Strongly graded P(z) products from restricted-dual images

For $w_1,w_2$ homogeneous of auxiliary degrees ${\beta }_1,{\beta }_2$, define $\kappa (w_1,w_2)$ componentwise by $⟨s,\kappa (w_1,w_2)⟩=s(w_1\otimes w_2)$. Only $s$ of degree $-{\beta }_1-{\beta }_2$ can contribute. Each fixed generalized-weight component is a functional on the finite-dimensional corresponding component of $S$, hence an element of $\overline{S^{\mathrm{\prime }}}$. For fixed $h$, the components at $h-m$, $m\to +\mathrm{\infty }$, vanish because the ONE degree ${\beta }_1+{\beta }_2$ integer string of $S^{\mathrm{\prime }}$ is lower truncated. Finite sums of degrees handle arbitrary inputs. Compatibility in F2 transposes to the full $P(z)$-Jacobi identity for $\kappa$ and the three transported $L^{\mathrm{\prime }}(j)$ identities transpose to its Möbius brackets. Thus $\kappa$ is an actual $P(z)$-map, with target $S^{\mathrm{\prime }}\in \mathcal{C}$.

Given $I:W_1\otimes W_2\to \bar{X}$ with $X\in \mathcal{C}$, its transpose $I^{\mathrm{\prime }}:X^{\mathrm{\prime }}\to S$ has its image in $S$ by the definition of that sum. Transposed Jacobi and the bracket identities make it a strong-module homomorphism. Its restricted transpose is a homomorphism $S^{\mathrm{\prime }}\to X^{\mathrm{\prime }\mathrm{\prime }}$, hence, by F1 and fullness, a morphism $f_I:S^{\mathrm{\prime }}\to X$. Pairing with $x^{\mathrm{\prime }}\in X^{\mathrm{\prime }}$ gives $I=\overline{f_I}\kappa$. Conversely, such a factorization forces $f_I^{\mathrm{\prime }}(x^{\mathrm{\prime }})$ to have exactly the values $⟨x^{\mathrm{\prime }},I(w_1,w_2)⟩$ on every input; injectivity of the inclusion $S\hookrightarrow (W_1\otimes W_2{)}^{\ast }$ makes this transpose unique. Restricted duals separate finite components, so $f_I$ is unique. The same pairings show naturality in $X$. This proves the representing property.

If the span of the homogeneous components of $\kappa (w_1,w_2)$ were proper, its intersection with some finite-dimensional $(\beta ,h)$-component would be proper. Choose a nonzero functional on that component vanishing on the intersection and extend it by zero to other components. It is an element $s\in (S^{\mathrm{\prime }}{)}^{\mathrm{\prime }}=S$, by F1, and it vanishes on every evaluation value, so $s(w_1,w_2)=0$ for all inputs. The inclusion of $S$ in the algebraic dual is injective, a contradiction. This proves the vector-space spanning assertion without an unlicensed assumption that every submodule or quotient belongs to $\mathcal{C}$.

Suppose $(T,\iota )$ already represents. Every $I$ is $\bar{f}\iota$ for a morphism $f:T\to X$, so $I^{\mathrm{\prime }}={\iota }^{\mathrm{\prime }}{f}^{\mathrm{\prime }}$ and every transposed image is contained in $\mathrm{im}{\iota }^{\mathrm{\prime }}$. The canonical map $\iota$ itself is one of the maps summed, giving the reverse inclusion. Thus $S=\mathrm{im}{\iota }^{\mathrm{\prime }}$. Once $S\in \mathcal{C}$, steps 1.1–2.1 show that $S^{\mathrm{\prime }}$ represents the same functor. Apply the two representing bijections to their canonical maps: the two comparison morphisms have compositions inducing the original universal maps, so uniqueness makes both compositions identities. This identifies $T$ with $S^{\mathrm{\prime }}$ and transports step 3.1 to $T$.