Graded Eilenberg–Watts and Shift Coherence — Examples

$F$ is a functor: for a degree-zero $k$-linear $u:X\to Y$ one has $u(X_0)\subseteq Y_0$, so the restriction $F(u):=u{\mathrm{\mid }}_{X_0}$ is a degree-zero $k$-linear map $F(X)\to F(Y)$; restrictions preserve identities and composites, so $F$ is a functor.

$F$ is additive and $k$-linear: on hom-groups the assignment $u\mapsto u{\mathrm{\mid }}_{X_0}$ satisfies $(u+v){\mathrm{\mid }}_{X_0}=u{\mathrm{\mid }}_{X_0}+v{\mathrm{\mid }}_{X_0}$ and $(\lambda u){\mathrm{\mid }}_{X_0}=\lambda u{\mathrm{\mid }}_{X_0}$, so the induced maps are $k$-linear [L10].

$F$ preserves every coproduct: the identity map gives $F({⨁}_i{X}_i)=({⨁}_i{X}_i{)}_0={⨁}_i(X_i{)}_0={⨁}_{i}F(X_i)$ with the same coordinate inclusions, so the canonical comparison is an isomorphism [L4].

On the regular module $F(k)=k_0=k$; on the internal shift $F(k\{1\})=(k\{1\}{)}_0=k_{-1}=0$ because $k$ is concentrated in degree $0$ [L1], while $T_k(k\{1\})=k{\otimes }_{k}k\{1\}\cong k\{1\}\mathrm{\ne }0$ by the unit isomorphism [L8], since $(k\{1\}{)}_1=k_0=k\mathrm{\ne }0$; a natural isomorphism $F\cong T_k$ would therefore induce an isomorphism $0\to k\{1\}$, which is impossible [L9].

$F$ preserves kernels and cokernels: for degree-zero $u:X\to Y$ the degreewise descriptions give $F(\ker u)=(\ker u{)}_0=\ker (u{\mathrm{\mid }}_{X_0})=\ker (Fu)$ and $F(\mathrm{coker}u)=(\mathrm{coker}u{)}_0=Y_0\mathrm{/}u(X_0)=\mathrm{coker}(Fu)$ with the induced maps [L3], so $F$ preserves the kernel and cokernel of every morphism; hence $F$ is exact, and in particular left and right exact, by [L11].

No coherent shift-comparison data exist for $F$: by [L12] such data would include a degree-zero isomorphism ${\theta }_{k,1}:F(k\{1\})\to F(k)\{1\}=k\{1\}$, but $F(k\{1\})=0$ while $(k\{1\}{)}_1=k\mathrm{\ne }0$ by steps 2.3 and [L2, L9], so no isomorphism exists. More generally, if $F\cong T_M$ for a graded $(k,k)$-bimodule $M$, evaluation at $k$ and [L8] give $M\cong F(k)=k$, whereas $T_M(k\{1\})\cong M\{1\}\mathrm{\ne }0$ by [L5], contradicting $F(k\{1\})=0$. With steps 2.1, 2.2 and 3.1 the functor $F$ is additive, right exact and coproduct preserving, hence cocontinuous by [L6, L7]; therefore $F$ is a $k$-linear exact cocontinuous functor with $F(k)=k$ that is not isomorphic to any graded tensor functor [L5] and does not lie in the coherent class, and the witness uses no choice.

$T_{A\{r\}}$ is $k$-linear, right exact and coproduct preserving with comparisons the identity on elementary tensors [L2]; the unit isomorphism $A{\otimes }_{A}X\cong X$ and the shift isomorphism $A\{r\}{\otimes }_{A}X\cong (A{\otimes }_{A}X)\{r\}$ of [L3] compose to a natural degree-zero isomorphism $A\{r\}{\otimes }_{A}X\cong X\{r\}$ compatible with the outer actions, so $T_{A\{r\}}\cong \{r\}$ as functors.

The shift $\{r\}$ is the identity functor precisely when $r=0$: for $r=0$ this is $\{0\}=\mathrm{i}\mathrm{d}$ [L5]; conversely, if $\{r\}$ is the identity functor then $\{r\}(A)=A\{r\}$ equals $A$, so the degree-zero pieces agree, $A_{-r}=(A\{r\}{)}_0=A_0$, and $1_A\in A_{-r}\cap A_0$ is nonzero by the standing hypothesis $1_A\mathrm{\ne }0$, so $-r=0$ because distinct homogeneous pieces intersect trivially [L4]. The excluded zero algebra is genuinely different: there the only graded left $A$-module is $0$, so every shift is the identity functor.

The functor $\{r\}$ is not the cochain shift $[1]$: $\{r\}$ changes only the internal grading and inserts no sign or differential [L4], while the cochain shift changes cochain placement and, by the convention of the bounded-complex page, its sign is the cochain Koszul sign carried by the differential alone, with internal degrees contributing no cochain sign [L6, L7]; the two are different operations on different structures, and the published counterexample cited in the statement records their non-interchangeability in the bounded homotopy category, where it is not a prerequisite of the present computation.

The kernel attached to the internal shift by the classification of [L1] is $A\{r\}$: the bimodule attached to a tensor functor $T_M$ is recovered as $T_M(A)=M{\otimes }_{A}A\cong M$, and here $T_{A\{r\}}(A)=A\{r\}{\otimes }_{A}A\cong A\{r\}$ by step 1.1 and [L3]; under the identification $T_{A\{r\}}\cong \{r\}$ of step 1.1 the comparisons ${\theta }^{A\{r\}}$ are the identity on elementary tensors [L2], which is the canonical comparison of the internal shift [L5].

Collecting steps 1.1, 1.2, 1.3 and 2.1: the internal shift is a $k$-linear right exact coproduct-preserving coherent functor, it is naturally isomorphic to $T_{A\{r\}}$, its kernel is the shifted regular bimodule $A\{r\}$ with the canonical comparisons, it is the identity exactly for $r=0$, and it is a different operation from the cochain shift $[1]$; all identifications used are the canonical ones, so no choice is made.

For every family $({\lambda }_d)$ the components ${\eta }_X(x):={\lambda }_{d}x$ for $x\in X_d$ extend uniquely over finite homogeneous sums to degree-zero $k$-linear maps ${\eta }_X:X\to X$, and these define a natural transformation $\eta :\mathrm{i}\mathrm{d}\Rightarrow \mathrm{i}\mathrm{d}$: for a degree-zero $u:X\to Y$ and $x\in X_d$ one has $u(x)\in Y_d$, so ${\eta }_Y(u(x))={\lambda }_{d}u(x)=u({\lambda }_{d}x)=u({\eta }_X(x))$.

Such an $\eta$ is coherent for the canonical comparisons of the identity functor if and only if ${\lambda }_{d+r}={\lambda }_d$ for all $d,r$: coherence reads ${\eta }_{X\{r\}}={\eta }_X\{r\}$ as underlying maps by [L2], and on $x\in X_{e-r}=(X\{r\}{)}_e$ this is ${\lambda }_{e}x={\lambda }_{e-r}x$; taking $X=k\{e-r\}$, whose degree-$(e-r)$ piece is nonzero [L3], gives ${\lambda }_e={\lambda }_{e-r}$ for all $e,r$, while a constant family clearly satisfies the square.

Hence the coherent endomorphisms of the identity functor are exactly the constant families, and they correspond bijectively to $k$ via $\eta \mapsto {\eta }_k$, in agreement with [L1], where the coherent endomorphisms correspond to ${\mathrm{Hom}}_{k\text{-}k}(k,k)\cong k$: the constant family with scalar $\lambda$ has component $\lambda {\mathrm{i}\mathrm{d}}_X$ at every $X$, and since $T_k\cong \mathrm{i}\mathrm{d}$ by the unit isomorphism [L5] the two descriptions agree.

The family ${\lambda }_d=1$ for $d=1$ and ${\lambda }_d=0$ for $d\mathrm{\ne }1$ is nonconstant, so by step 2.1 it is not coherent, but by step 1.1 it is a nonzero natural endomorphism of the identity functor: its component at $k$ is ${\lambda }_0{\mathrm{i}\mathrm{d}}_k=0$, the same as that of the zero transformation, while its component at $k\{1\}$ is ${\lambda }_1{\mathrm{i}\mathrm{d}}_{k\{1\}}={\mathrm{i}\mathrm{d}}_{k\{1\}}\mathrm{\ne }0$ because $(k\{1\}{)}_1\mathrm{\ne }0$ [L3]; hence two distinct natural transformations of $\mathrm{i}\mathrm{d}$ have the same component at the regular module, and after transporting along $T_k\cong \mathrm{i}\mathrm{d}$ [L5] there are distinct natural transformations $T_k\Rightarrow T_k$ with the same image under $\eta \mapsto {\eta }_k$.

Collecting steps 1.1 to 3.2: the unconstrained natural endomorphisms of the identity form a family strictly larger than the coherent ones, they are not determined by their component on the regular module, and only the constant families satisfy the equivariance square; so the restriction on 2-cells in Coherently shift-compatible functors and natural transformations is genuinely needed for the classification of Graded bimodule maps classify shift-compatible transformations, exactly as Hazrat's remark on transformations between shift-commuting functors warns, and no choice is used.