b)\) and \((0,c)\); coordinate composition recovers the displayed arrow in both cases. The empty product has the single empty tuple as object and one identity arrow. The binary product \(A\times\varnothing\) has no object because its second coordinate cannot be supplied, so it has no arrows. These categories differ in object count and cannot be isomorphic or equivalent. \(\square\)

Exercise 2 (intermediate: separate actions that fail to combine)

Let \(G=\{1,t\}\) with \(t^2=1\), and regard \(BG\) as its one-object category. On \(S=\{0,1,2\}\), let \(a\) swap \(0,1\) and let \(b\) swap \(1,2\). Use \(a\) for the first partial \(BG\)-action and \(b\) for the second. Prove that each is a functor to sets but they cannot be the partial actions of a bifunctor \(BG\times BG\to\mathsf{Set}\) with object value \(S\). Decide whether replacing \(b\) by \(a\) removes the obstruction, and give the joint action when it does.

Solution. Both permutations square to the identity, so \(1\mapsto1_S\), \(t\mapsto a\) and \(1\mapsto1_S\), \(t\mapsto b\) preserve every product in \(G\). Each is a functor. The two arrows \((t,1)\) and \((1,t)\) commute in the product category, since either composite is \((t,t)\). Their images in any bifunctor must therefore commute. But

\[ (ba)(0)=2,\qquad (ab)(0)=1. \tag{7.2} \]

They are unequal functions. Equation (3.2) fails, so no such bifunctor exists. Choosing one ordering to define an image of \((t,t)\) cannot fix this, because functoriality must respect both factorizations.

After replacing \(b\) by \(a\), every partial action commutes with the other. Define

\[ H(t^i,t^j)=a^{i+j},\qquad i,j\in\{0,1\}, \tag{7.3} \]

with exponents read modulo two. The identity pair has exponent zero. Composing pairs adds their two exponents, and composing their images does the same because powers of \(a\) multiply by addition and \(a^2=1_S\). Thus (7.3) preserves all composites. It has the required object and both partial actions, and is the unique bifunctor by (3.3). \(\square\)

Exercise 3 (advanced: how many transformations are constant?)

Let \(J\) be the disjoint union of the walking arrow and the one-object category with endomorphisms \(1,p\), where \(p^2=p\). Put \(T=\{0,1,2\}\). Compute \(\operatorname{End}(\Delta_T)\) and \(\operatorname{Aut}(\Delta_T)\), with their cardinalities, and compare them with the images of the corresponding maps induced by \(\Delta:\mathsf{Set}\to\operatorname{Fun}(J,\mathsf{Set})\). Then prove that for any ambient small category \(I\), this constant-diagram functor on sets is fully faithful exactly when \(I\) is nonempty and connected. Include empty \(I\).

Solution. For constants \(\Delta_X,\Delta_Y\), naturality at an arrow \(i\to j\) says that their components satisfy \(\alpha_i=\alpha_j\), since both diagram arrows are identities. The complete component construction in Zero maps, Section 4 identifies equality generated by such endpoints with equality along finite zigzags. Hence a transformation is exactly one map \(X\to Y\) for each component of \(I\):

\[ \begin{gathered} \operatorname{Nat}(\Delta_X,\Delta_Y)\\ \cong\prod_{c\in\pi_0(I)}\operatorname{Hom}(X,Y). \end{gathered} \tag{7.4} \]

The inverse assigns that component's map to every object in it. Every arrow stays in one component, so the inverse assignment satisfies all naturality equations. Restricting and then assigning recovers every object component, and assigning and restricting recovers every component map. Thus (7.4) is a bijection, including its empty product.

Each factor of \(J\) is nonempty and connected, and no arrow crosses between the two. Thus \(\pi_0(J)\) has two elements. There are \(3^3=27\) self-maps of \(T\), independently on each component, giving \(27^2=729\) endomorphisms of \(\Delta_T\). Composition is componentwise. A transformation is invertible exactly when each component map is invertible, by the complete inverse-components lemma in Natural transformations, Lemma 2.1. There are six permutations of \(T\), so its automorphism group is \(S_3\times S_3\), of order \(36\). The maps from \(\operatorname{End}(T)\) and \(\operatorname{Aut}(T)\) have diagonal images: they supply the same function or permutation twice. Their respective image sizes are \(27\) and \(6\).

For any nonempty \(I\), equality of constant-component transformations can be tested at one object, so \(\Delta\) is faithful. If \(I\) is connected, (7.4) has exactly one factor, and every transformation comes from exactly one underlying map; \(\Delta\) is fully faithful. If \(I\) is nonempty but not connected, take the two-element set and assign its identity to one component and its transposition to every other component. This is a transformation by (7.4), with unequal object components, so it is outside the diagonal image. Fullness fails.

If \(I\) is empty, every diagram and transformation is empty and its functor category has one object and one arrow. Distinct self-maps of a two-element set have the same image, so faithfulness fails. Fullness also fails: there is no function from a singleton to the empty set, whereas there is one transformation between the corresponding empty diagrams. These cases prove the claimed equivalence with the nonempty connectedness convention. \(\square\)

Exercise 4 (expert: an idempotent seen by tensor and Hom)

Let \(R\) be any central \(k\)-algebra and \(e\in R\) satisfy \(e^2=e\). For every left \(R\)-module \(M\), prove natural isomorphisms of \(k\)-modules

\[ \begin{gathered} eR\otimes_R M\cong eM\\ \cong\operatorname{Hom}_R(Re,M). \end{gathered} \tag{7.5} \]

Here \(eM=\{em:m\in M\}\); \(e\) need not be central. Show that (7.5) respects the natural left \(eRe\)-actions and that the resulting functor is exact. Prove also that multiplication \(eR\otimes_R Re\to eRe\) is an isomorphism. For \(R=M_2(k)\), \(e=E_{11}\), express this last map in the four matrix entries of a first-row matrix and a first-column matrix. Allow any commutative \(k\).

Solution. The multiplication rule \((er,m)\mapsto erm\) is \(k\)-bilinear and balanced, since \((er)s\,m=er(sm)\). The complete balanced tensor property supplies a unique \(k\)-linear map

\[ \begin{gathered} \mu:eR\otimes_R M\to eM,\\ er\otimes m\mapsto erm. \end{gathered} \tag{7.6} \]

Define \(\nu:eM\to eR\otimes_R M\) by \(x\mapsto e\otimes x\). It is \(k\)-linear, and \(\mu\nu(x)=ex=x\). On a pure tensor, balance and \(e^2=e\) give

\[ \begin{aligned} \nu\mu(er\otimes m) &=e\otimes erm\\ &=ee\otimes rm\\ &=e\otimes rm\\ &=er\otimes m. \end{aligned} \tag{7.7} \]

Pure tensors generate the tensor module: if they generated only a proper submodule, the quotient map would be a nonzero map with zero composite with the universal balanced map, contradicting uniqueness against the zero map. Therefore (7.7) proves \(\nu\mu=1\) on the whole module.

For an \(R\)-linear map \(h:Re\to M\), \(h(e)=h(e^2)=eh(e)\), so evaluation at \(e\) lands in \(eM\). For \(x=ex\in eM\), set

\[ h_x(re)=rx. \tag{7.8} \]

If \(re=r'e\), then \((r-r')x=(r-r')ex=0\); the formula is well-defined. It is additive and \(R\)-linear. Evaluation gives \(h_x(e)=ex=x\), while any \(h\) satisfies \(h(re)=rh(e)\), so evaluation and (7.8) are inverse. They are \(k\)-linear. For an \(R\)-linear map \(f:M\to M'\), \(f(erm)=erf(m)\), \(f(eh(e))=ef(h(e))\), and \(f(rx)=rf(x)\). These equations prove naturality of both comparisons and their inverses.

For \(a\in eRe\), let it act on \(eR\) and on \(eM\) by left multiplication. On \(\operatorname{Hom}_R(Re,M)\), set \((a\cdot h)(y)=h(ya)\). Right multiplication by \(a\) is \(R\)-linear on \(Re\). The equation \(a\cdot(b\cdot h)(y)=h(yab)\) proves the left multiplication law; \(e\) acts as the identity because \(ye=y\) on \(Re\). Evaluation intertwines it with left multiplication: \((a\cdot h)(e)=h(a)=ah(e)\). Multiplication in (7.6) likewise sends \(a(er\otimes m)\) to \(aerm\). Thus both isomorphisms respect these left \(eRe\)-actions, without making \(eM\) a left \(R\)-submodule when \(e\) is noncentral.

To prove exactness, take a short exact sequence \(0\to M'\xrightarrow{u}M\xrightarrow{v}M''\to0\). Restriction of injective \(u\) to \(eM'\) is injective. If \(y\in eM\) and \(v(y)=0\), write \(y=u(x)\); then \(y=ey=eu(x)=u(ex)\), so the restricted kernel equals the restricted image. If \(z\in eM''\), choose \(x\in M\) with \(v(x)=z\). Then \(v(ex)=ev(x)=ez=z\), proving restricted surjectivity. The restricted sequence is therefore short exact, and (7.5) gives exactness for both displayed tensor and Hom functors.

Taking \(M=Re\) gives \(eM=eRe\), and (7.6) is the asserted multiplication isomorphism. For the matrix example,

\[ \begin{aligned} \begin{pmatrix}a&b\\0&0\end{pmatrix} \begin{pmatrix}c&0\\d&0\end{pmatrix} &=(ac+bd)E_{11}. \end{aligned} \tag{7.9} \]

The corner \(eRe\) is isomorphic to \(k\) via \(\lambda E_{11}\mapsto\lambda\). Associativity of matrix multiplication makes the displayed bilinear rule balanced over \(M_2(k)\). The general inverse is \(\lambda E_{11}\mapsto E_{11}\otimes\lambda E_{11}\), already verified in (7.7). No division or field hypothesis occurs, so the calculation works over every commutative scalar ring. \(\square\)

References