Operators, matrix relations and internal tensor–Hom

Written and self-checked with GPT-6.1 Sol (OpenAI), Ultra reasoning effort. Original text: CC0. This reconstructed version has no independent review.

An operator gives two useful objects: its kernel and its cokernel. They can have the same dimension while the map naturally joining them fails to be an isomorphism. We begin with that computation, then use finite matrices of operators to interpret module presentations in an abelian category. Infinite presentations require additional limits; the finite calculation tells us precisely which ones.

Throughout, $k$ is a commutative unital ring, $R$ is an associative unital $k$-algebra with central scalars, and $\mathcal C$ is a $k$-linear abelian category. Thus its morphism groups are $k$-modules and composition is $k$-bilinear. Neither $R$ nor its internal actions need be commutative. Rings, ordinary modules and indexing sets belong to a fixed universe; morphism groups are small there, and categories may belong to a larger universe. A category need not be essentially small.

We use the full kernel/cokernel and coimage–image proofs in Kernels, cokernels and the abelian comparison. Section 6 gives the ordinary balanced-module construction needed for iteration. The constructions here do not require that an object of $\mathcal C$ have an underlying set.

1. A kernel and a quotient joined by a nonisomorphism

Worked problem. Let $R=k[t]$, first with $k$ a field, and let $t$ act on $X$ by an operator $T$. What does the relation $t^n=0$ select? A map into $X$ that respects this relation has image in $\ker T^n$. A map out of $X$ on which this relation vanishes factors through $\operatorname{coker}T^n$. Section 4 will prove that these are respectively $H_R(R/(t^n),X)$ and $R/(t^n)\otimes_R X$, using only a finite presentation.

Take $X=k^3$, with $Te_1=0$, $Te_2=e_1$, $Te_3=e_2$, and take $n=2$. The specified comparison is $\ker T^2\longrightarrow X\longrightarrow\operatorname{coker}T^2. \tag{1.1}$ The finite presentation $R\xrightarrow{t^n}R\to R/(t^n)\to0$ gives $\begin{gathered} R/(t^n)\otimes_R X=\operatorname{coker}T^n,\\ H_R(R/(t^n),X)=\ker T^n. \end{gathered}$ For the given operator, $\ker T^2$ has basis $e_1,e_2$, and $\operatorname{im}T^2=ke_1$. The cokernel has basis $[e_2],[e_3]$. The requested map sends $e_1$ to zero and $e_2$ to $[e_2]$, so its rank is one. Although the kernel and cokernel have equal dimensions here, their specified comparison is not an isomorphism.

There is a short exact sequence $0\longrightarrow R/(t) \xrightarrow{\;1\mapsto t\;}R/(t^2) \longrightarrow R/(t)\longrightarrow0.$ Tensoring with the right module $R/(t)$ means quotienting each term by the action of $t$. The first map becomes the zero map $k\to k$, losing monicity. Applying $\operatorname{Hom}_R(R/(t),-)$ means taking the elements killed by $t$. In $R/(t^2)$ these are the multiples of $t$, all of which map to zero in $R/(t)$. Thus the last induced map is zero $k\to k$, losing surjectivity. This proves that the general one-sided exactness assertions cannot be strengthened to exactness.

The maps matter as much as the objects: replacing (1.1) by some chosen vector-space isomorphism would erase the obstruction. This distinction will recur when a functor compares a matrix kernel or cokernel with the kernel or cokernel of its image.

2. Actions on objects and equivariant arrows

Write $\mathcal C_R$ for the category of internal left $R$-modules. An object is a pair $(X,\xi_X)$, where $X\in\mathcal C$ and $\xi_X: R\longrightarrow\operatorname{End}_{\mathcal C}(X)$ is a unital $k$-algebra homomorphism. An arrow $f: (X,\xi_X)\to(Y,\xi_Y)$ is an arrow in $\mathcal C$ satisfying $f\xi_X(r)=\xi_Y(r)f\qquad(r\in R).$ This definition is meaningful even when $X$ has no chosen underlying set.

Sums and scalar multiples of equivariant arrows are equivariant. The zero object carries its unique action, and a finite biproduct carries the diagonal action. The usual injections and projections are equivariant. Hence $\mathcal C_R$ is $k$-linear and additive. Forgetting the action is a faithful $k$-linear functor.

It need not be full. For a nonzero field $k$, give the one-dimensional space $k$ two $k[t]$-actions, with $t$ acting respectively as zero and as the identity. The underlying identity map is not equivariant: commuting with $t$ would assert $0=1$.

The center $Z(R)$ acts on equivariant morphism groups by $zf=\xi_Y(z)f=f\xi_X(z).$ For central $z$, this arrow remains equivariant; the action is compatible with composition. Thus $\mathcal C_R$ is $Z(R)$-linear, including $R$-linear when $R$ is commutative.

Two Hom groups have different sides: $\begin{gathered} \operatorname{Hom}_{\mathcal C}(Y,X)\text{ is a left }R\text{-module},\\ r\cdot h=\xi_X(r)h,\\ \operatorname{Hom}_{\mathcal C}(X,Y)\text{ is a right }R\text{-module},\\ h\cdot r=h\xi_X(r). \end{gathered}$ Here $Y$ is any object of $\mathcal C$, with no action required. For the right action, $(h\cdot r)\cdot s=h\xi_X(rs)$; this checks the multiplication order explicitly.

A $k$-linear functor $F: \mathcal C\to\mathcal D$ sends $(X,\xi_X)$ to $(FX,F\xi_X)$. It respects all algebra laws because it preserves composition, identities and $k$-linear sums. It sends equivariant arrows to equivariant arrows, and forgetting actions commutes with this lifted functor.

For $\mathcal C=k\text{-}\mathsf{Mod}$, this construction is precisely the usual category of left $R$-modules: a left action consists of $k$-linear operators with these algebra laws, and an $R$-linear map is exactly an equivariant $k$-linear map.

Taking opposites reverses the algebra as well: $(\mathcal C_R)^{\mathrm{op}}\simeq (\mathcal C^{\mathrm{op}})_{R^{\mathrm{op}}}. \tag{2.1}$ Indeed, $\operatorname{End}_{\mathcal C^{\mathrm{op}}}(X)$ is the opposite of $\operatorname{End}_{\mathcal C}(X)$. The same operators therefore define an $R^{\mathrm{op}}$-action there. Reversing an equivariant arrow gives exactly the equivariance equation in the opposite category. This operation reverses back to the original objects and arrows, so (2.1) is an equivalence without additional choices.

3. Kernels and cokernels retain the action

Theorem 3.1. If $\mathcal C$ is $k$-linear and abelian, then $\mathcal C_R$ is $k$-linear and abelian. The forgetful functor creates kernels and cokernels and is exact. A left exact or right exact $k$-linear functor lifts to a functor with the same exactness.

Proof. Let $f: X\to Y$ be equivariant, and let $i: K\to X$ be its kernel in $\mathcal C$. Since $f\xi_X(r)i=\xi_Y(r)fi=0$, there is a unique operator $\xi_K(r)$ with $i\xi_K(r)=\xi_X(r)i.$ Canceling the monomorphism $i$ verifies every additive, multiplicative, scalar and unit law. If an equivariant arrow $h: Z\to X$ satisfies $fh=0$, its unique kernel factorization $h=i\bar h$ is equivariant: compose the desired equation with $i$ and cancel $i$. Thus this is the kernel in $\mathcal C_R$.

Let $q: Y\to L$ be the cokernel of $f$. The equality $q\xi_Y(r)f=qf\xi_X(r)=0$ defines a unique $\xi_L(r)$ with $\xi_L(r)q=q\xi_Y(r).$ Canceling the epimorphism $q$ verifies the algebra laws. The same cancellation shows that every factorization of an equivariant arrow through $q$ is equivariant. Hence this is the internal cokernel.

The canonical coimage-to-image map is equivariant because all its constituent kernel and cokernel maps are. Its underlying map is an isomorphism in the abelian category $\mathcal C$. The inverse of an equivariant isomorphism is equivariant, by multiplying its commutation equation by the inverse on both sides. Thus $\mathcal C_R$ is abelian.

These constructions also show that exactness is detected on underlying objects: the image and kernel subobjects have the underlying image and kernel, and their comparison is an isomorphism exactly when its underlying map is. Finally, a left exact $F$ preserves the underlying kernel diagrams, whose induced actions are fixed by uniqueness; its lift preserves kernels. A right exact $F$ preserves the corresponding cokernel diagrams and their actions. Both lifts are additive, so these are the asserted exactness properties. $\square$

When a small limit or colimit exists in $\mathcal C$, the corresponding diagram of internal modules has one with the same underlying object. For a limit, define each operator by its composites with the limiting projections; for a colimit, define it by its composites with the coprojections. Joint uniqueness verifies the algebra laws and the equivariant universal property. In particular, the needed products or coproducts of internal modules exist under the corresponding assumptions on $\mathcal C$.

4. Finite matrices before infinite presentations

For a right module $N$ and a left module $M$, the desired objects have different universal properties: $\begin{gathered} \operatorname{Hom}_{\mathcal C}(N\otimes_R X,Y)\\ \simeq\operatorname{Hom}_{R^{\mathrm{op}}} (N,\operatorname{Hom}_{\mathcal C}(X,Y)), \end{gathered} \tag{4.1}$ $\begin{gathered} \operatorname{Hom}_{\mathcal C}(Y,H_R(M,X))\\ \simeq\operatorname{Hom}_{R} (M,\operatorname{Hom}_{\mathcal C}(Y,X)). \end{gathered} \tag{4.2}$ The inner Hom in (4.1) is a right module by precomposition; the inner Hom in (4.2) is a left module by postcomposition. The output is an object of $\mathcal C$, whereas the outer module Hom is an ordinary morphism group. These formulas specify the objects even before we know that they exist.

Theorem 4.1 (finite presentation). If $N$ is finitely presented as a right $R$-module, the object in (4.1) exists. If $M$ is finitely presented as a left $R$-module, the object in (4.2) exists. Their specified universal properties give canonical, mutually compatible presentation comparisons and additive $k$-linear bifunctors, covariant in both tensor variables and contravariant in the first Hom variable. No infinite product or coproduct is needed.

Proof. Take a finite right-module presentation $R^m\xrightarrow{d}R^q\longrightarrow N\longrightarrow0, \qquad d(e_j)=\sum_i e_i a_{ij}.$ Define $D_X:X^m\to X^q$ by these same operators $\xi_X(a_{ij})$, and put $T=\operatorname{coker}D_X$. A map $X^q\to Y$ is a family $h_i:X\to Y$. It kills $D_X$ exactly when $\sum_i h_i\xi_X(a_{ij})=0$ for each $j$. Evaluation on the free basis identifies these with right-linear maps $R^q\to\operatorname{Hom}(X,Y)$ killing $d$. Factoring through $N$ gives (4.1), naturally in $Y$.

For a finite left-module presentation $R^m\to R^q\to M\to0$, write instead $d(e_j)=\sum_i a_{ij}e_i$. Define $E_X:X^q\to X^m$ by $\pi_jE_X=\sum_i\xi_X(a_{ij})\pi_i$. A map $Y\to X^q$ lies in its kernel exactly when its components $h_i:Y\to X$ satisfy $\sum_i\xi_X(a_{ij})h_i=0$. These are the relations for a left-linear map $M\to\operatorname{Hom}(Y,X)$, proving (4.2) for $H=\ker E_X$.

Here is the presentation comparison immediately. Two objects with the same specified natural Hom bijections have unique comparison maps compatible with those bijections: evaluate one bijection on the identity of the other representing object. The composites correspond to identities, so the maps are inverse. For three presentations, a composite of two compatible comparisons is the third compatible comparison by uniqueness. Thus the comparisons satisfy the cocycle law.

A right-module arrow $N\to N'$ precomposes the module Hom in (4.1); it therefore gives $N\otimes_R X\to N'\otimes_R X$. An equivariant arrow $X\to X'$ precomposes the inner Hom and gives the same covariant direction in the object variable. In (4.2), precomposition in $M$ gives the reverse direction, and postcomposition in $X$ gives the forward direction. The transformations in the two variables commute. Uniqueness proves identities, composition, their mutual commutation and naturality of every presentation comparison. The bijections are $k$-linear, so sums and scalar multiples of input arrows induce the corresponding sums and scalar multiples of output arrows. Zero objects and finite direct sums represent the zero and direct-sum Hom functors; this also proves additivity. $\square$

For $k[t]/(t^n)$, the one-relation matrix is $T^n$. Both formulas in the first worked problem now follow in every $k$-linear abelian category. A finite free presentation also computes the two functors explicitly, without choosing elements of $X$.

The one-sided exactness proofs in Section 5 use only the selected presentation. Hence they already apply here to the object variable for every fixed finitely presented module, and to sequences whose ordinary module terms are all finitely presented. Over a general ring this does not assert that finitely presented modules form an abelian category.

5. Which infinite operations are actually needed

Assume first that $\mathcal C$ has all small coproducts. Every small right $R$-module has a presentation by small free modules, so the finite tensor matrix extends to a coproduct matrix. Each relation still has finite support.

Theorem 5.1 (arbitrary right modules). The object in (4.1) exists. These representing objects define a $k$-linear bifunctor $(-\otimes_R-): \mathsf{Mod}\text{-}R\times\mathcal C_R\longrightarrow\mathcal C$ that is additive and right exact in each variable. All choices of presentations give canonically compatible representing objects.

Proof of existence and functoriality. Choose a free right-module presentation $R^{(J)}\xrightarrow{d}R^{(I)}\longrightarrow N\longrightarrow0.$ Write $d(e_j)=\sum_i e_i a_{ij}$. For each $j$, only finitely many coefficients are nonzero. The map $D_X: X^{(J)}\longrightarrow X^{(I)}$ has, on the $j$-th summand, the finite sum of the injections of $\xi_X(a_{ij})$. Set $T=\operatorname{coker}D_X$.

A map $X^{(I)}\to Y$ is a family $h_i: X\to Y$. It kills $D_X$ exactly when $\sum_i h_i\xi_X(a_{ij})=0\qquad(j\in J). \tag{5.1}$ Evaluation at the basis vectors identifies these families with the right $R$-linear maps $R^{(I)}\to\operatorname{Hom}_{\mathcal C}(X,Y)$ that kill $d$. They are exactly the right-module maps from $N$. This proves (4.1), naturally in $Y$.

If $a: N\to N'$ is a right-module map, precomposition gives a transformation from the functor represented by $N'\otimes_R X$ to the functor represented by $N\otimes_R X$. The representing map has the direction $N\otimes_R X\to N'\otimes_R X$. If $b: X\to X'$ is equivariant, precomposition $\operatorname{Hom}(X',Y)\to\operatorname{Hom}(X,Y)$ is right $R$-linear and gives $N\otimes_R X\to N\otimes_R X'$. These transformations commute. Uniqueness of representing maps proves identity and composition laws, hence bifunctoriality.

The correspondence is $k$-linear. Therefore sums of input arrows induce sums of output arrows. A zero input represents the zero functor; a finite direct sum in either variable represents the direct sum of the two represented functors. Their canonical comparisons give the additive structure.

Two presentations represent the same specified functor (4.1). There is a unique isomorphism compatible with that representation. Between three choices, the composite of two such isomorphisms has the same compatibility as the third, so uniqueness proves its equality with the third. The same argument gives naturality in both variables. $\square$

Proof of right exactness. For $N_1\to N_2\to N_3\to0$, maps from $N_3$ to any right module are precisely the maps from $N_2$ killing $N_1$. By (4.1), this says that $N_3\otimes_R X$ is the cokernel of $N_1\otimes_R X\to N_2\otimes_R X$, with the specified structure map.

Now fix $N$ and its presentation, and let $X_1\xrightarrow{u}X_2\xrightarrow{q}X_3\to0$ be right exact in $\mathcal C_R$. A map $N\otimes_R X_2\to Y$ is a family $h_i: X_2\to Y$ satisfying (5.1). Its composite with $N\otimes_R X_1$ is zero exactly when all $h_i u$ vanish: the coproduct $X_1^{(I)}$ maps epimorphically to $N\otimes_R X_1$.

Each such $h_i$ uniquely descends to $\bar h_i: X_3\to Y$, because $q$ is the underlying cokernel of $u$. The relation for $\bar h_i$, composed with $q$, is the relation for $h_i$, since $q$ is equivariant. Canceling $q$ proves the relation for $\bar h_i$. Thus maps from $N\otimes_R X_3$ to $Y$ are exactly maps from $N\otimes_R X_2$ that kill $N\otimes_R X_1$. This is the cokernel property, for every $Y$, so the tensor sequence is right exact. No exactness of infinite products or filtered colimits was used. $\square$

Assume instead that $\mathcal C$ has all small products. Every small left module has a small free presentation. The corresponding product matrix exists because each output component uses only the finitely many coefficients of one relation; no infinite sum of endomorphisms is taken.

Theorem 5.2 (arbitrary left modules). The object in (4.2) exists and defines an additive $k$-linear bifunctor $H_R: \bigl(R\text{-}\mathsf{Mod}\bigr)^{\mathrm{op}} \times\mathcal C_R\longrightarrow\mathcal C.$ It is left exact in each variable, with contravariance in $M$.

Proof. Choose a left-module presentation $R^{(J)}\to R^{(I)}\to M\to0$, with $d(e_j)=\sum_i a_{ij}e_i$. Define $\begin{gathered} E_X: X^I\longrightarrow X^J,\\ \pi_jE_X=\sum_i\xi_X(a_{ij})\pi_i. \end{gathered}$ The sum for each $j$ is finite, so the map exists even though $I,J$ can be infinite. Set $H=\ker E_X$.

An arrow $Y\to X^I$ is a family $h_i: Y\to X$. It lands in $H$ exactly when $\sum_i\xi_X(a_{ij})h_i=0$ for every $j$. These are precisely the relations for a left-module map $M\to\operatorname{Hom}_{\mathcal C}(Y,X)$. This proves (4.2), naturally in $Y$. Precomposition in $M$ and postcomposition in $X$ define the indicated variance. The uniqueness argument in Section 5 proves all functor laws, additivity, $k$-linearity and compatibility of different presentations.

For $M_1\to M_2\to M_3\to0$, applying $\operatorname{Hom}_R(-,\operatorname{Hom}(Y,X))$ gives the kernel description $\begin{aligned} 0\longrightarrow H_R(M_3,X) &\longrightarrow H_R(M_2,X)\\ &\longrightarrow H_R(M_1,X). \end{aligned}$ Indeed, its Hom sequence from every $Y$ is the corresponding kernel sequence of module Homs. This is the required contravariant left exactness.

For $0\to X_1\xrightarrow{i}X_2\xrightarrow{v}X_3$ exact in $\mathcal C_R$, consider an arrow $Y\to H_R(M,X_2)$ whose composite into $H_R(M,X_3)$ vanishes. In the product description, every component $Y\to X_2$ is killed by $v$, so uniquely factors through $i$. The relation for these factors follows by composing with the monomorphism $i$, using equivariance, and canceling $i$. They therefore determine a unique arrow $Y\to H_R(M,X_1)$. This proves the kernel property and left exactness in $X$. $\square$

One can also obtain Theorem 5.2 from Theorem 5.1 in $\mathcal C^{\mathrm{op}}$, using (2.1). A left $R$-module is a right $R^{\mathrm{op}}$-module, and a cokernel in the opposite category is a kernel in $\mathcal C$. The direct proof above records the side and the variance.

Infinite inputs cannot in general be used in a category having only finite operations. For example, take finite-dimensional $k$-vector spaces and $X=k$, where $k$ is a field, with $R=k$. If $N$ or $M$ is an infinite-dimensional vector space, the value at $Y=k$ of the desired represented functor is $\operatorname{Hom}_k(N,k)$ or $\operatorname{Hom}_k(M,k)$, respectively. This vector space is infinite-dimensional: infinitely many coordinate functionals on a basis are linearly independent. A finite-dimensional representing object has a finite-dimensional Hom space at $k$, so it cannot represent that functor. The desired representing isomorphisms are $k$-linear, making the dimension obstruction decisive.

6. Matrix units, commuting actions and composition

Worked calculation (matrix units). Set $R=M_2(k)$ and $e=E_{11}$. We compute $eR\otimes_R X$ and $H_R(Re,X)$ and identify the entire internal-module category. Both ordinary input modules are finitely presented, so Theorem 4.1 suffices.

Indeed, the quotient $R\to eR$, $r\mapsto er$, has kernel $(1-e)R$, giving a one-generator relation presentation on the right. The quotient $R\to Re$, $r\mapsto re$, has kernel $R(1-e)$, giving the corresponding left presentation. These statements hold over every commutative scalar ring $k$.

The idempotent $\xi_X(e)$ splits in the abelian category: its image inclusion $i: eX\to X$ and image projection $p: X\to eX$ satisfy $ip=\xi_X(e)$ and $pi=1$. For example, $\xi_X(e)$ is the identity on its image by idempotence, which proves the second equality.

A right-module map $eR\to A$ is determined by $a=f(e)$, satisfying $ae=a$; its inverse sends $er$ to $ar$, which is well-defined because $a=ae$. With $A=\operatorname{Hom}_{\mathcal C}(X,Y)$, this condition is $a\xi_X(e)=a$. Such maps are exactly the arrows $eX\to Y$, by $a=\bar a p$ and $\bar a=ai$. Formula (4.1) therefore gives $eR\otimes_R X\simeq eX$.

A left-module map $Re\to B$ is likewise determined by $b=f(e)$ satisfying $eb=b$, with inverse $re\mapsto rb$. For $B=\operatorname{Hom}_{\mathcal C}(Y,X)$, these are exactly the maps $Y\to eX$. Formula (4.2) gives $H_R(Re,X)\simeq eX$. The left $eRe$-action on $eR$ and the right $eRe$-action on $Re$ induce the restriction of the given action to $eX$; this can be checked at the determining element $e$. As $eRe\cong k$, this is the original scalar action.

For the claimed equivalence, abbreviate $\xi_X(E_{ij})$ to $u_{ij}$. There are arrows $\begin{gathered} \alpha: eX\oplus eX\longrightarrow X,\qquad \alpha=(i,\ u_{21}i),\\ \beta: X\longrightarrow eX\oplus eX,\qquad \beta=\binom{p}{p u_{12}}. \end{gathered}$ Using $u_{ij}u_{\ell m}=\delta_{j\ell}u_{im}$ and $u_{11}+u_{22}=1_X$, compute $\alpha\beta=u_{11}+u_{21}u_{12}=u_{11}+u_{22}=1_X.$ The four components of $\beta\alpha$ are $1,0,0,1$, respectively: $pi=1$, $pu_{21}i=0$, $pu_{12}i=0$, and $pu_{12}u_{21}i=pi=1$. Thus $\alpha,\beta$ are inverse. The operators $u_{11},u_{22},u_{12},u_{21}$, in these coordinates, are the four standard matrix units. Hence the isomorphism is $R$-equivariant.

An equivariant map commutes with all four units. Commuting with the diagonal units forces a map in these coordinates to be diagonal; commuting with $E_{12}$ forces its two diagonal entries to be the same. Thus it is uniquely $f\oplus f$ for an arrow of $\mathcal C$. This proves full faithfulness and identifies all natural comparison maps. Conversely, $e(V\oplus V)\cong V$, and the preceding equivariant $\alpha$ identifies the other composite with the identity. These natural isomorphisms give the equivalence for every such $\mathcal C$, without requiring infinite products or sums.

This is the pointwise matrix-algebra calculation used by the later local module and Morita route. Sheaf gluing and descent require their separate proofs there.

The corner actions just computed are instances of the following general construction. It also makes iteration of tensor products possible.

Let $S$ be another unital $k$-algebra. An $(S,R)$-bimodule $N$ has a left $S$-action and a commuting right $R$-action, with matching scalar actions. Equivalently, it is a left module over $S\otimes_k R^{\mathrm{op}}$.

For each $s\in S$, multiplication $n\mapsto sn$ is a right $R$-linear endomorphism of $N$. Applying the covariant tensor functor gives an endomorphism of $N\otimes_R X$. Functoriality and additivity give the algebra laws and the required scalar action. Thus $N\otimes_R X\in\mathcal C_S. \tag{6.1}$

For a left $R$, right $S$ bimodule $M$, multiplication $\rho_s: m\mapsto ms$ is left $R$-linear. The Hom functor is contravariant in $M$, so define the $S$-action on $H_R(M,X)$ by $H_R(\rho_s,1_X)$. The multiplication order is correct: $\begin{gathered} \rho_t\rho_s=\rho_{st},\\ H_R(\rho_s,1)H_R(\rho_t,1) =H_R(\rho_t\rho_s,1),\\ H_R(\rho_t\rho_s,1)=H_R(\rho_{st},1). \end{gathered}$ In the represented Hom module, this action is $(s\cdot f)(m)=f(ms)$. Hence $H_R(M,X)\in\mathcal C_S. \tag{6.2}$ The unit, additive and scalar laws follow from the same functoriality and linearity. Bimodule maps and equivariant arrows in $X$ induce $S$-equivariant output maps, since they commute with the multiplication operators. These statements preserve the given output actions, not merely the underlying objects.

For the ordinary bimodule $P\otimes_S N$ used next, take the free $k$-module on the small set $P\times N$. Quotient by additivity in each variable, the relations $[ap,n]=a[p,n]=[p,an]$ for $a\in k$, and $[ps,n]=[p,sn]$ for $s\in S$. A $k$-linear map from this quotient is exactly a biadditive $k$-balanced, $S$-balanced function; assignment on generators proves both directions and uniqueness. The operators $t[p,n]=[tp,n]$ and $[p,n]r=[p,nr]$ preserve every relation because the original actions commute. They satisfy the left $T$, right $R$, unit and scalar laws on generators and hence on the quotient. This proves the needed ordinary tensor property and its $(T,R)$-bimodule structure directly.

Proposition 6.3 (associativity and coherence). Suppose $\mathcal C$ has small coproducts. Let $N$ be an $(S,R)$-bimodule, $P$ a $(T,S)$-bimodule, and $X\in\mathcal C_R$. There is a canonical natural $T$-equivariant isomorphism $\begin{gathered} (P\otimes_S N)\otimes_R X\\ \simeq P\otimes_S(N\otimes_R X). \end{gathered} \tag{6.3}$ For a further $(U,T)$-bimodule, all successive comparisons between parenthesizations agree.

Proof. Put $A=\operatorname{Hom}_{\mathcal C}(X,Y)$, a right $R$-module. The tensor representation identifies maps from the right side of (6.3) to $Y$ with $\operatorname{Hom}_{S^{\mathrm{op}}} \bigl(P,\operatorname{Hom}_{R^{\mathrm{op}}}(N,A)\bigr). \tag{6.4}$ The right $S$-action on the inner Hom is $(f\cdot s)(n)=f(sn)$: it comes from precomposing with the left $S$-action on $N$.

Define a bijection from right $R$-linear maps $v: P\otimes_S N\to A$ to (6.4) by $\widehat v(p)(n)=v(p\otimes n)$. Right $R$-linearity in $n$ is immediate. The equality $\widehat v(ps)(n)=v(ps\otimes n) =v(p\otimes sn)=\widehat v(p)(sn)$ is exactly right $S$-linearity in $p$. Conversely, a member $\psi$ of (6.4) gives the biadditive balanced map $(p,n)\mapsto\psi(p)(n)$. It respects the matching $k$-actions and the right $R$-action; the ordinary balanced tensor property gives a unique right $R$-linear map $P\otimes_S N\to A$. The constructions recover one another on every $p,n$, so they are inverse and natural.

By (4.1), the resulting natural Hom bijection represents the left side of (6.3), proving the isomorphism. The $T$-actions multiply $p$; the bijection commutes with that multiplication, so the representing comparison is $T$-equivariant. It is natural in both bimodules, in $X$, and in $Y$.

For an additional bimodule $L$, every sequence of the comparisons sends a represented map to the same function of $l,p,n$, obtained by successive evaluation. Ordinary tensor associativity is the unique comparison taking a nested pure tensor to the corresponding other parenthesization; balanced biadditivity gives this comparison and its inverse. Thus all parenthesized Hom descriptions agree on all triples and on their sums. Uniqueness of the representing map makes the resulting object comparisons equal. Iterating the same evaluation argument proves agreement for any finite list, including the usual pentagon with four factors. This establishes coherence of the specified comparisons, not just existence of unrelated isomorphisms.

7. Exact functors and a zero comparison between equal dimensions

The first worked problem warns us to test the canonical map. A finite presentation gives canonical comparisons after applying an additive functor, but their being isomorphisms requires the appropriate one-sided exactness.

Theorem 7.1 (finite constructions and exact functors). Let $F: \mathcal C\to\mathcal D$ be $k$-linear between $k$-linear abelian categories. If $F$ is right exact and $N$ is a finitely presented right $R$-module, then $N\otimes_R FX\simeq F(N\otimes_R X). \tag{7.1}$ If $F$ is left exact and $M$ is a finitely presented left $R$-module, then $F H_R(M,X)\simeq H_R(M,FX). \tag{7.2}$ Both comparisons are natural in the module and the internal object and compatible with second actions. A $k$-linear additive functor alone need not give either isomorphism.

Proof. In a finite presentation, the tensor object is the cokernel of the matrix map $X^m\to X^n$. A $k$-linear additive functor preserves finite biproducts with their specified injections and projections, and sends that matrix to the same matrix of operators on $FX$. If $F$ is right exact, it preserves the cokernel, giving (7.1). Its comparison is the canonical map induced from the cokernel diagram. A left exact $F$ preserves the kernel of the finite product matrix defining $H_R(M,X)$, giving the canonical comparison (7.2).

Here is also a verification that the comparisons do not depend on the chosen presentation or on a module arrow. For an arrow of finitely presented modules, lift its images on the finitely many degree-zero free generators to the target free presentation. The resulting matrix carries the source relations into the kernel of the target quotient. The finitely many relation generators lift through the target relation presentation, giving a commuting pair of presentation matrices. This works because finite free modules are projective in the ordinary module category. In the right-module case the induced cokernel map is exactly the representing tensor map from Section 5: precomposition with a family satisfying the target relations gives the family obtained by the degree-zero lift. Hence all lifts yield the same map. In the left-module case the induced map between the product kernels is exactly the representing Hom map, by the same evaluation of the degree-zero matrix; its direction reverses.

Applying $F$ to each such commuting pair proves naturality in the module. Taking its arrow to be the identity proves compatibility of different presentations. An equivariant arrow of internal objects commutes with all matrix coefficients, so its finite diagrams commute too, proving naturality in $X$. Second actions are particular module endomorphisms, and this same naturality proves equivariance. Thus (7.1) and (7.2) are canonical with all the stated compatibilities.

For the failures, let $k$ be a field, $A=k[\epsilon]/(\epsilon^2)$, and let $\mathcal C$ be finite-dimensional left $A$-modules. Take $R=k[t]$, $X=A$, with $t$ acting by multiplication by $\epsilon$, and take the finite module presentation $R\xrightarrow{t}R\to R/(t)\to0$.

The $k$-linear additive functor $F_1=\operatorname{Hom}_A(k,-)$ sends $X$ to $\epsilon A\cong k$, on which $t$ is zero. The tensor cokernel in $\mathcal C$ is $A/\epsilon A\cong k$, so both sides of the proposed (7.1) have dimension one. But its canonical comparison is zero: the quotient $q: A\to k$ kills $\epsilon A$, so $F_1q: \epsilon A\to k$ is zero. The comparison from the cokernel of $F_1(\epsilon)=0$ to $F_1(\operatorname{coker}\epsilon)$ is therefore zero. Thus even equality of dimensions does not repair lack of right exactness.

For (7.2), take the $k$-linear additive functor $F_2=k\otimes_A-$. The internal Hom object is $\ker(\epsilon: A\to A)=\epsilon A\cong k$. Both $F_2(\epsilon A)$ and the kernel of the zero operator on $F_2A=k$ are one-dimensional. The canonical comparison is again zero: tensoring the inclusion $\epsilon A\hookrightarrow A$ sends the class of $\epsilon$ to its zero class in $A/\epsilon A$. Thus $F_2$ does not preserve this kernel, and mere additivity cannot replace left exactness.

8. Universal scalars as a companion construction

An additive category $\mathcal C$ is $k$-linear when each morphism group is a $k$-module and composition is $k$-bilinear. A $k$-linear functor preserves these scalar multiplications.

There is a useful intrinsic description. Give the ring of natural endomorphisms of the identity functor pointwise addition and composition. For two such endomorphisms $\eta,\theta$, naturality of $\eta$ applied to the arrow $\theta_X$ gives $\eta_X\theta_X=\theta_X\eta_X.$ Thus this ring is commutative. Its elements are the endomorphisms that commute with every arrow of the category, rather than only with endomorphisms of one object.

Proposition 8.1. A $k$-linear structure on an additive category is equivalent to a unital ring homomorphism $k\longrightarrow\operatorname{End}(\operatorname{id}_{\mathcal C}).$ For every object $X$, it makes $\operatorname{End}_{\mathcal C}(X)$ a $k$-algebra with central scalars.

Proof. From a linear structure, assign to $a\in k$ the family $a1_X$. Bilinearity gives $f(a1_X)=(a1_Y)f$ for every $f: X\to Y$, so the family is natural. The module laws and bilinearity prove the additive, multiplicative and unit laws for the ring map.

Conversely, let the image of $a$ have components $\eta_X(a)$. Define $af=\eta_Y(a)f=f\eta_X(a)$. The equality is naturality. Additivity and multiplication of the ring map give the module laws on each morphism group; composition is bilinear by the same equality. Starting with either description and returning to it changes nothing: in a linear category $af=(a1_Y)f$, and in the constructed structure $a1_X=\eta_X(a)$. Centrality in each endomorphism ring follows by taking $f$ to be an endomorphism. $\square$

9. Graded exercises with full solutions

Exercise 1 (foundation: Jordan blocks). For the nilpotent block $J_m$ with $Te_1=0$ and $Te_i=e_{i-1}$, compute the rank of $\ker T^n\to\operatorname{coker}T^n$, for every $m,n\geq1$. When is it an isomorphism? Extend the calculation to a finite direct sum of such blocks.

Solution. Put $s=\min(m,n)$. The kernel is spanned by $e_1,\ldots,e_s$, and the image of $T^n$ is spanned by $e_1,\ldots,e_{m-s}$. A basis of the quotient consists of $[e_{m-s+1}],\ldots,[e_m]$. The surviving vectors from the kernel have indices $m-s+1\leq i\leq s$; therefore the rank is $\max(0,2s-m)$. Kernel and quotient both have dimension $s$. The comparison is invertible exactly when this rank equals $s$, which, since $s>0$, is equivalent to $s=m$, or $n\geq m$. On a finite direct sum, kernel, quotient and comparison all decompose by blocks. The ranks add; the map is invertible exactly when $n$ is at least the length of every block.

Exercise 2 (intermediate: larger matrix actions). Generalize the matrix-unit equivalence to $M_d(k)$ for $d\geq1$, in any $k$-linear abelian category. Give both comparison maps and determine all equivariant arrows.

Solution. Write $u_{ij}=\xi(E_{ij})$ and split $u_{11}=ip$ with $pi=1$, as in Section 6. Set $\alpha:(eX)^d\to X$ to have components $u_{i1}i$, and set $\beta:X\to(eX)^d$ to have components $pu_{1i}$. Then $\alpha\beta=\sum_i u_{i1}ip u_{1i} =\sum_i u_{ii}=1_X.$ The $(i,j)$-component of $\beta\alpha$ is $pu_{1i}u_{j1}i=\delta_{ij}pi=\delta_{ij}$. In these coordinates each $u_{ij}$ is the corresponding standard matrix operator, proving equivariance. An arrow commuting with all diagonal units has only diagonal entries, and commuting with every $E_{ij}$ identifies all of them with one arrow $eX\to eY$. Conversely, any such arrow repeated $d$ times commutes with every matrix. Thus $X\mapsto eX$ is fully faithful, and $V\mapsto V^d$ is its inverse up to the displayed natural comparisons. All operations are finite; $d=1$ gives the identity equivalence.

Exercise 3 (advanced: units and associativity). Prove canonical identifications $R\otimes_R X\simeq X$ and $H_R(R,X)\simeq X$. For an $(S,R)$-bimodule $N$, prove that removing a tensor unit before or after the associativity comparison of Proposition 6.3 gives the same map.

Solution. Evaluation at $1$ identifies a right-linear map $R\to\operatorname{Hom}(X,Y)$ with an arbitrary map $X\to Y$: the inverse sends $r$ to $h\xi_X(r)$. Likewise a left-linear map $R\to\operatorname{Hom}(Y,X)$ is determined by its value at $1$, with inverse $r\mapsto\xi_X(r)h$. Formulas (4.1) and (4.2) therefore represent $X$, and uniqueness gives the stated canonical identifications. They are natural, and any additional commuting action is respected because evaluation commutes with that action. Ordinary balanced tensors have the unit maps $s\otimes n\mapsto sn$ and $n\otimes r\mapsto nr$, with inverses $n\mapsto1\otimes n$ and $n\mapsto n\otimes1$. Each associativity comparison in Section 6 acts on represented maps by evaluation of the same balanced function. Evaluation followed by either unit map gives the same function, so uniqueness of the represented arrow proves equality of the object comparisons. This checks the unit triangles as actual maps.

Exercise 4 (expert: eliminate a matrix relation). For $R=k[t]$, present a right module by the columns $(t,0)$ and $(1,t)$, and a left module by the analogous left relations. Identify both with $R/(t^2)$. For $t$ acting as $T$ on an internal object, give explicit inverse comparisons between the tensor matrix cokernel and $\operatorname{coker}T^2$, and between the Hom matrix kernel and $\ker T^2$.

Solution. The relations are $e_1t=0$ and $e_1+e_2t=0$ on the right; hence $e_1=-e_2t$ and $e_2t^2=0$. Sending $e_2$ to $1$ and $e_1$ to $-t$ gives the quotient $R/(t^2)$; its inverse sends $1$ to $e_2$. The same elimination on the left gives the left-module identification.

The tensor matrix is $D(v,w)=(Tv+w,Tw)$, notation for the two components of a biproduct arrow. The arrow $(x,y)\mapsto[y-Tx]$ to $\operatorname{coker}T^2$ kills $D$, since its composite is $[-T^2v]$. It induces a map out of $\operatorname{coker}D$, whose inverse takes $[z]$ to the class of $(0,z)$. This is well defined: $(0,T^2v)=D(-v,Tv)$. The composites are identities because $(x,y)-(0,y-Tx)=(x,Tx)=D(0,x)$, while the composite on $[z]$ is visibly $[z]$. These are equations of biproduct arrows and descend by the cokernel property, so they work without elements in $\mathcal C$.

The Hom matrix is $E(x,y)=(Tx,x+Ty)$. Its kernel has $x=-Ty$ and $T^2y=0$. Projection to $y$ gives $\ker E\to\ker T^2$, with inverse $y\mapsto(-Ty,y)$. Substitution verifies both composites and both kernel factorizations. These are the specific comparisons induced by the module identifications, not merely some isomorphisms between objects of equal size.

10. References

The Stacks Project, Tensor products gives the ordinary commutative-module foundations. Local module categories and Morita equivalence develops sheaf-bimodule descent. The pointwise matrix calculation here has its stated finite scope.