Eilenberg–Watts Theorem and Natural Transformations — Examples

The associativity maps ${\alpha }_X:(N{\otimes }_{B}M){\otimes }_{A}X\to N{\otimes }_B(M{\otimes }_{A}X)=(T_N\circ T_M)(X)$ are natural isomorphisms in $X$ by [F1], and $N{\otimes }_{B}M$ is a $(C,A)$-bimodule by [F2], so $\alpha$ is a natural isomorphism of functors $T_{N{\otimes }_{B}M}\Rightarrow T_N\circ T_M$.

Witness data: take $A=B=C=k$, $M=N=k^2$, $M^{\mathrm{\prime }}=N^{\mathrm{\prime }}=k$, all regarded as bimodules via the field action by [F2]. Define $\beta :k^2\times k^2\to k$ by $\beta (m,n)=m_1{n}_1+m_2{n}_2$ for $m=m_1{e}_1+m_2{e}_2$, $n=n_1{e}_1+n_2{e}_2$. The map $\beta$ is bilinear, hence balanced by [F8], so by [F7] there is a unique group homomorphism $\bar{\beta }:k^2{\otimes }_k{k}^2\to k$ with $\bar{\beta }(m\otimes n)=\beta (m,n)$; it is $k$-linear because $\bar{\beta }(c(m\otimes n))=\beta (cm,n)=c\beta (m,n)$ on generators, and $\beta (e_i,e_j)={\delta }_{ij}$ since $e_1,e_2$ are the standard generators.

Conjugation by $\alpha$ and by the corresponding isomorphism ${\alpha }^{\mathrm{\prime }}$ for the primed bimodules is a bijection from $\mathrm{Nat}(T_N\circ T_M,T_{N^{\mathrm{\prime }}}\circ T_{M^{\mathrm{\prime }}})$ to $\mathrm{Nat}(T_{N{\otimes }_{B}M},T_{N^{\mathrm{\prime }}{\otimes }_B{M}^{\mathrm{\prime }}})$: for $\eta$ in the first set put ${\eta }_X^{\mathrm{\prime }}:=({\alpha }_X^{\mathrm{\prime }}{)}^{-1}\circ {\eta }_X\circ {\alpha }_X$, a natural transformation by [F3], and the assignment $\xi \mapsto ({\alpha }_X^{\mathrm{\prime }}\circ {\xi }_X\circ {\alpha }_X^{-1})$ is inverse to it by the componentwise cancellation of inverse natural isomorphisms.

A pair of bimodule maps $g:N\to N^{\mathrm{\prime }}$, $f:M\to M^{\mathrm{\prime }}$ gives the natural transformation $T_N\circ T_M\Rightarrow T_{N^{\mathrm{\prime }}}\circ T_{M^{\mathrm{\prime }}}$ with components $(g\otimes 1_{M^{\mathrm{\prime }}{\otimes }_{A}X})\circ (1_N\otimes (f\otimes 1_X))$, equal to ${\alpha }_X^{\mathrm{\prime }}\circ ((g\otimes f)\otimes 1_X)\circ {\alpha }_X^{-1}$ by [F1] and agreement on every $n\otimes (m\otimes x)$; this is natural by [F3] and [F4], and under the conjugations of step 2.1 it corresponds to the bimodule map $g\otimes f:N{\otimes }_{B}M\to N^{\mathrm{\prime }}{\otimes }_B{M}^{\mathrm{\prime }}$ with $(g\otimes f)(n\otimes m)=g(n)\otimes f(m)$ of [F4].

By [F5] the natural transformations $T_{N{\otimes }_{B}M}\Rightarrow T_{N^{\mathrm{\prime }}{\otimes }_B{M}^{\mathrm{\prime }}}$ correspond bijectively to $(C,A)$-bimodule maps $N{\otimes }_{B}M\to N^{\mathrm{\prime }}{\otimes }_B{M}^{\mathrm{\prime }}$; composing with the bijection of step 2.1 classifies the transformations between the composites, and step 3.1 identifies the image of every pair $(g,f)$.

The element $\tau :={\rho }^{-1}\circ \bar{\beta }\in {\mathrm{Hom}}_{k\text{-}k}(k^2{\otimes }_k{k}^2,k{\otimes }_{k}k)$ is a bimodule map by [F6] and step 1.2. Suppose that $\tau =g\otimes f$ for $k$-linear $g,f:k^2\to k$, i.e. $\tau (m\otimes n)=g(m)\otimes f(n)$ by [F4]; applying the isomorphism $\rho$ of [F6] and evaluating on the four elementary tensors $e_i\otimes e_j$ gives $g(e_i)f(e_j)=\rho (\tau (e_i\otimes e_j))=\beta (e_i,e_j)={\delta }_{ij}$. These four equations are contradictory: $g(e_1)f(e_1)=1$ forces $g(e_1)\mathrm{\ne }0$ and $f(e_1)\mathrm{\ne }0$, then $g(e_1)f(e_2)=0$ forces $f(e_2)=0$, and then $g(e_2)f(e_2)=0$ contradicts $g(e_2)f(e_2)=1$. Hence no pair $(g,f)$ produces $\tau$, while step 4.1 classifies $\tau$ as a genuine bimodule map $N{\otimes }_{B}M\to N^{\mathrm{\prime }}{\otimes }_B{M}^{\mathrm{\prime }}$.

Steps 1.1 and 4.1 identify $T_N\circ T_M$ with $T_{N{\otimes }_{B}M}$ and classify all natural transformations between the composites by all $(C,A)$-bimodule maps of kernels, step 3.1 gives the components ${\alpha }_X^{\mathrm{\prime }}\circ ((g\otimes f)\otimes 1_X)\circ {\alpha }_X^{-1}$ for pairs $(g,f)$, and steps 1.2 and 5.1 exhibit a bimodule map not of the form $g\otimes f$. No basis of an infinite-dimensional space and no presentation is chosen, and no commutativity of the rings is assumed.

The functor $T_S=S{\otimes }_R-$ is by definition the extension of scalars along $f$ by [F1]. It is additive and cocontinuous: it is additive by [F3], and it is cocontinuous because extension of scalars is left adjoint to restriction of scalars by [F2] and a left adjoint preserves every colimit that exists by [F2] (equivalently, additive and coproduct-preserving with right exactness gives cocontinuity directly by [F3]).

Kernel: apply the evaluation lemma [F5] to $F=T_S$ and the regular module $R$. For $a\in R$ one has $T_S(r_a)=1_S\otimes r_a$, so the reconstructed right action on $T_S(R)=S{\otimes }_{R}R$ is $(s\otimes x)a=s\otimes (xa)$; under the unit isomorphism ${\rho }_S$ of [F4] this corresponds to ${\rho }_S(s\otimes (xa))=sf(xa)=sf(x)f(a)$, since $f$ is a unital ring homomorphism, and the last term is ${\rho }_S(s\otimes x)\cdot a$ for the displayed right action $s^{\mathrm{\prime }}\cdot a=s^{\mathrm{\prime }}f(a)$ of [F1]. Hence the evaluation right action on the kernel $S{\otimes }_{R}R\cong S$ is exactly $s\cdot a=sf(a)$, the published action of ${}_S{S}_R$.

By step 1.1 the extension-of-scalars functor is the tensor functor $T_S$ with kernel the $(S,R)$-bimodule ${}_S{S}_R$, as computed in step 2.1, and by [F3] it is additive and right exact as the theorem requires; hence the Eilenberg-Watts classification reproduces extension of scalars together with its kernel bimodule.

The same bimodule ${}_S{S}_R$ and the same unit-isomorphism computation make sense for an arbitrary unital ring homomorphism, but the published definition of extension of scalars is stated only for commutative rings, so only the commutative case is claimed here; that caveat is recorded and not used. No element of an auxiliary set is chosen, so no choice is involved.

$F$ is an additive functor: the operations on ${\prod }_{n}V$ are coordinatewise by [F2], and for parallel maps $u,v$ one has $F(u+v)={\prod }_n(u+v)={\prod }_{n}u+{\prod }_{n}v=F(u)+F(v)$, with $F(\mathrm{i}\mathrm{d})=\mathrm{i}\mathrm{d}$ and $F(u^{\mathrm{\prime }}u)=F(u^{\mathrm{\prime }})F(u)$ coordinatewise.

$F$ is left exact: let $0\to V^{\mathrm{\prime }}\stackrel{u^{\mathrm{\prime }}}{\to }V\stackrel{u}{\to }{V}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ be a short exact sequence of $k$-modules. Coordinatewise, ${\prod }_n{u}^{\mathrm{\prime }}$ is injective, and an element $(v_n)$ of ${\prod }_{n}V$ lies in $\ker {\prod }_{n}u$ exactly when $u(v_n)=0$ for all $n$, i.e. $v_n\in \mathrm{im}{u}^{\mathrm{\prime }}$ for all $n$ by exactness at $V$; hence $\ker {\prod }_{n}u=\mathrm{im}{\prod }_n{u}^{\mathrm{\prime }}$ and $0\to {\prod }_n{V}^{\mathrm{\prime }}\to {\prod }_{n}V\to {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}$ is exact. By [F4] this proves left exactness.

Let $c:{⨁}_{j}F(k)\to F({⨁}_{j}k)$ be induced by the maps $F({\mathrm{ȷ}}_j)$. An element of its source is a family of scalar sequences $(s_j{)}_j$ supported on a finite set $J\subseteq \mathbb{N}$ of summand indices. Its image has $n$-th coordinate ${\sum }_{j\in J}{s}_j(n)e_j$, supported in $J$ for every $n$. Conversely, if $(v_n{)}_n$ has every $v_n$ supported in one finite set $J$, define $s_j(n)$ as the $j$-th coefficient of $v_n$ for $j\in J$ and put $s_j=0$ otherwise; then $(s_j{)}_j$ belongs to the source and maps to $(v_n{)}_n$. Thus the image consists exactly of families with supports contained in one fixed finite set of summand indices.

Under AC, $F$ is right exact: if $u:V\to V^{\mathrm{\prime }\mathrm{\prime }}$ is surjective, each fibre $u^{-1}(w_n)$ for $(w_n)\in {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}$ is nonempty, so by [F1] there is a choice function on the family $(u^{-1}(w_n){)}_{n\ge 0}$, whose values form $(v_n)\in {\prod }_{n}V$ with $u(v_n)=w_n$; hence ${\prod }_{n}u$ is surjective, and with the kernel computation of step 1.2 the sequence ${\prod }_n{V}^{\mathrm{\prime }}\to {\prod }_{n}V\to {\prod }_n{V}^{\mathrm{\prime }\mathrm{\prime }}\to 0$ is exact. By [F4], $F$ is right exact. The Axiom of Choice is used exactly here, to select one preimage in each of the countably many fibres; only this countable instance is used.

The element $w=(e_n{)}_{n\ge 0}\in {\prod }_n({⨁}_{j}k)$, where $e_n$ is the $n$-th standard basis vector, is not in the image of $c$: it would be the image of a source family supported on a finite set $S$ of summand indices, forcing $\{n\}=\mathrm{supp}(e_n)\subseteq S$ for every $n$, so $S\supseteq \mathbb{N}$, contradicting the finiteness of $S$. Hence $c$ is not surjective and $F$ does not preserve the coproduct of the family $(k{)}_{j\in \mathbb{N}}$.

By steps 1.1-1.3 and 2.1 the functor $F$ is additive, left exact and right exact, hence exact by [F4] and in particular right exact.

No tensor functor represents $F$: suppose $\sigma :F\Rightarrow T_N$ is a natural isomorphism. Naturality of $\sigma$ at the coordinate inclusions ${\mathrm{ȷ}}_j$ gives $T_N({\mathrm{ȷ}}_j)\circ {\sigma }_k={\sigma }_{{⨁}_{j}k}\circ F({\mathrm{ȷ}}_j)$ for every $j$, so by the universal property in [F2] the comparison maps satisfy ${\sigma }_{{⨁}_{j}k}\circ c=c^{\mathrm{\prime }}\circ ({⨁}_j{\sigma }_k)$, where $c^{\mathrm{\prime }}$ is the comparison of $T_N$; since ${\sigma }_{{⨁}_{j}k}$ and ${⨁}_j{\sigma }_k$ are isomorphisms, $c$ is an isomorphism if $c^{\mathrm{\prime }}$ is. But $c^{\mathrm{\prime }}$ is an isomorphism because $T_N$ preserves coproducts by [F3], while $c$ is not surjective by step 2.2; this is a contradiction. Hence $F$ is not naturally isomorphic to any tensor functor.

Therefore $F$ is additive and exact, hence right exact, but does not preserve coproducts and is not tensor: the coproduct-preservation hypothesis of the Eilenberg-Watts theorem cannot be dropped even for exact functors. The only choice used is the coordinatewise lifting in step 2.1; the rest of the argument is choice-free.

$F$ is additive: for parallel homomorphisms $u,v:A\to B$ and $\phi \in {\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,A)$, postcomposition satisfies $(u+v{)}_{\ast }(\phi )=(u+v)\circ \phi =u\circ \phi +v\circ \phi$ by [F1], so $F(u+v)=F(u)+F(v)$.

$F$ preserves arbitrary direct sums: the canonical map ${⨁}_i{\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,X_i)\to {\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,{⨁}_i{X}_i)$ induced by the coordinate inclusions is bijective. It is injective because distinct components differ on $[1]$ in distinct coordinates by [F3]; it is surjective because for $\phi$ the element $x=\phi ([1])$ has finite support $S$ by [F3] and satisfies $2x=0$ by [F4], so each $x_i$ satisfies $2x_i=0$ and the maps ${\phi }_i([1])=x_i$ defined for $i\in S$ and zero elsewhere are well-defined homomorphisms with $\phi ={\sum }_i{\phi }_i$.

$F$ is left exact by [F2].

$F$ is not right exact. The map $u:\mathbb{Z}\to \mathbb{Z}\mathrm{/}2$, $u(k)=[k]$, is surjective, hence an epimorphism: if $g\circ u=h\circ u$ then $g$ and $h$ agree on every class. If $F$ were right exact, $F(u)$ would be an epimorphism by [F5]. But ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,\mathbb{Z})=0$, since $2\phi ([1])=0$ in the torsion-free group $\mathbb{Z}$ forces $\phi ([1])=0$ by [F4]; so $F(u)$ is the zero map from $0$ to ${\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)$. That map is not an epimorphism, because the identity and zero endomorphisms of $H:={\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,\mathbb{Z}\mathrm{/}2)$ are distinct ($H$ contains the nonzero identity map of $\mathbb{Z}\mathrm{/}2$ by [F4]) and both have the same composite with $0\to H$. Hence $F$ is not right exact.

$F$ is not naturally isomorphic to any tensor functor $T_N=N{\otimes }_{\mathbb{Z}}-$: if $F\cong T_N$, then $F$ would be right exact, since $T_N$ is right exact by [F6] and right exactness is carried across a natural isomorphism, contradicting step 1.4.

Thus $F$ is an additive functor that is left exact and preserves arbitrary direct sums, but is not tensor; the statement is refuted. No choice is used: the supports occurring in steps 1.2 and 1.4 are determined by the elements involved, and no family of nonempty sets is selected from.