Homogeneous free presentations prove the graded comparison is an isomorphism

For $x\in X_d$ put ${\beta }_X(m,x)=F({\mathrm{\ell }}_x){\theta }_{A,d}^{-1}(m)$; since ${\mathrm{\ell }}_{x+x\mathrm{\prime }}={\mathrm{\ell }}_x+{\mathrm{\ell }}_{x\mathrm{\prime }}$ and ${\mathrm{\ell }}_0=0$, additivity of $F$ makes this additive in $x$ within each degree, and finite homogeneous decomposition extends it uniquely to all $x\in X$. It is additive in $m$ because the maps defining it are homomorphisms. For homogeneous $a\in A_i$, ${\mathrm{\ell }}_{ax}={\mathrm{\ell }}_x\circ (r_a\{d\})$, both maps sending $u$ to $uax$. Naturality of $\theta$ at $r_a$ with parameter $d$ and its cocycle give ${\theta }_{A,d}^{-1}F(r_a){\theta }_{A,i}^{-1}=F(r_a\{d\}){\theta }_{A,i+d}^{-1}$ on underlying maps. Thus ${\beta }_X(m\cdot a,x)=F({\mathrm{\ell }}_x)F(r_a\{d\}){\theta }_{A,i+d}^{-1}(m)={\beta }_X(m,ax)$; additivity extends balancing to arbitrary $a,x$.

By [L10] the balanced pairing ${\beta }_X$ induces a unique group homomorphism ${\tau }_X:M{\otimes }_{A}X\to F(X)$ with ${\tau }_X(m\otimes x)={\beta }_X(m,x)$ on homogeneous $x$; it is degree-zero because for $m\in M_g$ and $x\in X_d$ the tensor $m\otimes x$ has total degree $g+d$ [L7] while ${\theta }_{A,d}^{-1}(m)\in F(A\{d\}{)}_{g+d}$ and $F({\mathrm{\ell }}_x)$ is degree-zero, so ${\beta }_X(m,x)\in F(X{)}_{g+d}$; and it is $B$-linear because ${\beta }_X$ is $B$-linear in $m$ (both $F({\mathrm{\ell }}_x)$ and ${\theta }_{A,d}^{-1}$ are $B$-linear) and the $B$-action on the tensor is the unique one with $b(m\otimes x)=(bm)\otimes x$ [L12].

Naturality in $X$: for a degree-zero $u:X\to Y$ one has $u\circ {\mathrm{\ell }}_x={\mathrm{\ell }}_{u(x)}$ [L3], so $F(u){\tau }_X(m\otimes x)=F(u)F({\mathrm{\ell }}_x){\theta }^{-1}(m)=F({\mathrm{\ell }}_{u(x)}){\theta }^{-1}(m)={\tau }_Y(m\otimes u(x))={\tau }_Y(1\otimes u)(m\otimes x)$; elementary tensors generate $M{\otimes }_{A}X$, so $F(u)\circ {\tau }_X={\tau }_Y\circ (1\otimes u)$ by [L11] and [L17].

Shift compatibility: both sides of ${\theta }_{X,r}^F{\tau }_{X\{r\}}=({\tau }_X\{r\}){\theta }_{X,r}^M$ are degree-zero $B$-linear maps $M{\otimes }_{A}X\{r\}\to F(X)\{r\}$, so it suffices to compare them on $m\otimes y$ with $y\in (X\{r\}{)}_e=X_{e-r}$. The right-hand side gives $F({\mathrm{\ell }}_y^X){\theta }_{A,e-r}^{-1}(m)$ viewed in $F(X)\{r\}$, because ${\theta }^M$ is the identity on elementary tensors [L2]. The left-hand side is ${\theta }_{X,r}^{F}F({\mathrm{\ell }}_y^{X\{r\}}){\theta }_{A,e}^{-1}(m)$ with ${\mathrm{\ell }}_y^{X\{r\}}=({\mathrm{\ell }}_y^X)\{r\}$ the shift of ${\mathrm{\ell }}_y^X$ [L3, L4]; naturality of ${\theta }^F$ at ${\mathrm{\ell }}_y^X$ with parameter $r$ gives ${\theta }_{X,r}^{F}F(({\mathrm{\ell }}_y^X)\{r\})=(F({\mathrm{\ell }}_y^X)\{r\}){\theta }_{A\{e-r\},r}^F$ and the shift of a morphism is the same underlying map [L4], so the left-hand side equals $F({\mathrm{\ell }}_y^X){\theta }_{A\{e-r\},r}^F{\theta }_{A,e}^{-1}(m)$; the cocycle ${\theta }_{A,e}^F=({\theta }_{A,e-r}^F\{r\}){\theta }_{A\{e-r\},r}^F$ together with [L4] gives ${\theta }_{A\{e-r\},r}^F{\theta }_{A,e}^{-1}={\theta }_{A,e-r}^{-1}$ on underlying maps, so the left-hand side equals the right-hand side.

At $X=A\{d\}$ the generator map ${\mathrm{\ell }}_{1_A}:A\{d\}\to A\{d\}$ is the identity, so ${\tau }_{A\{d\}}(m\otimes 1_A)={\theta }_{A,d}^{-1}(m)$; under the tensor-unit and shift isomorphisms $M{\otimes }_{A}A\{d\}\cong (M{\otimes }_{A}A)\{d\}\cong M\{d\}$ of [L8] the map ${\tau }_{A\{d\}}$ corresponds to the degree-zero isomorphism ${\theta }_{A,d}^{-1}$, hence is itself an isomorphism.

Let $P={⨁}_{i}A\{d_i\}$ be a coproduct of shifts of $A$. Since $T_M$ preserves coproducts [L2] and $F$ does by hypothesis, naturality of $\tau$ at the coproduct inclusions identifies ${\tau }_P$ with the coproduct ${⨁}_i{\tau }_{A\{d_i\}}$ of the isomorphisms of step 3.3, under the canonical decompositions $T_M(P)\cong {⨁}_i{T}_M(A\{d_i\})$ and $F(P)\cong {⨁}_{i}F(A\{d_i\})$; a coproduct of isomorphisms is an isomorphism, so ${\tau }_P$ is an isomorphism.

For arbitrary $X$ the presentation $P_{K_X}\stackrel{d}{\to }{P}_X\stackrel{q_X}{\to }X\to 0$ of [L3] is exact at $P_X$ and at $X$ [L13], with $d$ of image $\ker q_X$ and $q_X$ an epimorphism; the functors $T_M$ and $F$ preserve cokernels because they are right exact [L2, L15, L16], so applying them gives two cokernel diagrams connected by $\tau$: ${\tau }_{P_X}{T}_M(d)=F(d){\tau }_{P_{K_X}}$ by naturality (step 3.1). The maps ${\tau }_{P_{K_X}}$ and ${\tau }_{P_X}$ are isomorphisms by step 4.1, so the unique map induced on the cokernels by cokernel universality [L14] is an isomorphism with inverse induced by the two inverses; this proves the graded counterpart of the ungraded comparison [L19] directly, and the first presentation map $d$ is never asserted to be monic.

Collecting steps 3.1, 3.2 and 5.1: $\tau :T_M\Rightarrow F$ is a natural transformation that is an isomorphism in every degree and hence a natural isomorphism [L18], it satisfies the equivariance identity of step 3.2, so it is a coherent morphism between the coherent functors $T_M$ and $F$ in the sense of [L6] and [L5], and consequently every such $F$ is coherently naturally isomorphic to $T_{F(A)}$; the displays of the statement record the two composites of the comparison matrix, which are equal by step 3.2.