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The left-to-right exact equivalence need not preserve the identity

Statement refuted

The equivalence $Γ^{r l} : Lex (A, A) \to Rex (A, A)$ of The left-to-right exact equivalence sends the identity to the Nakayama functor need not send the identity functor to a functor naturally isomorphic to the identity. Witness: let $A_{0}$ be the $k$ -algebra with $k$ -basis $e_{1}, e_{2}, u$ (Field, Vector space over a field, Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis), unit $1 = e_{1} + e_{2}$ (Algebras over a commutative ring, central structure maps, and algebra homomorphisms), orthogonal idempotents $e_{1}^{2} = e_{1}$ , $e_{2}^{2} = e_{2}$ , products $e_{1} u = u = u e_{2}$ , and all remaining products of basis elements zero; these are the upper triangular $2 \times 2$ matrices. Let $A = A_{0} - m o d$ . The Nakayama functor $N^{r}$ of Left and right Nakayama functors by finite kernel calculus satisfies $N^{r} ≅ A_{0}^{*} \otimes_{A_{0}} -$ by Nakayama kernels give well-defined adjoint functors, and evaluating on the projective left module $A_{0} e_{1}$ (Generated submodule, cyclic and finitely generated modules, module basis and free module, Unital left and right modules over a ring; unqualified module means left module) gives $\dim_{k} N^{r} (A_{0} e_{1}) = \dim_{k} A_{0}^{*} e_{1} = 2$ while $\dim_{k} A_{0} e_{1} = 1$ ; hence $N^{r}$ is not naturally isomorphic to the identity and the equivalence does not preserve the identity object (Natural isomorphism).

Facts & Assumptions

Given: A field $k$ , the $k$ -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 (Field, Vector space over a field, Algebras over a commutative ring, central structure maps, and algebra homomorphisms), the category $A = A_{0} - m o d$ of finite-dimensional left $A_{0}$ -modules, and the Nakayama functor $N^{r} = Γ^{r l} (1_{A}) ≅ A_{0}^{*} \otimes_{A_{0}} -$ (Left and right Nakayama functors by finite kernel calculus, Nakayama kernels give well-defined adjoint functors).

[F1]

A left $A_{0}$ -module is an abelian group with a scalar action satisfying $r (m + n) = r m + r n$ , $(r + s) m = r m + s m$ , $(r s) m = r (s m)$ and $1 m = m$ ; the submodule $A_{0} e_{1}$ is the image of the $A_{0}$ -linear map $a \mapsto a e_{1}$ , and $A_{0} = A_{0} e_{1} \oplus A_{0} e_{2}$ because $e_{1} + e_{2} = 1$ and $e_{1} e_{2} = e_{2} e_{1} = 0$ (Unital left and right modules over a ring; unqualified module means left module, Generated submodule, cyclic and finitely generated modules, module basis and free module, $(S, R)$ -bimodules and commuting left and right scalar actions).

[F2]

The $k$ -dual $A_{0}^{*} = {Hom}_{k} (A_{0}, k)$ is a right $A_{0}$ -module under $(λ \cdot a) (x) = λ (a x)$ , and $A_{0}^{*} e_{1}$ denotes the image of the right multiplication map $A_{0}^{*} \to A_{0}^{*}$ , $λ \mapsto λ \cdot e_{1}$ ; on the free left module $A_{0}$ the tensor product $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1}$ is generated by elementary tensors subject to $λ a \otimes x = λ \otimes a x$ (Linear map between vector spaces over the same field, $(S, R)$ -bimodules and commuting left and right scalar actions, Universal property of the tensor product for balanced maps into abelian groups, Module homomorphisms induce tensor-product homomorphisms functorially).

[F3]

For a finite-dimensional $k$ -vector space the dimension is the cardinality of a basis, and a $k$ -linear isomorphism preserves dimensions (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis, Vector space over a field, Linear map between vector spaces over the same field).

[F4]

The functor $Γ^{r l}$ of the categorical Eilenberg–Watts triangle sends the identity functor regarded as left exact to $N^{r} = Γ^{r l} (1_{A})$ ; a natural isomorphism $N^{r} ≅ 1_{A}$ would give an isomorphism $N^{r} (A_{0} e_{1}) ≅ A_{0} e_{1}$ of $k$ -vector spaces (The left-to-right exact equivalence sends the identity to the Nakayama functor, Natural isomorphism).

Counterexample

1.1givenF1

The algebra $A_{0}$ is well defined: the $k$ -linear assignment $e_{1} \mapsto (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ , $e_{2} \mapsto (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ , $u \mapsto (\begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array})$ identifies $A_{0}$ with the algebra of upper triangular $2 \times 2$ matrices, in which the listed products hold and multiplication is associative; consequently $e_{1}, e_{2}$ are orthogonal idempotents summing to $1$ , $e_{1} u = u = u e_{2}$ , $u e_{1} = 0$ , $e_{2} u = 0$ and $u^{2} = 0$ .

2.1step 1.1F1F3

The left ideal $A_{0} e_{1}$ equals the $k$ -span of $e_{1}$ : from $e_{1} e_{1} = e_{1}$ , $e_{2} e_{1} = 0$ and $u e_{1} = 0$ one gets $a e_{1} = κ (a) e_{1}$ for the coefficient functional $κ$ of $e_{1}$ , so $A_{0} e_{1} = span {e_{1}}$ has $\dim_{k} A_{0} e_{1} = 1$ . It is moreover projective in the lifting sense of Projective modules and the lifting property: by [F1] it is a direct summand of $A_{0}$ with projection $π : a \mapsto a e_{1}$ , the free module $A_{0}$ has the lifting property because a $A_{0}$ -linear map out of $A_{0}$ is determined by its value at the generator $1$ , which can be lifted along any epimorphism, and restricting a lift of $f \circ π$ to $A_{0} e_{1}$ lifts a given $f : A_{0} e_{1} \to M$ .

2.2step 1.1F2F3

The image $A_{0}^{*} e_{1}$ has dimension $2$ : for $λ \in A_{0}^{*}$ one computes $(λ \cdot e_{1}) (x) = λ (e_{1} x)$ , so $λ \cdot e_{1} = λ \circ φ$ for the $k$ -linear map $φ : A_{0} \to A_{0}$ , $φ (x) = e_{1} x$ , whose image is $e_{1} A_{0} = span {e_{1}, u}$ of dimension $2$ by step 1.1; the restriction map $A_{0}^{*} \to (e_{1} A_{0})^{*}$ , $λ \mapsto λ ∣_{e_{1} A_{0}}$ , is surjective since a functional on the direct summand $e_{1} A_{0}$ extends by zero on $span {e_{2}}$ , and composition with the surjection $φ$ is injective, so $A_{0}^{*} e_{1} = {μ \circ φ : μ \in (e_{1} A_{0})^{*}} ≅ (e_{1} A_{0})^{*}$ has dimension $2$ .

3.1step 2.1step 2.2F2given

By [F2] the multiplication map $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1} \to A_{0}^{*} e_{1}$ , $λ \otimes x \mapsto λ \cdot x$ , is a well-defined surjection, and it is injective with inverse $ν \mapsto ν \otimes e_{1}$ : indeed $(λ \cdot e_{1}) \otimes e_{1} = λ \otimes (e_{1} e_{1}) = λ \otimes e_{1}$ , and for $x = c e_{1} \in A_{0} e_{1}$ one has $λ \cdot x = (λ \cdot c) \cdot e_{1}$ and $((λ \cdot c) \cdot e_{1}) \otimes e_{1} = (λ \cdot c) \otimes e_{1} = λ \otimes c e_{1}$ . Hence $A_{0}^{*} \otimes_{A_{0}} A_{0} e_{1} ≅ A_{0}^{*} e_{1}$ , and since $N^{r} ≅ A_{0}^{*} \otimes_{A_{0}} -$ by the given data, step 2.2 gives $\dim_{k} N^{r} (A_{0} e_{1}) = \dim_{k} A_{0}^{*} e_{1} = 2$ .

4.1step 3.1F3F4∎

Since $\dim_{k} N^{r} (A_{0} e_{1}) = 2$ while $\dim_{k} A_{0} e_{1} = 1$ , the vector spaces $N^{r} (A_{0} e_{1})$ and $A_{0} e_{1}$ are not isomorphic, so by [F4] there is no natural isomorphism $N^{r} ≅ 1_{A}$ ; equivalently $N^{r}$ is not naturally isomorphic to the identity functor. By [F4] the equivalence $Γ^{r l}$ sends the identity functor, regarded as left exact, to $N^{r}$ , so it does not send the identity to a functor naturally isomorphic to the identity, and the equivalence does not preserve the identity object.

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