Internal shifts are autoequivalences and commute with the graded tensor product

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.