Graded Eilenberg–Watts and Shift Coherence

Give the direct sum $X:={⨁}_{i\in I}{X}_i$ of the underlying left $A$-modules its componentwise action $a\cdot (x_i{)}_{i\in I}:=(a{x}_i{)}_{i\in I}$; this is a left $A$-module structure preserving finite supports, and with $X_d:={⨁}_{i\in I}(X_i{)}_d$ one has $X={⨁}_{d\in \mathbb{Z}}{X}_d$: every element of $X$ is a finite sum of elements of the subgroups $X_d$, and if ${\sum }_d{x}^{(d)}=0$ with $x^{(d)}\in X_d$ then in each coordinate ${\sum }_d{x}_i^{(d)}=0$ with $x_i^{(d)}\in (X_i{)}_d$, so $x_i^{(d)}=0$ for all $i,d$ by the directness of each $X_i={⨁}_d(X_i{)}_d$.

The shifted regular module $A\{d\}$ has the same underlying left $A$-module as $A$, which is free on the single standard basis element $1_A$, and a left $A$-module map out of it is uniquely determined by the image of $1_A$; for every $e$ the map ${\mathrm{\ell }}_x(a):=ax$ sends $(A\{d\}{)}_e=A_{e-d}$ into $X_e$ because $A_{e-d}x\subseteq X_e$, so it is degree-zero $A$-linear with ${\mathrm{\ell }}_x(1_A)=x$, and every degree-zero $A$-linear $\phi :A\{d\}\to X$ with $\phi (1_A)=x$ satisfies $\phi (a)=a\phi (1_A)=ax$.

Every family of degree-zero $A$-linear maps $f_i:X_i\to Y$ assembles by [L3] to the unique $A$-linear $f:X\to Y$ with $f\circ {\mathrm{ȷ}}_i=f_i$, namely $f((x_i{)}_{i\in I})={\sum }_i{f}_i(x_i)$; it is degree-zero because for $x=(x_i)\in X_d$ all components satisfy $x_i\in (X_i{)}_d$, whence $f_i(x_i)\in Y_d$ and $f(x)\in Y_d$, and it is the unique degree-zero $A$-linear map with the prescribed composites, so $X$ is the coproduct of the family.

Since $A_i(X_j{)}_d\subseteq (X_j{)}_{i+d}$ for all $i,d,j$, the componentwise action satisfies $A_i{X}_d\subseteq X_{i+d}$, so $X$ is a graded left $A$-module; the coordinate inclusion ${\mathrm{ȷ}}_j:X_j\to X$ maps $(X_j{)}_d$ into $X_d$, so it is degree-zero $A$-linear.

For $I=\mathrm{\varnothing }$ the direct sum is the zero module by [L2], and the unique map from a zero module to $Y$ is $A$-linear and degree-zero, so the zero module is initial and is the coproduct of the empty family; together with step 2.1 this shows that the degreewise direct sum is the coproduct of every family, so ${\mathrm{GrMod}}_0(A)$ has all small coproducts.

Suppose $I$ is finite. Then $X$ is also the product of the family: given degree-zero $A$-linear maps $g_i:Y\to X_i$, the elementwise map $g(y):=(g_i(y){)}_{i\in I}$ has finite support in $X$, is $A$-linear and degree-zero, and is the unique such map with ${\pi }_i\circ g=g_i$ for the coordinate projections ${\pi }_i$; for two summands $X$ has pieces $X_d=(X_1{)}_d\oplus (X_2{)}_d$ with the same coordinate inclusions and projections as the binary biproduct of [L6], and the finite case follows by iterating that identification, so the coproducts here agree with the finite biproducts computed in [L6].

Let $H_X$ be the set of nonzero homogeneous elements of $X$; each $x\in H_X$ has a unique degree $\mathrm{deg}x$ with $x\in X_{\mathrm{deg}x}$, since $x\in X_d\cap X_e$ with $x\mathrm{\ne }0$ and $d\mathrm{\ne }e$ would exhibit $x$ as two different finite decompositions of one element. Hence $P_X:={⨁}_{x\in H_X}A\{\mathrm{deg}x\}$ is a graded left $A$-module by steps 1.1 and 2.2, and by step 2.1 and step 1.2 the maps ${\mathrm{\ell }}_x:A\{\mathrm{deg}x\}\to X$ assemble to the unique degree-zero $A$-linear $q_X:P_X\to X$ with $q_X(e_x)=x$ for each $x\in H_X$.

The map $q_X$ is surjective, hence an epimorphism: every $x^{\mathrm{\prime }}\in X$ is the finite sum of its nonzero homogeneous components $x^{(d)}\in H_X$, and $q_X(e_{x^{(d)}})=x^{(d)}$; and two degree-zero $A$-linear maps out of $X$ agreeing after composition with a surjection agree everywhere.

The kernel $K_X=\ker q_X$ is ${⨁}_d\ker ((q_X{)}_d)$ by [L6], hence a graded submodule of $P_X$ and therefore a graded left $A$-module with pieces $K_X\cap (P_X{)}_d$; applying the construction of steps 3.3 and 4.1 to $K_X$ produces $\mathrm{Free}(K_X):=P_{K_X}$ and a degree-zero $A$-linear epimorphism $q_{K_X}:P_{K_X}\to K_X$, whose composite $d$ with the inclusion $K_X\to P_X$ is degree-zero $A$-linear with image $K_X$.

Therefore $\mathrm{im}d=K_X=\ker q_X$ while $\mathrm{im}{q}_X=X$ by step 4.1, so the sequence $\mathrm{Free}(K_X)\stackrel{d}{\to }{P}_X\stackrel{q_X}{\to }X\to 0$ is exact at $P_X$ and at $X$; the map $d$ is not asserted monic, the indexing set $H_X$ and the degrees $\mathrm{deg}x$ are determined by $X$, and the direct sums are indexed by those elements, so no generator, basis, resolution or other choice is made.

By [L1] and [L9] the category ${\mathrm{GrMod}}_0(A)$ is abelian, hence additive [L10] and preadditive [L11], and by [L2] it has all small coproducts; for a small diagram $D$ the coproducts $R$ and $S$ and the coequalizer of $(d,c)$ therefore exist, and by [L5] that coequalizer, which by [L6] is the cokernel of $c-d$, is a colimit of $D$. In particular every small diagram has a colimit in ${\mathrm{GrMod}}_0(A)$.

Assume $F$ preserves cokernels and all coproducts, and let $Q=\mathrm{coker}(c-d)$ be the colimit of $D$ from step 1.1 with its canonical cocone. Since $F$ preserves coproducts, the maps $F({\mathrm{ȷ}}_u)$ exhibit $F(R)$ as a coproduct of the $F(D(j))$ and the maps $F({\mathrm{ȷ}}_j)$ exhibit $F(S)$ as a coproduct of the $F(D(j))$, so $F(d)$ and $F(c)$ are the canonical maps of the same recipe for the composite $FD$; since $F$ preserves cokernels, $F(Q)$ with $F$ of the canonical map is a cokernel of $F(c-d)$, and $F(c-d)=F(c)-F(d)$ by additivity [L12]; by [L6] that cokernel is a coequalizer of $(F(d),F(c))$, so by [L5] applied to $FD$ the object $F(Q)$ with the image cocone is a colimit of $FD$.

Conversely, if $F$ is cocontinuous then it preserves every coproduct, because a coproduct of a family is the colimit of the discrete diagram on its index set, and it preserves every cokernel, because the cokernel of a morphism $f$ is the coequalizer of $(f,0)$ by [L6] and hence a colimit over a parallel pair; both index categories are small.

Since the small diagram $D$ of step 2.1 was arbitrary, $F$ preserves every small colimit, that is, $F$ is cocontinuous; together with step 2.2 this shows that preservation of all small colimits is equivalent to preservation of cokernels and all coproducts.

For the right-exact reformulation: a right exact functor preserves cokernels, since a cokernel is a finite colimit [L4]; conversely, if $F$ preserves cokernels and all coproducts, then it preserves the colimit of every finite diagram, because for finite $\mathcal{J}$ the coproducts $R$ and $S$ are finite and are finite biproducts of ${\mathrm{GrMod}}_0(A)$ [L2, L10] and all finite colimits of the source exist [L8], so the computation of step 2.1 with finite index sets applies verbatim; hence such an $F$ is right exact.

Combining steps 3.1 and 3.2: $F$ preserves all small colimits if and only if $F$ preserves cokernels and all coproducts if and only if $F$ is right exact and coproduct preserving, so cocontinuity of $F$ is exactly right exactness together with coproduct preservation; the coproducts and cokernels used are the canonical ones of step 1.1, so no choice is made.

Let $\{r\}$ send an object $X$ to the graded module $X\{r\}$ and a morphism $u:X\to Y$ to the same underlying map $u$. This is well-defined: for $x\in (X\{r\}{)}_d=X_{d-r}$ one has $u(x)\in Y_{d-r}=(Y\{r\}{)}_d$, so $u$ is degree-zero and $A$-linear as a map $X\{r\}\to Y\{r\}$; identities and composites are inherited from ${\mathrm{GrMod}}_0(A)$, so $\{r\}$ is a functor [L6]. On objects and on morphisms the shift formula gives $(X\{r\})\{s\}=X\{r+s\}$ and $X\{0\}=X$ literally, because both sides have the same underlying set and the same homogeneous pieces; hence $\{r\}\circ \{s\}=\{r+s\}$ and $\{0\}=\mathrm{i}\mathrm{d}$ as equalities of functors, and $\{-r\}$ is inverse to $\{r\}$.

For fixed $X,Y$ the sets $\mathrm{Hom}(X,Y)$ and $\mathrm{Hom}(X\{r\},Y\{r\})$ are equal: a function $X\to Y$ is degree-zero $A$-linear for the shifted pair exactly when $u(X_{d-r})\subseteq Y_{d-r}$ for all $d$, which is the same condition as $u(X_e)\subseteq Y_e$ for all $e$, and the addition, the $k$-scalar action and the composition law are pointwise and unchanged by the shift. Hence the induced map on hom-groups is the identity of one and the same $k$-module, so it is additive and $k$-linear; when $k$ is a field this is exactly $k$-linearity in the sense of [L7], since then the hom-modules are $k$-vector spaces [L8] and every field is a commutative ring [L9, L10].

For each degree $d$ the identity map gives $(({⨁}_i{X}_i)\{r\}{)}_d=({⨁}_i{X}_i{)}_{d-r}={⨁}_i(X_i{)}_{d-r}={⨁}_i(X_i\{r\}{)}_d$, and the coordinate inclusions of the two sides correspond under this identification, so the identity on the underlying module is a degree-zero $A$-linear isomorphism $({⨁}_i{X}_i)\{r\}\cong {⨁}_i(X_i\{r\})$; it is natural because it is the identity on underlying sets and intertwines every family of maps.

For a degree-zero $u:X\to Y$ the same identification gives $(\ker (u\{r\}){)}_d=\{x\in X_{d-r}:u(x)=0\}=(\ker u)\cap X_{d-r}=((\ker u)\{r\}{)}_d$, and likewise $(u\{r\})((X\{r\}{)}_d)=u(X_{d-r})=(\mathrm{im}u{)}_{d-r}=((\mathrm{im}u)\{r\}{)}_d$ and $(\mathrm{coker}(u\{r\}){)}_d=Y_{d-r}\mathrm{/}u(X{)}_{d-r}=((\mathrm{coker}u)\{r\}{)}_d$, using that kernels, images and cokernels in ${\mathrm{GrMod}}_0(A)$ are degreewise [L2]; the resulting degreewise equalities are equalities of graded submodules and quotients, so the identity maps are the asserted degree-zero $A$-linear isomorphisms, natural in $u$ because all constructions agree with the underlying maps.

Part 3 of [L3] states precisely the natural degree-zero isomorphism $M\{r\}{\otimes }_{A}N\{s\}\cong (M{\otimes }_{A}N)\{r+s\}$ compatible with the outer actions, for the graded balanced tensor of [L4]; no further verification of the isomorphism is needed, and the outer-action compatibility is the one recorded there.

Collecting steps 2.1, 2.2 and 2.3: the internal shift is an autoequivalence inverting $\{-r\}$ with strict composition and unit equalities, it preserves degreewise coproducts, kernels, images and cokernels, and it commutes with the graded balanced tensor product by a natural isomorphism compatible with outer actions; since the shift leaves elements, actions, maps and differentials as they are, it inserts no sign [L1] and is a relabelling of degrees rather than the cochain shift $[1]$ of a complex, and no selection of bases, generators or lifts is made anywhere.

The identity functor $1$ of ${\mathrm{GrMod}}_0(A)$ with ${\theta }_{X,r}:=1_{X\{r\}}:1(X\{r\})=X\{r\}\to X\{r\}=1(X)\{r\}$ is coherently shift-compatible: each ${\theta }_{X,r}$ is a natural degree-zero $A$-linear isomorphism (it is an identity), ${\theta }_{X,0}=1_X$ because $X\{0\}=X$, and $({\theta }_{X,r}\{s\})\circ {\theta }_{X\{r\},s}=1_{X\{r\}\{s\}}=1_{X\{r+s\}}={\theta }_{X,r+s}$ by the strict shift identities.

Given coherent $F$ and $G$, the composite comparison ${\theta }_{X,r}^{GF}:={\theta }_{F(X),r}^G\circ G({\theta }_{X,r}^F)$ is a composite of natural degree-zero isomorphisms and hence a natural degree-zero $C$-linear isomorphism $(GF)(X\{r\})\to (GF)(X)\{r\}$; its unit is ${\theta }_{F(X),0}^G\circ G({\theta }_{X,0}^F)=1\circ G(1)=1$, and its cocycle follows by expanding ${\theta }_{X,r+s}^{GF}={\theta }_{F(X),r+s}^G\circ G({\theta }_{X,r+s}^F)$, rewriting the inner factor with the cocycle of ${\theta }^F$, applying the functor $G$, substituting the naturality of ${\theta }^G$ at the morphism ${\theta }_{X,r}^F$ with parameter $s$, and finally the cocycle of ${\theta }^G$ at $F(X)$; the comparisons are strictly associative and unital, ${\theta }^{(HG)F}={\theta }^{H(GF)}$ and ${\theta }^{1\circ F}={\theta }^F={\theta }^{F\circ 1}$, because both sides are the same composites of ${\theta }^H$, $H({\theta }^G)$ and $H(G({\theta }^F))$ and $H$ preserves composition and identities.

If $\eta :F\Rightarrow G$ and $\nu :G\Rightarrow H$ are coherent, then $\nu \circ \eta$ is natural and ${\theta }_{X,r}^H(\nu \circ \eta {)}_{X\{r\}}={\theta }_{X,r}^H{\nu }_{X\{r\}}{\eta }_{X\{r\}}=({\nu }_X\{r\}){\theta }_{X,r}^G{\eta }_{X\{r\}}=({\nu }_X\{r\})({\eta }_X\{r\}){\theta }_{X,r}^F=((\nu \circ \eta {)}_X\{r\}){\theta }_{X,r}^F$ by the coherence of $\nu$ and $\eta$, so $\nu \circ \eta$ is coherent; the identity transformations are coherent by the condition (i) of the definition.

For coherent $\eta ,{\eta }^{\mathrm{\prime }}:F\Rightarrow G$ and $\lambda \in k$, the pointwise sum $\eta +{\eta }^{\mathrm{\prime }}$ and scalar multiple $\lambda \eta$ are natural transformations, and they satisfy the equivariance square because both sides are additive in the component: ${\theta }_{X,r}^G(\eta +{\eta }^{\mathrm{\prime }}{)}_{X\{r\}}={\theta }_{X,r}^G{\eta }_{X\{r\}}+{\theta }_{X,r}^G{\eta }_{X\{r\}}^{\mathrm{\prime }}=({\eta }_X\{r\}){\theta }_{X,r}^F+({\eta }_X^{\mathrm{\prime }}\{r\}){\theta }_{X,r}^F=((\eta +{\eta }^{\mathrm{\prime }}{)}_X\{r\}){\theta }_{X,r}^F$, and likewise ${\theta }_{X,r}^G(\lambda \eta {)}_{X\{r\}}=\lambda {\theta }_{X,r}^G{\eta }_{X\{r\}}=(\lambda {\eta }_X\{r\}){\theta }_{X,r}^F$; the zero transformation is coherent and $-\eta =(-1)\eta$, so the coherent transformations $F\Rightarrow G$ form a subgroup of all natural transformations closed under the pointwise $k$-action.

Let $F,G$ be $k$-linear, right exact and coproduct preserving, and let $\kappa ,{\kappa }^{\mathrm{\prime }}:F\Rightarrow G$ be coherent with ${\kappa }_{P_X}={\kappa }_{P_X}^{\mathrm{\prime }}$ for every graded module $X$. For each $X$ the cover $q_X:P_X\to X$ of [L3] is an epimorphism, hence $F(q_X)$ and $G(q_X)$ are epimorphisms because right exact functors preserve epimorphisms [L16]; naturality gives ${\kappa }_{X}F(q_X)=G(q_X){\kappa }_{P_X}=G(q_X){\kappa }_{P_X}^{\mathrm{\prime }}={\kappa }_X^{\mathrm{\prime }}F(q_X)$, and cancelling the epimorphism $F(q_X)$ gives ${\kappa }_X={\kappa }_X^{\mathrm{\prime }}$; since $X$ was arbitrary, $\kappa ={\kappa }^{\mathrm{\prime }}$.

If $\kappa ,{\kappa }^{\mathrm{\prime }}:F\Rightarrow G$ agree on every component at a shifted regular module $A\{d\}$, they agree on every $P_X$: writing $P_X={⨁}_{x\in H_X}A\{\mathrm{deg}x\}$ with coordinate inclusions ${\mathrm{ȷ}}_x$ [L3], the modules $F(A\{\mathrm{deg}x\})$ with the maps $F({\mathrm{ȷ}}_x)$ present $F(P_X)$ as a coproduct, so a morphism out of $F(P_X)$ is determined by its composites with all $F({\mathrm{ȷ}}_x)$; naturality of $\kappa ,{\kappa }^{\mathrm{\prime }}$ at ${\mathrm{ȷ}}_x$ gives ${\kappa }_{P_X}F({\mathrm{ȷ}}_x)=G({\mathrm{ȷ}}_x){\kappa }_{A\{\mathrm{deg}x\}}$ and the same with ${\kappa }^{\mathrm{\prime }}$, and the right sides agree, so ${\kappa }_{P_X}={\kappa }_{P_X}^{\mathrm{\prime }}$; with step 1.5, $\kappa ,{\kappa }^{\mathrm{\prime }}$ then agree everywhere.