Quiver stack integration is multiplicative with the Euler twist

By [F5], $H_Q$ with its Hall convolution is an associative unital $R$-algebra. Bilinearity of ${\chi }_Q$ gives ${\chi }_Q(\mathbf{d},\mathbf{e})+{\chi }_Q(\mathbf{d}+\mathbf{e},\mathbf{f})={\chi }_Q(\mathbf{e},\mathbf{f})+{\chi }_Q(\mathbf{d},\mathbf{e}+\mathbf{f})$. Hence the two products of $x^{\mathbf{d}},x^{\mathbf{e}},x^{\mathbf{f}}$ have the same coefficient and degree; the zero vector gives the unit. Thus $T_Q$ is also an associative unital $R$-algebra.

Let $\alpha =[X\to {\mathfrak{M}}_{\mathbf{d}}]$ and $\beta =[Y\to {\mathfrak{M}}_{\mathbf{e}}]$ be generators. Their Hall product is the class of the exact-sequence stack over $X\times Y$, pushed to the middle-object stack by [F1]. By [F2], its absolute motive is ${\mathbb{L}}^{-{\chi }_Q(\mathbf{d},\mathbf{e})}[X\times Y]$. Therefore ${\mathcal{I}}_Q(\alpha \ast \beta )={\mathbb{L}}^{-{\chi }_Q(\mathbf{d},\mathbf{e})}[X][Y]x^{\mathbf{d}+\mathbf{e}}={\mathcal{I}}_Q(\alpha ){\mathcal{I}}_Q(\beta )$ by [F3]. The unit from [F5] is the class of ${\mathfrak{M}}_{\mathbf{0}}(Q)=\mathrm{Spec}k$, whose image is $[\mathrm{Spec}k]x^{\mathbf{0}}=x^{\mathbf{0}}$, the target unit.

Isomorphisms, scissor decompositions, and vector-bundle-stack relations in each relative group map to the same relations in $R=\mathcal{K}(\mathrm{Spec}k)$ under $[X\to {\mathfrak{M}}_{\mathbf{d}}]\mapsto [X]x^{\mathbf{d}}$. Thus ${\mathcal{I}}_Q$ is well-defined and additive. For $r\in R$, the coefficient-system product gives ${\mathcal{I}}_Q(r\alpha )=r{\mathcal{I}}_Q(\alpha )$, so it is $R$-linear. Every element of $H_Q$ is a finite sum of homogeneous generators, so bilinearity and step 1.2 extend the multiplication identity to all elements. Together with the unit preservation in step 1.2 and the algebra structures in step 1.1, ${\mathcal{I}}_Q$ is the claimed algebra homomorphism.