The Ringel assignment is a bialgebra homomorphism onto the extended composition subalgebra

The torus generators map to commuting group-like elements with inverse $k_{-e_i}$. By [F3], $k_{e_i}[S_j]k_{-e_i}={\nu }^{(e_i,e_j)}[S_j]={\nu }^{a_{ij}}[S_j]$, which is the Cartan action relation in the Borel presentation.

Fix $i\mathrm{\ne }j$, and put $r=c_{ij}$, $s=c_{ji}$, $t=r+s=-a_{ij}$ and $m=t+1$. The ordinary Serre relation in [F7] is equivalent to the divided-power form ${\sum }_{\mathrm{\ell }=0}^m(-1{)}^{\mathrm{\ell }}[S_i{]}^{(\mathrm{\ell })}\star [S_j]\star [S_i{]}^{(m-\mathrm{\ell })}=0:$ multiply this expression by $[m{]}_{\nu }!$, reindex $r=m-\mathrm{\ell }$, use ${\left(\genfrac{}{}{0px}{}{m}{\mathrm{\ell }}\right)}_{\nu }={\left(\genfrac{}{}{0px}{}{m}{r}\right)}_{\nu }$, and note that the two sign conventions differ by the common factor $(-1{)}^m$.

Since there are no loops, every extension of $S_i$ by $S_i$ splits. For $\mathrm{\ell }\ge 0$, the coefficient of $[S_i^{\oplus \mathrm{\ell }}]$ in the $\mathrm{\ell }$-fold Hall product counts complete flags in ${\mathbb{F}}_q^{\mathrm{\ell }}$, namely $[\mathrm{\ell }{]}_q!$, and the iterated Euler twist is ${\nu }^{\mathrm{\ell }(\mathrm{\ell }-1)\mathrm{/}2}$. The identity $[j{]}_q={\nu }^{j-1}[j{]}_{\nu }$ gives $[S_i{]}^{\mathrm{\ell }}={\nu }^{\mathrm{\ell }(\mathrm{\ell }-1)}[\mathrm{\ell }{]}_{\nu }![S_i^{\oplus \mathrm{\ell }}],[S_i{]}^{(\mathrm{\ell })}={\nu }^{\mathrm{\ell }(\mathrm{\ell }-1)}[S_i^{\oplus \mathrm{\ell }}]\mathrm{.}$ The same formula holds for $n=m-\mathrm{\ell }$.

Consider a representation $M$ of dimension $m{e}_i+e_j$, so $M_j$ is one-dimensional. Define $U_M={\bigcap }_{a:i\to j}\ker (x_a:M_i\to M_j),V_M={\sum }_{b:j\to i}\mathrm{im}(x_b:M_j\to M_i),$ and let $u=\dim U_M$, $w=\dim V_M$. Every composition $M_j\stackrel{x_b}{\to }{M}_i\stackrel{x_a}{\to }{M}_j$ is nilpotent and acts on a one-dimensional space, hence is zero. Therefore $V_M\subseteq U_M$. A flag with successive layers $S_i^{\oplus n},S_j,S_i^{\oplus (m-n)}$ is specified by a subspace $W\subseteq M_i$ of dimension $n$ with $V_M\subseteq W\subseteq U_M$: the first inclusion makes the intermediate object a subrepresentation, and the second makes the bottom object a subrepresentation. Thus the number of such flags is zero unless $V_M\subseteq U_M$, and otherwise it is $\mathrm{\#}\mathrm{Gr}(n-w,u-w{)}_q$.

The flag count for the three Hall factors, together with the twist $⟨\mathrm{\ell }{e}_i,e_j⟩=-\mathrm{\ell }r$, $⟨\mathrm{\ell }{e}_i+e_j,n{e}_i⟩=n(\mathrm{\ell }-s)$ and the two divided-power scalars from step 1.3, shows that the coefficient of $[M]$ in $[S_i{]}^{(\mathrm{\ell })}\star [S_j]\star [S_i{]}^{(n)}$ is zero unless $V_M\subseteq U_M$, and otherwise equals ${\nu }^{-ns-\mathrm{\ell }r+n\mathrm{\ell }+\mathrm{\ell }(\mathrm{\ell }-1)+n(n-1)}\mathrm{\#}\mathrm{Gr}(n-w,u-w{)}_q\mathrm{.}$ This is obtained by first choosing the bottom subspace $W$, then using the exact-sequence Hall coefficients for the middle and top quotients; the displayed pairings are the only cross terms between the three graded layers.

For $a=u-w$ and $b=n-w$, counting ordered linearly independent $b$-tuples and dividing by $\mathrm{\mid }{\mathrm{GL}}_b({\mathbb{F}}_q)\mathrm{\mid }$ gives $\mathrm{\#}\mathrm{Gr}(b,a{)}_q={\prod }_{j=0}^{b-1}\frac{q^a-q^j}{q^b-q^j}={\nu }^{b(a-b)}{\left(\genfrac{}{}{0px}{}{a}{b}\right)}_{\nu }\mathrm{.}$ The second equality follows from the factorial definitions in [F6] and $q={\nu }^2$. Hence the coefficient in step 2.1 can be written ${\nu }^{-ns-\mathrm{\ell }r+n\mathrm{\ell }+\mathrm{\ell }(\mathrm{\ell }-1)+n(n-1)+(u-n)(n-w)}{\left(\genfrac{}{}{0px}{}{u-w}{n-w}\right)}_{\nu }\mathrm{.}$

Put $\mathrm{\ell }=m-n$ and $d=2s+1-u-w$. Expanding the exponent in step 3.1 and using $m=r+s+1$ gives $-ns-\mathrm{\ell }r+n\mathrm{\ell }+\mathrm{\ell }(\mathrm{\ell }-1)+n(n-1)+(u-n)(n-w)=ms-uw-dn\mathrm{.}$ Thus the coefficient of $[M]$ in the divided Serre sum of step 1.2 is zero if $V_M⊈U_M$, and otherwise equals ${\nu }^{ms-uw}(-1{)}^{m-w}{\nu }^{-dw}{\sum }_{k=0}^{u-w}(-1{)}^k{\nu }^{-dk}{\left(\genfrac{}{}{0px}{}{u-w}{k}\right)}_{\nu }\mathrm{.}$ Here $k=n-w$; Gaussian coefficients outside $0\le n-w\le u-w$ vanish.

There are $r$ maps from the $m$-dimensional $M_i$ to the one-dimensional $M_j$, so $u\ge m-r=s+1$. There are $s$ reverse-arrow images, each of dimension at most one, so $w\le s$. Hence $u-w\ge 1$ and $d-(1-(u-w))=2(s-w)\ge 0,(u-w-1)-d=2(u-s-1)\ge 0,$ while $d-(u-w-1)=2(s+1-u)$ is even. Therefore $d$ satisfies the range and parity hypotheses of [F5] with $m$ there replaced by $u-w$. The preceding Gaussian sum vanishes. Every degree-$(m{e}_i+e_j)$ object has zero coefficient, so the divided Serre relation holds in the Hall algebra.

Step 1.1 verifies the torus action, and step 5.1 proves every quantum Serre relation; the Borel presentation [F7] therefore gives an algebra homomorphism with the stated images. By [F4], it intertwines coproducts on generators; since both coproducts are algebra maps, it intertwines them on every product.

On the Borel presentation define ${ϵ}_B(E_i)=0$ and ${ϵ}_B(K_i^{\pm 1})=1$. These assignments respect the torus relations, the Cartan relations (both sides have value zero) and the Serre relations (every monomial has value zero), so define an algebra map. Both counit identities follow on generators from the coproduct formulas in [F7], and extend to all products because the coproduct and ${ϵ}_B$ are algebra maps. For the Hall counit in [F4], the group-like counit identity gives ${ϵ}_H(k_{\pm e_i})=1$; applying ${ϵ}_H\otimes \mathrm{i}\mathrm{d}$ to the simple coproduct then gives ${ϵ}_H([S_i])k_i^{-1}+[S_i]=[S_i]$, hence ${ϵ}_H([S_i])=0$. Thus the assignment preserves counits on generators and hence everywhere. Its image is exactly the subalgebra generated by the simple classes and torus, the extended composition subalgebra. This proves the stated surjective bialgebra homomorphism.