The untwisted quiver CoHA is associative and unital

For dimension vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, the left parenthesization first chooses an exact sequence $0\to M_{\mathbf{a}}\to M_{\mathbf{a}+\mathbf{b}}\to M_{\mathbf{b}}\to 0$ and then an exact sequence $0\to M_{\mathbf{a}+\mathbf{b}}\to M_{\mathbf{a}+\mathbf{b}+\mathbf{c}}\to M_{\mathbf{c}}\to 0$. The resulting data are exactly a flag $0\subset M_{\mathbf{a}}\subset M_{\mathbf{a}+\mathbf{b}}\subset M_{\mathbf{a}+\mathbf{b}+\mathbf{c}}$ whose successive quotients have dimensions $\mathbf{a},\mathbf{b},\mathbf{c}$, in that order. This identification is 2-Cartesian on families: the fibre product of ${\mathfrak{E}}_{\mathbf{a}+\mathbf{b},\mathbf{c}}$ with ${\mathfrak{E}}_{\mathbf{a},\mathbf{b}}\times {\mathfrak{M}}_{\mathbf{c}}$ over ${\mathfrak{M}}_{\mathbf{a}+\mathbf{b}}\times {\mathfrak{M}}_{\mathbf{c}}$ includes an isomorphism of the two middle representations. Transporting the subbundle through that isomorphism gives exactly the flag; extracting the two extensions from the flag is inverse on objects and compatible isomorphisms, and commutes with every family base change. Thus proper/smooth base change and composition in [F3] identify the left-associated convolution with $(p_{\mathbf{a},\mathbf{b},\mathbf{c}}{)}_{\ast }(q_{\mathbf{a},\mathbf{b},\mathbf{c}}{)}^{!}$ of the ordered external product on this flag stack.

If either dimension vector is $\mathbf{0}$, the corresponding exact sequence has a unique zero subobject or a unique zero quotient. Hence both maps in the correspondence reduce to the identity on ${\mathfrak{M}}_{\mathbf{d}}$; the class $1$ acts as the identity on either side.

In the right parenthesization, the first exact sequence is $0\to M_{\mathbf{b}}\to M_{\mathbf{b}+\mathbf{c}}\to M_{\mathbf{c}}\to 0$; pulling its subobject back along the quotient $M_{\mathbf{a}+\mathbf{b}+\mathbf{c}}\to M_{\mathbf{a}+\mathbf{b}+\mathbf{c}}\mathrm{/}{M}_{\mathbf{a}}$ gives the unique intermediate subobject $M_{\mathbf{a}+\mathbf{b}}$. Conversely quotienting the two-step flag of step 1.1 by $M_{\mathbf{a}}$ recovers that sequence. Thus the right composite uses the same flag stack. The construction is inverse on families and their isomorphisms, so the right-associated square is 2-Cartesian as well. The actual Cartesian base-change, external-product and composition identities in [F3] make both induced homology maps equal to that single flag pull-push map with its positive complex orientation.

The degree shifts agree: bilinearity of ${\chi }_Q$ gives $-{\chi }_Q(\mathbf{b},\mathbf{a})-{\chi }_Q(\mathbf{c},\mathbf{a}+\mathbf{b})=-{\chi }_Q(\mathbf{c},\mathbf{b})-{\chi }_Q(\mathbf{b}+\mathbf{c},\mathbf{a})\mathrm{.}$ So the equality in step 2.1 is an equality in the same graded component.

Steps 2.1 and 3.1 show that the two parenthesizations agree in the same graded component, while step 1.2 supplies a two-sided unit. Bilinearity extends these identities to the direct sum ${\mathcal{H}}_Q$. AC is used only through [A1].