Euler-Twisted Hall Products, Green Coproducts, and Quantum Groups

Substituting ${\nu }^{-1}$ for the parameter in [F1] gives ${\prod }_{j=0}^{m-1}(1+{\nu }^{-2j}z)={\sum }_{n=0}^m{\nu }^{-n(m-1)}{C}_{m,n}{z}^n\mathrm{.}$ This identity is valid at the real number $\nu >1$ because both sides are Laurent polynomials. Set $z=-{\nu }^{m-1-d}$; the right side becomes exactly ${\sum }_{n=0}^m(-1{)}^n{\nu }^{-dn}{C}_{m,n}$.

The parity assumption makes $j=(m-1-d)\mathrm{/}2$ an integer, and the bounds on $d$ give $0\le j\le m-1$. The factor indexed by this $j$ on the left side of step 1.1 is $1-{\nu }^{-2j}{\nu }^{m-1-d}=0$. Therefore the right side, which is the sum in the Statement, is zero.

Call a filling of the displayed two-extension an object $T$ with an exact sequence $0\to B\stackrel{b}{\to }T\stackrel{t}{\to }X\to 0$ and a map $r:T\to D$ such that $rb=jp$ and $qr=t$, for which $0\to A\stackrel{bi}{\to }T\stackrel{r}{\to }D\to 0$ is exact. A lift $G$ in the statement is exactly such a filling: its pushout along $p$ is $T\mathrm{/}A$, and an endpoint-fixed equivalence to $F$ is an isomorphism by [F2]. Conversely the quotient map $r$ identifies that pushout with $F$. Thus it suffices to prove that existence of a filling is invariant under each direction of an endpoint-fixed chain map of two-extensions.

Consider any such chain map with middle maps $u:B\to B^{\mathrm{\prime }}$ and $v:D\to D^{\mathrm{\prime }}$. Factoring the differentials through their images gives $c:C\to C^{\mathrm{\prime }}$ satisfying $ui=i^{\mathrm{\prime }}$, $p^{\mathrm{\prime }}u=cp$, $vj=j^{\mathrm{\prime }}c$, and $q^{\mathrm{\prime }}v=q$. The natural maps $B\to B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C$ and $C^{\mathrm{\prime }}{⨿}_{C}D\to D^{\mathrm{\prime }}$ induce identities on the endpoints of the respective short exact sequences. They are isomorphisms by [F2]. In particular $B=B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C$ and $F^{\mathrm{\prime }}=c_{\ast }F$, up to these specified isomorphisms.

If the unprimed two-extension has a filling $T$, push $0\to B\to T\to X\to 0$ out along $u$ to obtain $0\to B^{\mathrm{\prime }}\to T^{\mathrm{\prime }}\to X\to 0$. The maps $T\stackrel{r}{\to }D\stackrel{v}{\to }{D}^{\mathrm{\prime }}$ and $B^{\mathrm{\prime }}\stackrel{p^{\mathrm{\prime }}}{\to }{C}^{\mathrm{\prime }}\stackrel{j^{\mathrm{\prime }}}{\to }{D}^{\mathrm{\prime }}$ agree on $B$, so induce $r^{\mathrm{\prime }}:T^{\mathrm{\prime }}\to D^{\mathrm{\prime }}$. Quotienting the pushout by the common copy of $A$ gives $T^{\mathrm{\prime }}\mathrm{/}A\cong C^{\mathrm{\prime }}{⨿}_C(T\mathrm{/}A)\cong C^{\mathrm{\prime }}{⨿}_{C}D\cong D^{\mathrm{\prime }}$. These isomorphisms follow directly from the universal property of a cokernel and a pushout: a map out of either side is precisely a compatible pair of maps annihilating $A$. Consequently $\ker r^{\mathrm{\prime }}=A$ and $r^{\mathrm{\prime }}$ is epic. This is a filling of the primed extension.

If instead the primed extension has a filling $T^{\mathrm{\prime }}$, form $T=T^{\mathrm{\prime }}{\times }_{D^{\mathrm{\prime }}}D$ using its quotient $r^{\mathrm{\prime }}:T^{\mathrm{\prime }}\twoheadrightarrow D^{\mathrm{\prime }}$ and $v$. The projection $r:T\to D$ is epic with kernel $A$ by pullback exactness. Set $t=qr$, which is epic. Its kernel is $T^{\mathrm{\prime }}{\times }_{D^{\mathrm{\prime }}}C$, where $C\to D^{\mathrm{\prime }}$ is $j^{\mathrm{\prime }}c$. Since the inverse image of $C^{\mathrm{\prime }}\subset D^{\mathrm{\prime }}$ in $T^{\mathrm{\prime }}$ is $B^{\mathrm{\prime }}$, this kernel is $B^{\mathrm{\prime }}{\times }_{C^{\mathrm{\prime }}}C\cong B$ by step 1.2. The induced $B\to T$ satisfies $rb=jp$ and identifies its copy of $A$ with $\ker r$. Thus $T$ is a filling of the unprimed extension. Every object and morphism used here is a finite pullback or pushout; the argument uses neither resolutions nor a choice principle.

The zero representative $Z$ has filling $A\oplus X$, with its two evident split short exact sequences. If the given splice is zero, [F1] supplies an actual finite zigzag from it to $Z$. Traverse this zigzag backwards starting with that filling, applying step 2.1 when the traversal follows a chain map and step 2.2 when it opposes one. Finite induction produces a filling of the given splice, hence the required $G$. This explicitly supplies the factorization data through every switch, including switches whose intermediate kernels change.

Conversely let a filling $T$ be given. Form $H=B{\times }_{C}B$, with first projection $h:H\to B$ and second projection $s:H\to B$. The sequence $0\to A\stackrel{(i,0)}{\to }H\stackrel{s}{\to }B\to 0$ is split by the diagonal and has a retraction $\rho :H\to A$ specified by $i\rho =h-s$. Splice this split extension with $0\to B\stackrel{b}{\to }T\to X\to 0$ to obtain $L:0\to A\to H\stackrel{bs}{\to }T\to X\to 0$. There is an endpoint-fixed chain map $L\to EF$, with middle maps $h$ and $r$, since $jph=jps=rbs$. There is also an endpoint-fixed chain map $L\to Z$, with middle maps $\rho$ and $t$; the differential square commutes because $tbs=0$. Hence $EF\leftarrow L\to Z$ is a concrete finite zigzag, proving the splice is zero.

Every displayed map has the asserted direction: Hom uses composition; the two maps between short extension classes are $i_{\ast }$ and $p_{\ast }$; $\delta (f)=f^{\ast }E$; and $\mathrm{\partial }[F]=[EF]$. Universal properties send endpoint-fixed maps to endpoint-fixed maps, so pullback, pushout, and splicing respect the generated equivalence relation. These constructions commute with finite direct sums. Pullbacks and pushouts in independent endpoints commute, as is seen by their common universal property. Consequently the maps between short extension classes preserve Baer addition.

Exactness at the first three Hom terms follows from the kernel property of $i$ and the cokernel property of $p$: a morphism $X\to B$ is killed by $p$ exactly when it factors uniquely through $i$. At $\mathrm{Hom}(X,C)$, a map $f$ belongs to $\ker \delta$ exactly when $f^{\ast }E$ splits. A section of that pullback is exactly a map $s:X\to B$ satisfying $ps=f$. Thus $\ker \delta =\mathrm{im}(p\circ -)$.

The connecting maps are additive as well. For $f,g:X\to C$, form the pullback of $E\oplus E$ along $(f,g):X\to C\oplus C$ and then push out along $A\oplus A\stackrel{\mathrm{\nabla }}{\to }A$. Addition $B\oplus B\to B$ induces an endpoint-fixed map from this short extension to $(f+g{)}^{\ast }E$, proving $\delta (f+g)=\delta (f)+\delta (g)$. For the other connecting map, use the two-fold Baer construction: direct sum, diagonal pullback at $X$, codiagonal pushout at $A$. Write $F_0={\mathrm{\Delta }}_X^{\ast }(F\oplus F^{\mathrm{\prime }})$, whose kernel is $C\oplus C$. Then $F+F^{\mathrm{\prime }}={\mathrm{\nabla }}_{C\ast }{F}_0$. Put $H=B{\times }_C(C\oplus C)$, with pullback along ${\mathrm{\nabla }}_C$. The map $B\oplus B\to H$ induced by ${\mathrm{\nabla }}_B$ and $p\oplus p$ induces an isomorphism ${\mathrm{\nabla }}_{A\ast }(E\oplus E)\cong {\mathrm{\nabla }}_C^{\ast }E$ of extensions of $C\oplus C$ by $A$. Finally the defining pushout map $F_0\to {\mathrm{\nabla }}_{C\ast }{F}_0$ and pullback map ${\mathrm{\nabla }}_C^{\ast }E\to E$ give an endpoint-fixed chain map between their splices. These explicit maps identify the two-fold Baer representative of $[EF]+[E{F}^{\mathrm{\prime }}]$ with $[E(F+F^{\mathrm{\prime }})]$. Thus $\mathrm{\partial }$ preserves addition wherever the extension-class collections are sets; the same formulas hold on classes.

Let $H:0\to A\stackrel{a}{\to }T\stackrel{t}{\to }X\to 0$. If $i_{\ast }H$ splits, its middle pushout $P=B{⨿}_{A}T$ admits a retraction to its subobject $B$. Composing $T\to P$ with this retraction gives $h:T\to B$ with $ha=i$. Hence $ph$ factors uniquely through $t$ as $f:X\to C$. The pair $(h,t)$ gives an endpoint-fixed map $H\to f^{\ast }E$, so $[H]=\delta (f)$. Conversely $i_{\ast }E$ splits: on its pushout $B{⨿}_{A}B$, the two identity maps to $B$ induce a retraction onto the first copy of $B$. Pulling back along $f$ shows $i_{\ast }\delta (f)=0$. This proves exactness at ${\mathrm{YExt}}^1(X,A)$.

Let $H:0\to B\stackrel{b}{\to }T\stackrel{t}{\to }X\to 0$ and suppose $p_{\ast }H$ splits. A retraction from its middle pushout to $C$, composed with $T$'s pushout map, yields $h:T\to C$ satisfying $hb=p$. Let $K=\ker h$. The kernel of $t{\mathrm{\mid }}_K$ is $A$, and $t{\mathrm{\mid }}_K:K\to X$ is epic. For the last assertion without elements, if $z:X\to Y$ satisfies $zt{\mathrm{\mid }}_K=0$, then $zt$ factors through $h$, since $h$ is epic and $K=\ker h$. Write $zt=wh$; restricting to $B$ gives $wp=0$, so $w=0$ and then $z=0$. Thus the cokernel of $t{\mathrm{\mid }}_K$ is zero. The resulting extension $0\to A\to K\to X\to 0$ pushes out along $i$ to $H$: the map $B{⨿}_{A}K\to T$ induced by $b$ and the inclusion of $K$ is an endpoint-fixed short-extension isomorphism. Conversely $p_{\ast }{i}_{\ast }{H}^{\mathrm{\prime }}=0$ for every $H^{\mathrm{\prime }}$ by $pi=0$, since pushout along the zero map is split. Therefore exactness holds at ${\mathrm{YExt}}^1(X,B)$.

At ${\mathrm{YExt}}^1(X,C)$, [F3] states precisely that $\mathrm{\partial }[F]=0$ if and only if $[F]$ lies in the image of $p_{\ast }$. This proves every asserted exactness in the covariant sequence, with no assertion of surjectivity onto the final term.

The opposite of an abelian category is abelian: kernels and cokernels interchange, biproducts remain biproducts, and the coimage-image isomorphism dualizes. Reversing all arrows therefore identifies its short and two-fold extensions, endpoint-fixed zigzags, pullbacks, pushouts, and Baer operations with the corresponding dual constructions in the original category. Apply steps 1.1–3.1 in that opposite category to $0\to C\to B\to A\to 0$ and $X$. The resulting covariant sequence is exactly the second displayed sequence; its first connecting map pushes $E$ out along $A\to X$, and its last connecting map splices an extension of $A$ by $X$ on the left of $E$. This proves all its asserted exactness.

Fix orderings of the finite sets $Q_0$, $Q_1$, and $k$. Kernels, images and cokernels of a morphism of representations are formed at each vertex, with the induced arrow maps; the vector-space kernel and cokernel identities imply the abelian-category axioms vertexwise. For each dimension vector, enumerate the finitely many arrow-matrix tuples on the fixed spaces $k^{d_i}$ and retain the lexicographically least tuple in each orbit under ${\prod }_i{\mathrm{GL}}_{d_i}(k)$. Their union ${\mathcal{R}}_Q$ is an explicit small skeleton; no representatives from a proper class are selected.

For a vertex $i$, let $P_i$ have at vertex $j$ the vector space with basis the directed paths $i\to j$, with arrows acting by path extension. Acyclicity and finiteness of $Q$ imply there are finitely many paths, so $P_i$ is finite-dimensional. Evaluation gives a natural isomorphism $\mathrm{Hom}(P_i,W)\cong W_i$; since evaluation is exact, $P_i$ is projective in ${\mathrm{Rep}}_k(Q)$. Its chosen representative in ${\mathcal{R}}_Q$ is projective as well, since the skeleton is equivalent to the full representation category.

The simple $S_i$ is $k$ at vertex $i$ and zero elsewhere. If a representation is simple, choose a sink $i$ in its nonzero support; a one-dimensional subspace at that vertex is a nonzero subrepresentation, so simplicity forces the representation to equal $S_i$. A composition series then shows $[V]={\sum }_i(\dim V_i)[S_i]$ in $K_0$ by [F3]. The dimension-vector map $K_0\to {\mathbb{Z}}^{Q_0}$ sends $[S_i]$ to the standard basis vector; it is inverse to the map sending that vector to $[S_i]$.

Every Hom set is a subset of the finite product ${\prod }_i{\mathrm{Hom}}_k(V_i,W_i)$ and is therefore finite. For fixed endpoints $V,W$, an extension has vertex dimensions $\dim V_i+\dim W_i$; there are only finitely many arrow-matrix collections and endpoint maps in those dimensions. Yoneda classes are a quotient of this finite set, so ${\mathrm{YExt}}^1(V,W)$ is finite.

For a representation $V$, define $P_0(V)={⨁}_i{P}_i{\otimes }_k{V}_i,P_1(V)={⨁}_{a:i\to j}{P}_j{\otimes }_k{V}_i\mathrm{.}$ The augmentation $P_0(V)\to V$ sends a path $p:i\to r$ tensored with $v$ to the image of $v$ under the composite map $V_p$. For $a:i\to j$ and a path $p:j\to r$, the differential sends $p\otimes v$ to $(pa)\otimes v-p\otimes V_a(v)$. These maps commute with every arrow.

In an extension $0\to S_j\to E\to S_i\to 0$, the vertex dimensions are fixed. If $i\mathrm{\ne }j$, all arrow maps vanish except the maps on arrows $i\to j$, each of which is a scalar in $k$. Endpoint-fixed equivalence leaves these scalars unchanged, and every scalar tuple defines an extension. Under Baer sum, the direct sum is pulled back along the diagonal and pushed out along the codiagonal; on each arrow coordinate these two operations add the two scalars. Pushing out along multiplication by $\lambda \in k$ multiplies each coordinate by $\lambda$. If $i=j$, looplessness leaves every arrow map zero and the extension splits. Thus ${\mathrm{YExt}}^1(S_i,S_j)\cong k^{c_{ij}}$ as $k$-spaces.

More generally, in an extension $0\to W\to E\to V\to 0$, choose a vector-space splitting at each vertex. Each arrow map on $E_i=W_i\oplus V_i$ then has block form $\left(\begin{array}{cc}{W}_a& {\xi }_a\\ 0& V_a\end{array}\right)$ with ${\xi }_a:V_i\to W_j$. Changing the vertex splittings by maps ${\theta }_i:V_i\to W_i$ changes the family $({\xi }_a)$ by the image of $\mathrm{\partial }(\theta {)}_a=W_a{\theta }_i-{\theta }_j{V}_a$. Conversely every family $({\xi }_a)$ defines such an extension, and an endpoint-fixed isomorphism changes it by exactly such a splitting change. Hence ${\mathrm{YExt}}^1(V,W)\cong \mathrm{coker}\mathrm{\partial }$, while $\mathrm{Hom}(V,W)=\ker \mathrm{\partial }$ because a morphism is precisely a family of vertex maps commuting with every arrow. Rank–nullity gives $\dim \mathrm{Hom}(V,W)-\dim {\mathrm{YExt}}^1(V,W)={\sum }_i\dim V_i\dim W_i-{\sum }_{a:i\to j}\dim V_i\dim W_j\mathrm{.}$ The right side is bilinear in the two dimension vectors.

At vertex $r$, every positive-length path ending at $r$ has a unique first arrow $a:i\to j$ followed by a path $p:j\to r$. The relation in step 2.2 reduces $(pa)\otimes v$ to $p\otimes V_a(v)$; induction on path length shows the augmentation is surjective and its kernel is generated by those relations. A linear combination of relations has a unique maximal path length; its terms of that length have distinct first-arrow/path/vector indices, so they cannot cancel unless their coefficients vanish. Descending on path length proves the differential is injective. Hence $0⟶P_1(V)⟶P_0(V)⟶V⟶0$ is a projective resolution of length one.

To see directly that every Yoneda two-extension of $V$ by $W$ is zero, write it as $0\to W\to E_1\to E_0\to V\to 0$. Projectivity of $P_0(V)$ gives a lift $s:P_0(V)\to E_0$ of its augmentation. The composite of $P_1(V)\to P_0(V)\stackrel{s}{\to }{E}_0$ lands in $J=\ker (E_0\to V)=\mathrm{im}(E_1\to E_0)$. Projectivity of $P_1(V)$ lifts it through $E_1\twoheadrightarrow J$ to $t:P_1(V)\to E_1$. The resulting map from the split two-extension $0\to W\to W\oplus P_1(V)\to P_0(V)\to V\to 0$ to the given extension is the identity on $W,V$ and commutes with all arrows. The split two-extension is zero in Yoneda ${\mathrm{Ext}}^2$, so the given extension is zero.

Step 1.1 constructs the small abelian model, step 2.1 verifies Hom and extension finiteness, and step 4.1 proves Yoneda ${\mathrm{Ext}}^2=0$. Steps 1.3, 2.3 and 2.4 give the asserted $K_0$, simple extension dimensions and Euler-form formula. Thus all clauses of [F2] hold for ${\mathcal{R}}_Q$, and ${\mathrm{Rep}}_k(Q)$ is essentially small with this finite-field hereditary Hall skeleton.

Kernels and cokernels of representation maps are formed at each vertex; the induced arrow maps satisfy the abelian identities vertexwise. Subrepresentations and quotients of a representation killed by all paths of length $n$ have the same property. If $0\to U\to V\to W\to 0$ is vertexwise exact and all length-$r$ paths kill $U$ while all length-$s$ paths kill $W$, then any length-$s$ path sends $V$ into $U$, and the next $r$ arrows kill it. Thus all length-$(r+s)$ paths kill $V$. Consequently nilpotent finite-dimensional representations are closed under kernels, cokernels, and extensions, and form an abelian $k$-linear category. This quantitative bound applies to every extension constructed below.

Fix an ordering of the finite field and of the finite vertex and arrow sets. For each dimension vector $d$, use the fixed spaces $k^{d_i}$ and list the finitely many arrow-matrix tuples. Keep the nilpotent tuples and the lexicographically least member of each orbit under the finite group ${\prod }_i{\mathrm{GL}}_{d_i}(k)$. Their union over $d\in {\mathbb{N}}^{Q_0}$, with all commuting families of matrices as morphisms, is the supplied small skeleton ${\mathcal{A}}_Q$. Isomorphisms to representatives may also be chosen as the least in the corresponding finite matrix sets, so transporting kernels, cokernels, and biproducts to it needs no infinite choice. We use this small abelian model when applying [F1]. Hom spaces are subspaces of the finite-dimensional vector space ${⨁}_i{\mathrm{Hom}}_k(V_i,W_i)$.

For fixed nilpotent $V,W$, split a short extension vertexwise, using bases in finitely many finite-dimensional vector spaces. Each arrow $a:i\to j$ then acts on $W_i\oplus V_i$ by the block matrix $\left(\begin{array}{cc}{W}_a& {\xi }_a\\ 0& V_a\end{array}\right)$ with ${\xi }_a:V_i\to W_j$. Conversely every tuple $({\xi }_a)$ defines a representation extension because a quiver representation has no relations beyond the specified compositions of arrows. It is nilpotent by step 1.1, independently of the tuple. Changing the vertex sections by maps ${\theta }_i:V_i\to W_i$ changes the tuple by $d(\theta {)}_a=W_a{\theta }_i-{\theta }_j{V}_a$. Every endpoint-fixed isomorphism is exactly such a change of sections. Hence ${\mathrm{YExt}}^1(V,W)$ is identified with $\mathrm{coker}(d:{⨁}_i{\mathrm{Hom}}_k(V_i,W_i)\to {⨁}_{a:i\to j}{\mathrm{Hom}}_k(V_i,W_j))$. Under [F3], Baer sum adds arrow tuples and pushout along $\lambda {\mathrm{id}}_W$ multiplies them by $\lambda$. This proves that the prescribed extension operations give a finite-dimensional $k$-space.

The vertex representations $S_i$ are simple. Every nonzero nilpotent representation has a subobject isomorphic to some $S_i$: among all nonzero images of nonzero vectors under paths choose one of greatest path length, which exists because sufficiently long paths vanish. Its terminal vector is killed by every outgoing arrow, so its one-dimensional span at that vertex and zero at the other vertices is the required subrepresentation. In a simple representation this is the whole object. Total vector-space dimension strictly decreases on a proper quotient, giving a finite composition series with factors among the $S_i$. Counting vertex dimensions along it and using [F3] gives $[V]={\sum }_i{\dim }_k(V_i)[S_i]$. The dimension-vector map is additive on short exact sequences and sends these classes to the standard basis, so it is inverse to the map ${\mathbb{Z}}^{Q_0}\to K_0$ taking that basis to $[S_i]$.

Consider an arbitrary two-extension $0\to A\to B\to D\to X\to 0$ in the nilpotent category. Its image $C\subset D$ gives short extensions $E:0\to A\to B\stackrel{p}{\to }C\to 0$ and $F:0\to C\to D\to X\to 0$. Choose a vector-space section of $D_i\to X_i$ and of $p_i:B_i\to C_i$ at every vertex. There are finitely many choices, so this does not use the Axiom of Choice. In the resulting coordinates $D_i=C_i\oplus X_i$, its arrow maps have form $\left(\begin{array}{cc}{C}_a& {\xi }_a\\ 0& X_a\end{array}\right)$. If $s_j:C_j\to B_j$ is the chosen section of $p_j$, put ${\tilde{\xi }}_a=s_j{\xi }_a$. Define $T_i=B_i\oplus X_i$ and $T_a=\left(\begin{array}{cc}{B}_a& {\tilde{\xi }}_a\\ 0& X_a\end{array}\right)$. These maps define a finite-dimensional quiver representation and an exact sequence $0\to B\to T\to X\to 0$.

If length-$r$ paths kill $B$ and length-$s$ paths kill $X$, then length-$(r+s)$ paths kill $T$ by step 1.1. Thus the constructed intermediate object belongs to the nilpotent category, including when $Q$ has oriented cycles. The maps $T_i\to D_i$ given by $(b,x)\mapsto (p_{i}b,x)$ commute with arrows because $p_j{\tilde{\xi }}_a={\xi }_a$ and $p_j{B}_a=C_a{p}_i$. They give a surjection $T\to D$ with kernel $A$, and the pushout of $0\to B\to T\to X\to 0$ along $p$ is precisely $F$. By [F2] the given two-extension is zero within the nilpotent category itself: the finite zigzag in that lemma involves only finite sums, subquotients, and extensions of these objects, which remain finite-dimensional and nilpotent by step 1.1. Since the original two-extension was arbitrary, all its Yoneda two-extension classes vanish.

Apply step 2.2 to $V=S_i,W=S_j$. If $i\mathrm{\ne }j$, the source of $d$ is zero, and its target has one scalar coordinate for each arrow $i\to j$. If $i=j$, that target is zero because $Q$ has no loops. Thus ${\dim }_k{\mathrm{YExt}}^1(S_i,S_j)=c_{ij}$ in all cases. Steps 1.1–4.1 verify every condition in [F1], while step 2.3 and this computation prove the remaining assertions for the chosen skeleton. If $Q$ is acyclic then every path has length less than $\mathrm{\mid }{Q}_0\mathrm{\mid }$, so all finite-dimensional representations are nilpotent; hence the same argument covers the chosen small skeleton of the acyclic category in the statement, consistently with Finite-dimensional representations of an acyclic quiver form a hereditary finitary Hall category.

Fix $X$. The covariant sequence from [F2] ends $\mathrm{Hom}(X,A)\to \mathrm{Hom}(X,B)\to \mathrm{Hom}(X,C)\to {\mathrm{YExt}}^1(X,A)\to {\mathrm{YExt}}^1(X,B)\to {\mathrm{YExt}}^1(X,C)\to 0,$ because its next term is ${\mathrm{YExt}}^2(X,A)=0$ by [F1]. It is a finite exact sequence of finite-dimensional vector spaces.

Apply the contravariant sequence in [F2]. Its finite exact segment is $\mathrm{Hom}(C,X)\to \mathrm{Hom}(B,X)\to \mathrm{Hom}(A,X)\to {\mathrm{YExt}}^1(C,X)\to {\mathrm{YExt}}^1(B,X)\to {\mathrm{YExt}}^1(A,X)\to 0.$ Alternating dimensions give $⟨[B],[X]⟩=⟨[A],[X]⟩+⟨[C],[X]⟩$.

The alternating-dimension identity for the finite exact sequence of the previous step gives $\dim \mathrm{Hom}(X,B)-\dim {\mathrm{YExt}}^1(X,B)=(\dim \mathrm{Hom}(X,A)-\dim {\mathrm{YExt}}^1(X,A))+(\dim \mathrm{Hom}(X,C)-\dim {\mathrm{YExt}}^1(X,C))\mathrm{.}$ Thus $⟨[X],[B]⟩=⟨[X],[A]⟩+⟨[X],[C]⟩$.

The two additivity identities in steps 2.1 and 1.2 respect every defining relation of $K_0$ in [F3]. The universal property of the quotient therefore extends the formula uniquely and additively in both variables to $K_0(\mathcal{A})\times K_0(\mathcal{A})$. All entries are integers, proving the statement.

Fix a subobject class counted by [F1], represented by $i:N\to R$. Its quotient is a quotient object isomorphic to $M$. The number of isomorphisms $N\to \mathrm{im}(i)$ is $a_N$, and the number of isomorphisms $R\mathrm{/}\mathrm{im}(i)\to M$ is $a_M$. Each pair of isomorphisms determines exactly one endpoint-labelled short exact sequence with this subobject; every such sequence arises from that pair.

Distinct subobject classes give disjoint sets of labelled sequences, because the image of the first map recovers the subobject class. Therefore each counted subobject contributes exactly $a_N{a}_M$ sequences, and summing over the $g_{M,N}^R$ classes proves the formula.

The total set of labelled sequences is a subset of $\mathrm{Hom}(N,R)\times \mathrm{Hom}(R,M)$ and hence finite by [F2]. The count $P_{M,N}^R$ is therefore a well-defined natural number.

For fixed $M,N$, [F1] gives a finite set of isomorphism classes $[L]$ with nonzero Hall coefficient. Multiplication by the single scalar in [F2] does not enlarge that support, so the displayed basis product is a finite linear combination in the direct sum.

Every Hall vector has finite support. Bilinearity therefore expands any product of two vectors into a finite sum of the finite basis products in step 1.1, producing an element of the same direct sum.