Graded associativity, units, and internal-shift tensor isomorphisms

Let $X,Y$ be graded abelian groups and $f:X\to Y$ a bijective degree-zero homomorphism. Then $f$ is an isomorphism of graded abelian groups, i.e. $f^{-1}$ is degree-zero: for $y\in Y_d$ write $y=f(x)$ with $x={\sum }_e{x}_e$ the finite homogeneous decomposition, so $y={\sum }_{e}f(x_e)$ with $f(x_e)\in Y_e$; uniqueness of the homogeneous decomposition in $Y$ gives $f(x_d)=y$ and $f(x_e)=0$ for $e\mathrm{\ne }d$, hence $x=x_d\in X_d$.

For homogeneous $m\in M_i$, $n\in N_j$, $p\in P_l$ the tensor $(m\otimes n)\otimes p$ has degree $(i+j)+l$ and $m\otimes (n\otimes p)$ has degree $i+(j+l)$, the same integer, so ${\alpha }_{M,N,P}$ carries the homogeneous part of degree $d$ into degree $d$ on elementary tensors and, being additive, on all of $(M{\otimes }_{R}N){\otimes }_{S}P$. It is a bijective group homomorphism by [L3], so step 1.1 makes it a degree-zero isomorphism; its naturality and compatibility with outer actions are those of the published associator.

For homogeneous $r\in R_i$ and $n\in N_j$ one has ${\lambda }_N(r\otimes n)=rn\in N_{i+j}$, so ${\lambda }_N$ is degree-zero, and it is bijective by [L4]; step 1.1 makes it a degree-zero isomorphism, and its compatibility with outer actions is the published one. The same computation with ${\rho }_M(m\otimes r)=mr\in M_{i+j}$ for $m\in M_i$ treats ${\rho }_M$.

The pairing $M\{r\}\times N\{s\}\to (M{\otimes }_{R}N)\{r+s\}$, $(x,y)\mapsto x\otimes y$, is balanced with respect to $R$: the underlying $R$-actions of $M\{r\}$ and $N\{s\}$ are those of $M$ and $N$, so $(xa)\otimes y$ and $x\otimes (ay)$ are equal in $M{\otimes }_{R}N$; it is additive in each variable. By [L5] it induces a group homomorphism $\phi$ with $\phi (x\otimes y)=x\otimes y$. For $x\in (M\{r\}{)}_i=M_{i-r}$ and $y\in (N\{s\}{)}_j=N_{j-s}$ the element $x\otimes y$ lies in $(M{\otimes }_{R}N{)}_{i+j-r-s}=((M{\otimes }_{R}N)\{r+s\}{)}_{i+j}$, so $\phi$ is degree-zero, and the same construction in the reverse direction gives $\psi$ with $\psi (x\otimes y)=x\otimes y$; the two composites fix all elementary tensors and hence are identities. By step 1.1, $\phi$ is a degree-zero isomorphism.

The outer actions on both sides of $\phi$ are the unique actions with the elementary-tensor formulas of [L5], and the shifts change no action, so $\phi$ is compatible with the outer actions; it is natural because it is induced from the universal property of the pairing of underlying modules.

Collecting steps 2.1, 2.2, 2.3 and 3.1: the associator, the two unit maps and the shift comparison are degree-zero isomorphisms of graded modules, with the naturality and outer-action compatibility stated.

Therefore the ordinary balanced associator and unit maps are degree-zero graded isomorphisms, and the identity on elementary tensors induces the natural degree-zero isomorphism $M\{r\}{\otimes }_{R}N\{s\}\cong (M{\otimes }_{R}N)\{r+s\}$ compatible with outer actions.