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The kernel end and coend distinguish the regular and co-regular bimodules

Statement

For the identity functor of a finite $k$ -linear abelian category $A ≃ A - m o d$ , the two kernel formulas of the categorical Eilenberg–Watts triangle give the regular and co-regular bimodules, which need not be isomorphic: the end $\int_{a \in A} \overset{ˉ}{a} ⊠ 1 (a)$ is the regular bimodule $A$ , while the coend $\int^{a \in A} \overset{ˉ}{a} ⊠ 1 (a)$ is the co-regular bimodule $A^{*}$ (Finite Eilenberg–Watts kernels: explicit end and coend universal maps, The end and the coend of a functor $C^{o p} \times C \to D$ ; Nakayama kernels give well-defined adjoint functors). These are generally non-isomorphic as $A$ -bimodules, so an end and a coend of the same functor need not agree; both are the identity's images under the Nakayama calculus of Left and right Nakayama functors by finite kernel calculus. Witness: for the upper triangular algebra $A_{0}$ of the companion counterexample (Algebras over a commutative ring, central structure maps, and algebra homomorphisms), $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1}$ has dimension $2$ while $A_{0} e_{1}$ has dimension $1$ ; hence $A_{0}^{*} ≇ A_{0}$ as bimodules ( $(S, R)$ -bimodules and commuting left and right scalar actions, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ , Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Vector space over a field) and the regular and co-regular kernels are distinguished. Under the equivalences of the triangle these two objects correspond to the identity functor as an object of $Rex$ and of $Lex$ respectively (The left-to-right exact equivalence sends the identity to the Nakayama functor).

Example

Facts & Assumptions

Given: A finite $k$ -linear abelian category with module model $A ≃ A - m o d$ for a finite-dimensional unital $k$ -algebra $A$ , the regular $(A, A)$ -bimodule $A$ and the co-regular bimodule $A^{*} = {Hom}_{k} (A, k)$ , together with the Eilenberg–Watts functors $Φ^{l}, Φ^{r}, Ψ^{l}, Ψ^{r}$ of the triangle (Left and right Nakayama functors by finite kernel calculus, $(S, R)$ -bimodules and commuting left and right scalar actions, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ); and the upper triangular $k$ -algebra $A_{0}$ with $k$ -basis $e_{1}, e_{2}, u$ , unit $1 = e_{1} + e_{2}$ , $e_{1}^{2} = e_{1}$ , $e_{2}^{2} = e_{2}$ , $e_{1} u = u = u e_{2}$ and all remaining products of basis elements zero (Algebras over a commutative ring, central structure maps, and algebra homomorphisms, Vector space over a field, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

[F1]

For a finite $(B, A)$ -bimodule $M$ with $F = Φ^{l} (M) = {Hom}_{A} (M^{*}, -)$ and $G = Φ^{r} (M) = M \otimes_{A} -$ , the coend $\int^{a} \overset{ˉ}{a} ⊠ F (a)$ is the coend of $a \mapsto F (a) \otimes_{k} a^{*}$ with universal cowedge $ρ_{a} : F (a) \otimes_{k} a^{*} \to M$ , $ρ_{a} (f \otimes λ) = λ \circ f$ , and the end $\int_{a} \overset{ˉ}{a} ⊠ G (a)$ is the end of $a \mapsto G (a) \otimes_{k} a^{*}$ with universal wedge $ω_{a} : M \to {Hom}_{k} (a, G (a))$ , $ω_{a} (m) (x) = m \otimes x$ ; in particular the (co)end object is $M$ itself in the bimodule model (Finite Eilenberg–Watts kernels: explicit end and coend universal maps, The end and the coend of a functor $C^{o p} \times C \to D$ ).

[F2]

The identity functor satisfies $1_{A} ≅ Φ^{r} (A)$ , since $Φ^{r} (A) (X) = A \otimes_{A} X ≅ X$ by the unit isomorphism, and $1_{A} ≅ Φ^{l} (A^{*})$ , since $Φ^{l} (A^{*}) (X) = {Hom}_{A} (A^{* *}, X) ≅ {Hom}_{A} (A, X) ≅ X$ by double duality and evaluation at $1_{A}$ (Nakayama kernels give well-defined adjoint functors, Natural isomorphism, Unital left and right modules over a ring; unqualified module means left module, Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

[F3]

The tensor product over $A$ is functorial in the first variable, so an isomorphism of right $A$ -modules, in particular an isomorphism of $(A, A)$ -bimodules $A^{*} \to A$ , induces a natural isomorphism $A^{*} \otimes_{A} - ≅ A \otimes_{A} -$ , and $A \otimes_{A} X ≅ X$ naturally in $X$ ; for the algebra $A_{0}$ the element $e_{1}$ satisfies $e_{1}^{2} = e_{1}$ , so $A_{0} e_{1}$ is a $k$ -subspace and $A_{0}^{*} e_{1}$ is the image of right multiplication by $e_{1}$ (Module homomorphisms induce tensor-product homomorphisms functorially, Universal property of the tensor product for balanced maps into abelian groups, $(S, R)$ -bimodules and commuting left and right scalar actions, Linear map between vector spaces over the same field).

Verification

1.1givenF1F2

The end is the regular bimodule: apply [F1] with $A = B$ and $M = A$ , the regular $(A, A)$ -bimodule. Then $G = Φ^{r} (A) ≅ 1_{A}$ by [F2], so the end $\int_{a} \overset{ˉ}{a} ⊠ 1 (a)$ of the identity diagram is the end of the diagram $a \mapsto G (a) \otimes_{k} a^{*}$ and equals the (co)end object $M = A$ of [F1], with universal wedge $ω_{a} (m) (x) = m \otimes x$ .

1.2givenF3

For $A_{0}$ one has $A_{0} e_{1} = span {e_{1}}$ of dimension $1$ , because $e_{1} e_{1} = e_{1}$ , $e_{2} e_{1} = 0$ and $u e_{1} = 0$ ; and $A_{0}^{*} e_{1}$ has dimension $2$ , because $(λ \cdot e_{1}) (x) = λ (e_{1} x)$ expresses $λ \cdot e_{1}$ as the composite of $λ$ with the map $x \mapsto e_{1} x$ , whose image is $e_{1} A_{0} = span {e_{1}, u}$ of dimension $2$ , and every functional on that direct summand of $A_{0}$ extends to $A_{0}$ .

2.1step 1.1F1F2

The coend is the co-regular bimodule: apply [F1] with $A = B$ and $M = A^{*}$ . Then $F = Φ^{l} (A^{*}) ≅ 1_{A}$ by [F2], so the coend $\int^{a} \overset{ˉ}{a} ⊠ 1 (a)$ is the coend of the diagram $a \mapsto F (a) \otimes_{k} a^{*}$ and equals the (co)end object $M = A^{*}$ of [F1], with universal cowedge $ρ_{a} (f \otimes λ) = λ \circ f$ ; under the identification $A^{o p} ⊠ A ≃ (A, A) - b i m o d$ the object $A$ is the regular and $A^{*}$ the co-regular bimodule.

2.2step 1.2F3

Consequently $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1} ≅ A_{0}^{*} e_{1}$ has dimension $2$ , by the multiplication isomorphism $λ \otimes x \mapsto λ \cdot x$ with inverse $ν \mapsto ν \otimes e_{1}$ , while $A_{0} \otimes_{A_{0}} A_{0} e_{1} ≅ A_{0} e_{1}$ has dimension $1$ by the unit isomorphism of [F3].

3.1step 2.1step 2.2F3

The bimodules $A_{0}^{*}$ and $A_{0}$ are not isomorphic: an isomorphism would by [F3] induce an isomorphism $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1} ≅ A_{0} \otimes_{A_{0}} A_{0} e_{1}$ , hence equality of dimensions, contradicting step 2.2. Hence the regular kernel $A_{0}$ and the co-regular kernel $A_{0}^{*}$ of steps 1.1 and 2.1 are distinguished, so an end and a coend of the same functor need not agree.

4.1step 1.1step 2.1F1F2∎

Finally, the end $A = Ψ^{r} (1_{A})$ is the image of the identity functor regarded as an object of $Rex (A, A)$ under $Ψ^{r}$ , and the coend $A^{*} = Ψ^{l} (1_{A})$ is the image of the identity regarded as an object of $Lex (A, A)$ under $Ψ^{l}$ , by steps 1.1 and 2.1; applying $Φ^{r}$ and $Φ^{l}$ recovers the Nakayama functors $N^{r} = Φ^{r} Ψ^{l} (1_{A}) ≅ A^{*} \otimes_{A} -$ and $N^{l} = Φ^{l} Ψ^{r} (1_{A}) ≅ {Hom}_{A} (A^{*}, -)$ of the Nakayama calculus (Left and right Nakayama functors by finite kernel calculus, The left-to-right exact equivalence sends the identity to the Nakayama functor).

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