Proof. By the definition of \(\upcat {v}\) we know \(\sub(V) \subseteq \upcat {v}.\) Note that \(\soc {w_0^k} Q_i\) is supported at vertex \(k\) so it is a submodule of \(Q_k.\) In particular, \[\soc {w_0^k} Q_i \subseteq \soc {w_0^k} Q_k = Q_k.\] Since \(v \in \quotmax\) implies \(v = w_0^k u\) then \[{\soc {v^{-1}} Q_i} \subseteq {\soc {v^{-1}} Q_k}.\] In particular, there is an embedding \(\oplus_i \soc {v^{-1}} Q_i \hookrightarrow V^n.\) ◻

Proof. In the case where \(v = w_0^k\) this follows from [3]. Now suppose by as an inductive hypothesis that \(v = ws_i\) is a length additive expression and the lemma is true for \(w.\) Consider the commutative diagram \[\begin{tikzcd} {M_{w^{-1}[k]}} & {\cmcpro{n-k}} & {\soc {w^{-1}} Q_k} \\ {M_I} & {\cmcpro{n-k}} & {\soc {v^{-1}} Q_k.} \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[shift right, no head, from=1-2, to=2-2] \arrow[shift left, no head, from=1-2, to=2-2] \arrow[two heads, from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\]

The first vertical map is injective by the snake lemma. Likewise the last vertical arrow is surjective by the snake lemma. If \(\soc {v^{-1}} Q_k = \soc {w^{-1}} Q_k\) the snake lemma implies \(I = w^{-1}[k].\) But in this case either \(i \notin w^{-1}[k]\) or \(i, i+1 \in w^{-1}[k]\) since \(\soc {s_iw^{-1}} Q_k = \soc {w^{-1}} Q_k\) and so \(s_i \notin \Soc \soc {w^{-1}} Q_k.\) Therefore \(w^{-1}[k] = v^{-1}[k].\) If \(\soc {v^{-1}} Q_k \neq \soc {w^{-1}} Q_k\) the snake lemma says we have \[M_{w^{-1}[k]} \hookrightarrow M_I \twoheadrightarrow S_i.\] This implies that \(I = (w^{-1}[k]\backslash \{i\}) \cup \{i+1\} = s_i(w^{-1}[k]) = v^{-1}[k].\) ◻

Proof. Given \[M_I^r \hookrightarrow M\] which surjects at vertex \(i\) then this clearly induces an injective and surjective map at vertex \(i.\) In particular the embedding \[M \hookrightarrow M_I^{r}\] that surjects at \(i\) gives an isomorphism \[M_i \cong (M_I^{r})_i.\] Conversely suppose \(f: M_I^r \rightarrow M\) induces an isomorphism at vertex \(i.\) This immediately implies it surjects at vertex \(i\) and since \(\cm(C)\) is closed under submodules and \(f\) is injective at vertex \(i\) then \(f\) is injective at all vertices. This implies \(f\) is an injective map which surjects at vertex \(i.\) ◻

Proof. Consider \(N \in \fa(I,n).\) By the definition of \(\fa(I,n)\) there is a \(r \in \mathbb{N}\) and a morphism \[M_{I(V)}^r \hookrightarrow N\] which surjects at vertex \(n.\) By composing this with an embedding \[N \hookrightarrow \cmcpro{n-k}^r\] that surjects at vertex \(n\) we get an embedding \[M_{I(V)}^r \hookrightarrow \cmcpro{n-k}^r\] which surjects at vertex \(n.\) By the universality of cokernels it follows that we have a surjection \[\begin{tikzcd} & N \\ {M_I^r} & {\cmcpro{n-k}^r} & {V^r} \\ & {\omega_{n-k}(N).} \arrow[hook, from=1-2, to=2-2] \arrow[dashed, hook, from=2-1, to=1-2] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \arrow[two heads, from=2-2, to=3-2] \arrow[dashed, two heads, from=2-3, to=3-2] \end{tikzcd}\] This implies \(\omega_{n-k}(N) \in \fac(V).\) Conversely, given \(X \in \fac(V)\) with quotient \[V^r \twoheadrightarrow X,\] take the kernel \(N_X\) of the composition \(\cmcpro{n-k}^r \twoheadrightarrow V^r \twoheadrightarrow X.\) By the universality of kernels we have the factorisation \[\begin{tikzcd} {M_{I(V)}^r} & {N_X} & {\cmcpro{n-k}^r}. \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-2, to=1-3] \end{tikzcd}\] In particular both embeddings surject at vertex \(n.\) It follows that \(N_{X} \in \fa(I,n)\) and \(\omega_{n-k}(N_X) = X.\) ◻

Proof. Since \(M\) and \(M_I\) are free \(\mathbb{C}[[t]]\)-modules so is \(\Hom_{C}(M_I,M).\) Furthermore its rank is equal to \(\rank M = r\) since \(M_I\) is a rank one module. Let \(f_1,\dotsc,f_r\) be a \(\mathbb{C}[[t]]\)-basis of \(\Hom_{C}(M_I,M).\) It suffices to show that the map \[\rho: \bigoplus_{i=1}^{r} M_I \rightarrow \eta_{M_I}(M)\] given by \(\rho(a_1,\dotsc,a_r) = f_1(a_1) + \dotsb + f_r(a_r)\) is an isomorphism. By the definition of \(\eta_{M_I}(M)\) it is easy to see \(\rho\) is surjective. Since \(\cm(C)\) is closed under submodules and \(\rho\) is a surjective map between modules of the same rank its kernel is a rank zero module in \(\cm(C).\) Therefore \(\rho\) is injective. ◻

Proof. For \(N \in \fa(I,n)\) there is an embedding \[M_{I}^r \hookrightarrow N\] which surjects at vertex \(n.\) We can compose this with an embedding \[N \hookrightarrow \cmcpro{n-k}^r\] which surjects at \(n.\) We see from the diagram \[\begin{tikzcd} {M_{I}^r} & N & {\pi_{I}(N)} \\ {M_{I}^r} & {\cmcpro{n-k}^r} & {V^r} \arrow[hook, from=1-1, to=1-2] \arrow[shift right, no head, from=1-1, to=2-1] \arrow[shift left, no head, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[hook, from=1-2, to=2-2] \arrow[dashed, hook, from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\] that \(\pi_{I}(N)\) embeds into \(V^r.\) Conversely suppose we have \(X \in \sub(V).\) Choose an embedding \[X \hookrightarrow V^r\] and perform the pullback in \(\Mod C\) \[\begin{tikzcd} {M_{I}^r} && {M_{I}^r} \\ \\ {\tilde{X}} && {\cmcpro{n-k}^r} \\ \\ X && {V^r}. \arrow[shift right, no head, from=1-1, to=1-3] \arrow[shift left, no head, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-1, to=5-1] \arrow[two heads, from=3-3, to=5-3] \arrow[hook, from=5-1, to=5-3] \end{tikzcd}\] The pullback exists since \(\Mod C\) is an abelian category. Since \(\tilde{X}\) is a submodule of \(\cmcpro{n-k}^r\) then it is in \(\cm(C)\) as \(\cm(C)\) is closed under kernels in \(\Mod C.\) The injection \(M_I^r \hookrightarrow \tilde{X}\) comes from the snake lemma. Further, since \[M_{I}^r \hookrightarrow \tilde{X}\] surjects at vertex \(n\) then \(\tilde{X} \in \fa(I,n).\) Again, since this embedding surjects at vertex \(n\) we have \(\pi_{I}(\tilde{X}) = X.\) ◻

Proof. We have the diagram \[\begin{tikzcd} {\eta_{M_{I}}(M)} && M && {\pi_{I}(M)} \\ {\eta_{M_{I}}(N)} && N && {\pi_{I}(N).} \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \arrow[shift right, no head, from=1-5, to=2-5] \arrow[hook, from=2-1, to=2-3] \arrow[two heads, from=2-3, to=2-5] \arrow[no head, from=2-5, to=1-5] \end{tikzcd}\] We apply \(\Hom_C(M,-)\) to the bottom sequence to obtain the exact sequence \[0 \rightarrow \Hom_C(M,\eta_{M_{I}}(N)) \rightarrow \Hom_C(M,N) \rightarrow \Hom_C(M,\pi_{I}(N)) \rightarrow 0.\] Here we have use the fact that \(\eta_{M_{I}}(N) \in \mathrm{add}(M_{I})\) and \(\ext^{1}_{C}(M,M_{I}) = 0.\) Using this final surjection and the universality of kernels we recover the commuting diagram \[\begin{tikzcd} {\eta_{M_{I}}(M)} && M && {\pi_{I}(M)} \\ {\eta_{M_{I}}(N)} && N && {\pi_{I}(N).} \arrow["l_1", hook, from=1-1, to=1-3] \arrow["y", from=1-1, to=2-1] \arrow["p_1", two heads, from=1-3, to=1-5] \arrow["w", from=1-3, to=2-3] \arrow[shift right, no head, from=1-5, to=2-5] \arrow["l_2", hook, from=2-1, to=2-3] \arrow["p_2", two heads, from=2-3, to=2-5] \arrow[no head, from=2-5, to=1-5] \end{tikzcd}\] We will now show that the first square is a pullback square. It suffices to show it has the universal property. Suppose we have another commuting square \[\begin{tikzcd} A && M \\ {\eta_{M_{I}}(N)} && N \arrow["a", from=1-1, to=1-3] \arrow["b"', from=1-1, to=2-1] \arrow["w", from=1-3, to=2-3] \arrow["l_2", hook, from=2-1, to=2-3] \end{tikzcd}\] By commutativity we have that \(a: A \rightarrow M\) composed with the quotient \(p_1\) is zero. So by the universality of kernels it follows that there is a unique morphism \(p : A \rightarrow \eta_{M_{I}}(M)\) such that \(a\) factors as \(a = l_1p.\) We observe \[l_2b = wa = wl_1p = l_2yp\] so we obtain universality of the commuting square in the sense of pullbacks.

This pullback square gives rise to an exact sequence \[\eta_{M_{I}}(M) \hookrightarrow M \oplus \eta_{M_{I}}(N) \rightarrow N.\] We claim the last map is actually surjective. Observe that we could also have applied \(\Hom_C(N,-)\) to our first sequence to get the diagram \[\begin{tikzcd} {\eta_{M_I}(M)} && M && {\pi_I(M)} \\ {\eta_{M_I}(N)} && N && {\pi_I(N).} \arrow["{l_1}"', hook, from=1-1, to=1-3] \arrow["y", shift left, from=1-1, to=2-1] \arrow["{p_1}", two heads, from=1-3, to=1-5] \arrow["w", shift left, from=1-3, to=2-3] \arrow[shift right, no head, from=1-5, to=2-5] \arrow[shift left, no head, from=1-5, to=2-5] \arrow["h", shift left, from=2-1, to=1-1] \arrow["{l_2}", from=2-1, to=2-3] \arrow["j", shift left, from=2-3, to=1-3] \arrow["{p_2}"', two heads, from=2-3, to=2-5] \end{tikzcd}\] We can write any element \(n \in N\) as \(n = (n-wj(n)) + wj(n).\) Since \(p_2wj = p_2\) then \((n-wj(n)) \in \ker(p_2) = \mathrm{Im}(l_2).\) This implies we have our surjection and exact sequence \[\begin{tikzcd} {\eta_{M_I}(M)} && {M\oplus \eta_{M_I}(N)} && N. \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \end{tikzcd}\] Now we use the fact that \(\ext^{1}_{C}(N,M_{I}) = 0\) and \(\eta_{M_{I}}(M) \in \mathrm{add}(M_{I})\) to get \[N \oplus \eta_{M_{I}}(M) \cong M \oplus \eta_{M_{I}}(N).\] ◻

Proof. The result will follow by an adaptation of the argument from the proof of Lemma 1.9 of Section 6 in [21]. It follows by applying \(\Hom_C(X,-)\) to the sequence that \[\begin{tikzcd} {\Hom_C(X,N)} && {\Hom_C(X,Y)} && {\Hom_C(X,M)} \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \end{tikzcd}\] is exact. After tensoring with \(X\) over \(\End(X),\) this gives a commutative diagram with exact rows \[\begin{tikzcd} {\Hom_C(X,N)\otimes_{\End(X)} X} & {\Hom_C(X,Y)\otimes_{\End(X)} X}& {\Hom_C(X,M)\otimes_{\End(X)} X} \\ N & Y & M. \arrow[from=1-1, to=1-2] \arrow[from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[from=1-2, to=2-2] \arrow[from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\] since tensoring is right exact. The vertical morphisms here are the canonical morphisms \[\begin{align} \Hom_C(X,N)\otimes_{\End(X)} X &\longrightarrow N \\ (f, x) &\longmapsto f(x). \end{align}\] By applying the snake lemma we get an exact sequence \[\pi_{X} N \rightarrow \pi_{X} Y \twoheadrightarrow \pi_{X} M.\] Since \(N,M \in \overline{F}(X,J)\) then \(\pi_{X} N\) and \(\pi_{X} M\) are not supported at the vertices in \(J.\) This implies \(\pi_{X} Y\) is not supported at the vertices in \(J\) and so \(Y \in \overline{F}(X,J).\) ◻

Proof. For \(X \in \su(I,n)\) there is an isomorphism \(\eta_{M_{I}}(X) \cong M_{I} \otimes_{\mathbb{C}[[t]]} \Hom(M_{I},X)\) by the proof of Lemma 19. Now consider an exact sequence \(X \hookrightarrow Y \twoheadrightarrow Z\) in \(\su(I,n).\) The same argument as in Lemma 25 gives an exact sequence \[\pi_{I}X \hookrightarrow \pi_{I} Y \twoheadrightarrow \pi_{I} Z.\] ◻

Proof. By Proposition 20 we have \(\pi_{I}(\fa(I,n)) = \sub(V).\) Now consider \(X \in \sub(V)\) and take \(N \in \fa(I,n)\) with \(\pi_{I}(N) \cong X.\) If \(N \in \su(I,n)\) we are done in this case. Otherwise we have \[\dim\ext^{1}_{C}(N,M_{I}) \neq 0.\] Since \(M_{I} \in \su(I,n),\) by Lemma 25 applied to a non-split extension \[\begin{tikzcd} {M_{I}} && L && N \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \end{tikzcd}\] all of the terms are in \(\fa(I,n).\) Apply \(\Hom(-,M_{I})\) to get \[\begin{tikzcd} {\Hom(L,M_{I})} & {\mathrm{End}(M_{I})} & {\ext^1(N,M_{I})} & {\ext^1(L,M_{I})} \arrow["f", from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[two heads, from=1-3, to=1-4] \end{tikzcd}\] Since we have taken a non-split extension \(f\) cannot surject. This implies \(\dim\ext^{1}(L,M_{I}) < \dim\ext^1(N,M_{I}).\) We see from the diagram

\[\begin{tikzcd} {\Hom(M_{I},M_{I})\otimes_{\mathbb{C}[[t]]}M_{I}} & {\Hom(M_{I},L)\otimes_{\mathbb{C}[[t]]}M_{I}} & {\Hom(M_{I},N)\otimes_{\mathbb{C}[[t]]}M_{I}} \\ {M_{I}} & L & N \\ 0 & {\pi_{I}(L)} & {\pi_{I}(N)} \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[hook, from=1-2, to=2-2] \arrow[hook, from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-2, to=2-3] \arrow[two heads, from=2-2, to=3-2] \arrow[two heads, from=2-3, to=3-3] \arrow[hook, from=3-1, to=3-2] \arrow[two heads, from=3-2, to=3-3] \end{tikzcd}\] with exact rows and columns that \(\pi_{I}(L) \cong \pi_{I}(N) \cong X.\) Since \(\dim\ext^1(N,M_{I})\) is finite we iterate this process to find a lift in \(\su(I,n).\) ◻

Proof. For \(t \in [n]\) then if \(I \leq J\) and \(J \cap [t] = \{J_1, \dots, J_l\}\) then since \(I_l \leq J_l\) it follows that \(\{I_1,\dots, I_l\} \subseteq I\cap[t]\) so \(|i: i \in I \cap [t]| \geq |i: i \in J \cap [t]|.\) Conversely suppose \(|i: i \in I \cap [t]| \geq |i: i \in J \cap [t]|\) for all \(t \in [n].\) Then for \(t = J_l\) since \(J_l \in [t]\) it follows that \(I_l \in [t]\) so \(I_l \leq J_l.\) ◻

Proof. We will prove the above lemma in the case where \(i = n\) in order to simplify notation. The cases with \(i \neq n\) can be proved in the same way but require more care with notation. As noted above, a basis for \(\Hom_C(M_I,M_J)\) is given by \(t^\alpha = (t^{\alpha_i})_{i \in [n]}\) where \[\alpha_{h(a)} - \alpha_{t(a)} = \begin{cases} 1 & \text{ if } a \in I/J, \\ -1 & \text{ if } a \in J/I, \\ 0 & \text{ otherwise.} \end{cases}\] Notice that \[\displaystyle\sum_{1 \leq i \leq t} (\alpha_i - \alpha_{i-1}) = |I \cap [t]| - | J\cap [t]|.\] Consider a pair of embeddings \[(t^{\alpha_j})_{j \in [n]}: \cmcpro n \hookrightarrow M_I,\] \[(t^{\beta_j})_{j \in [n]}: \cmcpro n \hookrightarrow M_J\] which surjects at vertex \(n.\) This implies \(\alpha_n = \beta_n = 0\) so for \(t \in [n]\) we see \[\begin{align} \alpha_t &= \alpha_t - \alpha_n \\ &= \sum_{1 \leq i \leq t} (\alpha_i - \alpha_{i-1}). \end{align}\] Since \(\cmcpro n = M_{[k]}\) it follows that \[\alpha_t = |[k] \cap [t]| - |I\cap [t]|.\] Likewise \[\beta_t = |[k]\cap [t]| - |J \cap [t]|.\] Therefore \[\beta_t - \alpha_t = |I \cap [t]| - |J \cap[t]|.\] Since \(J \geq I\) then we see \(\beta_t \geq \alpha_t\) for all \(t \in [n]\) by Lemma 28. In particular the map \[(t^{\beta_j})_{j \in [n]}: \cmcpro n \hookrightarrow M_J\] factorises via \[(t^{\alpha_j})_{j \in [n]}: \cmcpro n \hookrightarrow M_I.\]

We now show equivalence of the second and third conditions. First suppose \(\cmcpro n \hookrightarrow M_J\) factors via \(M_I.\) By [3] \(\pi_n\) is the quotient by \(\cmcpro n\) so it immediately follows that \(\pi_n(M_I) \hookrightarrow \pi_n(M_J).\) Now suppose there is an embedding \(\pi_n(M_I) \hookrightarrow \pi_n(M_J).\) This gives an embedding \(M_I \hookrightarrow M_J\) which doesn’t factor via \(\cmcpro n\) but surjects at vertex \(n.\) Consider an embedding \[\cmcpro n \hookrightarrow M_I\] which surjects at \(n\) and compose with the above embedding to get an embedding \[\cmcpro n \hookrightarrow M_J.\] Since both embeddings surject at vertex \(n\) this must also surject at \(n.\) A morphism \(\cmcpro n \rightarrow M_J\) which surjects at vertex \(n\) is then a \(\mathbb{C}[[t]]\)-generator for \(\Hom(\cmcpro n ,M_J).\)

Finally, we show that the second and fourth conditions are equivalent. If the embedding \(\cmcpro n \hookrightarrow M_J\) surjecting at \(n\) factors via \(M_I\) then since the embedding of \(\cmcpro n\) surjects at \(n\) then so must the morphism \(M_I \hookrightarrow M_J.\) Now suppose we have a morphism \[(t^{\alpha_j})_{j \in [n]} : M_I \hookrightarrow M_J\] which surjects at vertex \(n.\) In particular this means the map \(t^{\alpha_n} : e_nM_I \cong \mathbb{C}[[t]] \rightarrow e_nM_J \cong \mathbb{C}[[t]]\) must have \(\alpha_n = 0.\) Now consider \[(t^{\beta_j})_{j \in [n]} : \cmcpro n \hookrightarrow M_I\] an embedding surjecting at \(n.\) This also surjects at \(n\) so \(\beta_n = 0.\) The composition gives a morphism \[(t^{\gamma_j})_{j \in [n]}: \cmcpro j \hookrightarrow M_J\] with \(\gamma_n = 0.\) In particular this also surjects at \(n.\) This again gives a generator for \(\Hom(\cmcpro n, M_J).\) ◻

Proof. By Lemma 3.4 of [5] we have \(\ker(Q_k \twoheadrightarrow \soc {v^{-1}} Q_k) = \head {v^{-1}w_0} Q_k.\) The result then follows from the commuting diagram \[\begin{tikzcd} 0 & {M_{I}} & {\cmcpro{n-k}} & V & 0 \\ 0 & {\pi_n(M_{I})} & {Q_k} & {\soc {v^{-1}} Q_k} & 0. \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[two heads, from=1-2, to=2-2] \arrow[from=1-3, to=1-4] \arrow[two heads, from=1-3, to=2-3] \arrow[from=1-4, to=1-5] \arrow[shift right, no head, from=1-4, to=2-4] \arrow[shift left, no head, from=1-4, to=2-4] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \arrow[from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \end{tikzcd}\] ◻

Proof. Since \(J \geq I\) there is a map \(M_{I} \hookrightarrow M_J\) which surjects at vertex \(n\) by Lemma 29. There is also a map \[\tilde{P}_n \hookrightarrow M_{I}\] which surjects at \(n.\) We can compose these maps to give the commuting diagram with exact rows \[\begin{tikzcd} {\tilde{P}_n} & {\tilde{P}_n} & 0 \\ {M_I} & {M_J} & {\pi_{I}(M_J).} \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[hook, from=1-2, to=2-2] \arrow[from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\] Applying the snake lemma gives us the exact sequence \[Y \hookrightarrow \pi_n(M_J) \twoheadrightarrow \pi_{I}(M_J).\] ◻

Proof. In [5] it is shown that for \(X \in \Mod \propi\) there is a surjection \[X/t_{v}(X) \twoheadrightarrow \soc {v^{-1}}(X).\] Apply this to \(\overline{Q}_j\) and take the exact sequence \[Z \hookrightarrow \overline{Q}_j/t_{v}(\overline{Q}_j) \twoheadrightarrow \soc {v^{-1}}(\overline{Q}_j) = \soc {v^{-1}}(Q_j).\] Since the middle term is in \(\upcat v\) and \(\upcat v\) is closed under submodules \(Z \in \upcat v.\) We know \(\soc {v^{-1}}(Q_j)\) is projective injective in \(\upcat v\) so \[\overline{Q}_j/t_{v}(\overline{Q}_j) \cong \soc {v^{-1}}(Q_j) \oplus Z.\] But \(\overline{Q}_j/t_{v}(\overline{Q}_j)\) has a simple top as does \(\soc {v^{-1}}(Q_j)\) so it follows \(\Top(Z) = 0\) and therefore \(Z = 0.\) ◻

Proof. Firstly observe that \(t_v(\pi_n(\tilde{P}_{j}))\) is a submodule of \(\pi_n(\tilde{P}_{j})\) by definition. To see that \(t_v(\pi_n(\tilde{P}_{j}))\) is a submodule of \(Y,\) notice that by [5] we have an exact sequence \[\head {v^{-1}w_0} Q_k \hookrightarrow Q_k \twoheadrightarrow \soc {v^{-1}} Q_k.\] We have \(t_v(\pi_n(\tilde{P}_{j})) \hookrightarrow Q_k\) but \(t_v(\pi_n(\tilde{P}_{j})) \in \downcat v\) and \((\downcat v, \upcat v)\) is a torsion pair so this map must factor via the kernel \(\head {v^{-1}w_0} Q_k = Y.\)

To see that \(Y \cap \pi_n(\tilde{P}_{j}) \subseteq t_v(\pi_n(\tilde{P}_{j}))\) we observe that \(t_v(\pi_n(\tilde{P}_{j}))\) is the kernel of the quotient \[f \colon \overline{Q}_{n-j} \twoheadrightarrow \soc {v^{-1}} \overline{Q}_{n-j}\] by Lemma 32. Since \(Y \in \downcat v\) then \(\soc {v^{-1}} Y = 0\) by [5] so it follows that \(Y \cap \pi_n(\tilde{P}_{j})\) is in the kernel of \(f.\) ◻

Proof. Consider the following pushout in \(\Mod C\) \[\begin{tikzcd} {t_v(\pi_n(\cmcpro j))} && Y \\ \\ {\pi_n(\cmcpro j)} && {Z.} \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[hook, from=3-1, to=3-3] \end{tikzcd}\]

This gives us an exact sequence \[t_v(\pi_n(\cmcpro j)) \hookrightarrow Y \oplus \pi_n(\cmcpro j) \twoheadrightarrow Z.\] Universality of the push out gives a morphism \(f\) such that the following diagram commutes \[\begin{tikzcd} {t_v(\pi_n(\cmcpro j))} & {\pi_n(\cmcpro j)\oplus Y} & Z \\ {Q_k} & {Q_k^{\oplus 2}} & {Q_k.} \arrow[hook, from=1-1, to=1-2] \arrow["{i_1}", hook, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow["{i_2}", hook, from=1-2, to=2-2] \arrow["f", from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\] Since \(t_v(\pi_n(\cmcpro j)) = Y \cap \pi_n(\cmcpro j)\) then \(\mathrm{coker}(i_1) \hookrightarrow \mathrm{coker}(i_2)\) so by the snake lemma \(f\) is injective and \(Z \subseteq Q_k.\)

In particular there is a rank one module \(M_{L_j}\) with \(Z = \pi_n (M_{L_j}).\) The lift of the embedding \(Y \hookrightarrow Z\) implies that there is an embedding \(M_{I} \hookrightarrow M_{L_j}\) which surjects at vertex \(n.\) Therefore \(M_{L_j} \in \fa(I,n).\)

By extending the pushout square and using Lemma 32 we see \[\begin{tikzcd} {t_v(\pi_n(\cmcpro j))} && Y \\ \\ {\pi_n(\cmcpro j)} && {\pi_n(M_{L_j})} \\ \\ {\soc {v^{-1}} {Q}_{n-j}} && {\pi_Y \pi_n (M_{L_j}).} \arrow[hook, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-1, to=5-1] \arrow[two heads, from=3-3, to=5-3] \arrow[shift right, no head, from=5-1, to=5-3] \arrow[shift left, no head, from=5-1, to=5-3] \end{tikzcd}\] Since \[M_{I} \hookrightarrow M_{L_j}\] surjects at vertex \(n\) we have \({L_j} \geq I\) by Lemma 29. We can then use Proposition 31 to see \[\pi_{I}(M_{L_j}) = \pi_Y(\pi_n(M_{L_j})).\] ◻

Proof. To see that \(L_j = I(j+1)\) we can use Lemma 29 and prove that for any \(T \geq I\) there is a morphism \[M_{L_j} \hookrightarrow M_T\] that surjects at vertex \(j.\)

Consider \(M_T\) with \(T \geq I.\) Consider an embedding \[\phi: \cmcpro j \hookrightarrow M_T\] that surjects at \(j\) and \(M_R\) the minimal lift of \(t_v(\pi_n(\cmcpro j))\) via \(\pi_n\) we can form the following commuting diagram: \[\begin{tikzcd} {M_R} && {M_{I}} \\ &&& {M_T} \\ {\cmcpro j} && {M_T} \arrow["f", from=1-1, to=1-3] \arrow["h"', from=1-1, to=3-1] \arrow["g", from=1-3, to=2-4] \arrow["{t^l}", from=2-4, to=3-3] \arrow["\phi"', from=3-1, to=3-3] \end{tikzcd}\] where \(f, g\) and \(h\) all surject at vertex \(n.\) To see that such a diagram exists note that \(f,h\) are lifts of the morphisms in the pushout diagram defining \({L_j}.\) Since \(gf\) surjects at \(n\) and \(\Hom(M_R,M_T) \cong \mathbb{C}[[t]]\) there exists \(l \in \mathbb{N}\) such that \(\phi h = t^l(gf).\)

We lift the exact sequence \[\begin{tikzcd} 0 & {\pi_n(M_R)} & {\pi_n(\cmcpro j)\oplus Y} & {\pi_n(M_{L_j})} & 0 \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] via \(\pi_n\) to get an exact sequence \[\begin{tikzcd} 0 & {M_R} & {\cmcpro j\oplus M_{I}} & {M_{L_j}} & 0. \arrow[from=1-1, to=1-2] \arrow["\left(\begin{matrix} f \\ h \end{matrix}\right)", from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\]

By the universality of cokernels we get a commuting diagram

\[\begin{tikzcd} 0 & {M_R} & { M_{I}\oplus \cmcpro j} & {M_{L_j}} & 0. \\ && {M_{T}} \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow["{(-t^lg, \phi)}"', from=1-3, to=2-3] \arrow[from=1-4, to=1-5] \arrow[from=1-4, to=2-3] \end{tikzcd}\]

Since \(\phi\) surjects at vertex \(j\) then \(M_{L_j} \rightarrow M_{T}\) surjects at vertex \(j.\) This ensures that \[I(j) = \min_{\leq_j} \{T:T\geq I\}\] exists and \(L_j = I(j+1).\) That is to say, there is a unique minimal element in \(\{T: T \geq I\}\) with respect to the order \(\leq_j\) and this unique minimal element coincides with \(L_j.\) ◻

Proof. We know \(M_{I(j+1)}\) is in \(\fa(I,n).\) We need to show that \(\ext^1(M_{I(j+1)},M) = 0\) for all \(M \in \su(I,n).\) Since \(M_{I} \in \su(I,n)\) this proves both conditions at once. Consider the exact sequence \[M_R \hookrightarrow \tilde{P}_{j}\oplus M_{I} \twoheadrightarrow M_{I(j+1)}\] produced in the proof of Proposition 34. Apply \(\Hom(-,M)\) for \(M \in \su(I,n).\) This gives us an exact sequence \[\Hom(\tilde{P}_j\oplus M_{I},M) \rightarrow \Hom(M_R,M) \twoheadrightarrow \ext^{1}(M_{I(j+1)},M).\] We now show that \(\Hom(\tilde{P}_j\oplus M_{I},M) \rightarrow \Hom(M_R,M)\) is surjective. Let \(t = \rank M\) and \(g_1,\cdots,g_t\) be a basis of \(\Hom(M_{I},M).\) Since the morphism \(f:M_R \hookrightarrow M_{I}\) surjects at vertex \(n\) then \[(g_1f,\cdots,g_tf):M_R^{\oplus t} \rightarrow M\] surjects at vertex \(n\) and factors via \(M_{I}^{\oplus t}.\) This implies that \(\Hom(\tilde{P}_j\oplus M_{I},M) \rightarrow \Hom(M_R,M)\) surjects. Therefore \(\ext^1(M_{I(j+1)},M) = 0.\) ◻

Proof. This proof is essentially the same as the proof of Theorem 3 in [20] however we will reproduce the details here for the sake of completeness. By Proposition 22 we know that the induced functor \[\pi_{I} : \su(I,n)/\langle M_{I}\rangle \rightarrow \upcat v\] is a bijection between isoclasses of indecomposables. So we need to show that this restricted functor is full and faithful. Suppose we have a morphism \[f: M_1 \rightarrow M_2\] such that \(\pi_{I}(f) = 0.\) This implies that \(Im(f) \subseteq \eta_{M_{I}}M_2.\) In particular it factors via \(\mathrm{add}(M_{I})\) and so \(\overline{f} = 0\) in \(\su(I,n)/\langle M_{I}\rangle.\) Now consider \(h: X \rightarrow Y\) in \(\upcat v.\) Let \(\widetilde{X} \in \su(I,n)\) be the minimal lift of \(X.\) By applying \(\Hom(\widetilde{X},-)\) to the exact sequence \[\begin{tikzcd} {\eta_{M_{I}}\widetilde{Y}} & {\widetilde{Y}} & Y \arrow[hook, from=1-1, to=1-2] \arrow[two heads, from=1-2, to=1-3] \end{tikzcd}\] we see, using \(\ext^{1}_{C}(\widetilde{X},M_{I}) = 0\), that there is a morphism \(\tilde{h}\) making the following diagram commute \[\begin{tikzcd} {\eta_{M_{I}}\widetilde{X}} & {\widetilde{X}} & X \\ {\eta_{M_{I}}\widetilde{Y}} & {\widetilde{Y}} & Y. \arrow[hook, from=1-1, to=1-2] \arrow[from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow["{\tilde{h}}"', from=1-2, to=2-2] \arrow["h", from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-2, to=2-3] \end{tikzcd}\] In particular \(\pi_{I}(\tilde{h}) = h.\) ◻

Proof. If \(M \in \cm(B_I)\) there is a morphism \(f: B_I^r \rightarrow M\) that surjects. In particular, it surjects at vertex \(n.\) Since \(I(j) \geq I,\) by Lemma 29 there is an embedding \[g: M_{I}^t\rightarrow B_I^r\] that surjects at vertex \(n.\) The composition gives a morphism \[fg: M_{I}^t \rightarrow M\] that surjects at vertex \(n.\) In particular we have \(M \in \fa(I,n).\) Conversely, suppose we have \(M \in \fa(I,n).\) Let \[P_M \twoheadrightarrow \pi_{I}M\] be the projective cover of \(\pi_{I}M \in \upcat v.\) By adding in some extra projective summands where needed, we can use Lemma 34 to recover a surjection \[\pi_{I}(B_I^r) \twoheadrightarrow \pi_{I}M.\] Since \(\pi_{I}(B_I)\) is a quotient of \(B_I\) and \(\ext^{1}(B_I,M_{I}) = 0\) we can use the same argument as in Lemma 1 applied to the sequence \[0 \rightarrow \eta_{M_{I}}M \rightarrow M \rightarrow \pi_{I}(M) \rightarrow 0,\] to see that the canonical morphism \[\Hom_C(B_I,M)\otimes_{\mathbb{C}[[t]]}B_I \rightarrow M\] is surjective. In particular we have \(M \in \cm(B_I).\)

For \(M \in \gp(B_I),\) since \(\ext^1(M,B_I) = 0\) then by additivity we have \(\ext^{1}(M,M_{I}) = 0.\) We see \(M \in \su(I,n).\)

To see that \(M \in \gp(B_I)\) we observe from Proposition 36 that for \(M \in \fa(I,n)\) we have \[\ext^{1}_C(M,M_{I}) = 0 \Leftrightarrow \ext^1_C(M,B_I) = 0.\] ◻

Proof. We already know the \(M_{I(j)}\) satisfy these conditions. The equivalence with \(\gp(B_I)\) tells us there are not any more projective-injective indecomposables. ◻

Proof. This follows the proof of Corollary 4.6 in [3]. Since we know that \(\su(I,n)\) is Frobenius then, due to a theorem of Happel [22], its stable category is triangulated. The functor \(\pi_{I}\) is exact by Proposition 26, maps projectives to projectives (see Proposition 35, Corollary 3), and is quotienting by a projective (see Proposition 37). In particular this implies it induces a triangle equivalence between the stable categories. ◻

Proof. Using the above endofunctors this follows from the case where \(i = n.\) ◻

Proof. Firstly to see \(M_I^{\vee} = M_{I^{c}}\) we note that the duality \((-)^{\vee}\) swaps the \(x\) and \(y\) arrows in the underlying quiver. The label of a rank one module is determined by the \(x\) arrows so the label of \(M_I^{\vee}\) are determined by the \(y\) arrows. This gives us the label \(I^{c}.\) We recall from Lemma 14 that the categories \(\su(I^c,i)\) and \(\opsu(I,i)\) can be characterised in terms of maps inducing isomorphisms at vertices. In particular the embedding \[M \hookrightarrow M_I^{r}\] that surjects at \(i\) gives an isomorphism \[e_iM \cong e_i(M_I^{r}).\] This gives a map \[(M_I^{r})^{\vee} = M_{I^c}^{r} \rightarrow M^{\vee}\] which restricts to an isomorphism \[e_i(M_{I^c}^{r}) \cong e_iM^{\vee}.\] As in the proof of Lemma 14 this gives an embedding which surjects at vertex \(i.\) In particular we see that \((\opfa(I,i))^{\vee} \cong (\fa(I^{c},i))^{op}.\) Finally, using the duality again we see \[\ext^{1}_{C_{k,n}}(M,M_I) = 0 \Leftrightarrow \ext^{1}_{C_{n-k,n}}(M_{I^c}, M^{\vee}) = 0.\] ◻

Proof. We will use the relation between \(\opsu(I,i)\) and \(\su(I^c,i).\) Notice that the functor \(\pi_{I^c}:\su(I^c,i) \rightarrow \Mod \propi^{op}\) has essential image \(\upcat v \subseteq \Mod \propi^{op}\) where \(v \in \left(W^{n-k}\backslash W\right)^{max}\) such that \(v^{-1}[n-k] = I^c.\) We use the diagram \[\begin{tikzcd} {\opsu(I,i)/\langle M_I\rangle} &&& {\su(I^{c},i)^{op}/\langle M_{I^c}\rangle} \\ \\ {\fac(D(V))} &&& {(\upcat v )^{op} = \sub(V)^{op}} \arrow["{(-)^{\vee}}", from=1-1, to=1-4] \arrow[dashed, from=1-1, to=3-1] \arrow["{\pi_{I^c}}", from=1-4, to=3-4] \arrow["{D = \Hom(-,\mathbb{C})}", from=3-4, to=3-1] \end{tikzcd}\] where each arrow is an equivalence. We note that the top and bottom arrows in this diagram would typically be considered dualities without passing to the opposite categories. In order to consider everything as covariant equivalences we have passed to the opposite categories on the right and consider the top and bottom arrows to be equivalences rather than dualities. The \(\mathbb{C}\)-linear dual gives an equivalence from \(\sub(V)^{op}\) to \(\fac(D(V))\) since \(\propi\) is finite dimensional and \(\Mod \mathbb{C}\) is a semisimple category. It is worth explaining where each of these categories lives. We have \(\opsu(I,i) \subseteq \cm(C_{k,n})\) and \(\su(I^c,i) \subseteq \cm(C_{n-k,n}).\) The category \(\left(\upcat v\right)^{op} \subseteq \left(\Mod \propi^{op}\right)^{op}\) so \(\fac(D(V)) \subseteq \Mod \propi.\) We observe that \(D(V)\) has a simple socle at vertex \(i-k\) since \(V\) has a simple top at vertex \(i-k.\) This means there is a \(w \in \quotmin\) such that \[D(V) = \emb i (\head {w^{-1}w_0} Q_k).\] More precisely \(w\) is the unique element of \(\quotmin\) such that \(w^{-1}w_0[k] = z_i^{-1}I.\) Further, the dual version of Lemma 9 tells us that \(\fac(D(V)) = \downcat w .\) The triangle equivalence follows from the same argument as for Theorem 41. ◻

Proof. The second result follows immediately from the first. Since \(M_J \in \fa(I,j)\) there is an embedding \(M_I^{\rank M} \hookrightarrow M_J^{\rank M}\) that surjects at vertex \(j.\) It follows that for \(M \in \rij(I,J)\) then \(M \in \fa(I,j)\) and \(M \in \opfa(J,j).\)

Conversely if \(M \in \fa(I,j) \cap \opfa(J,j)\) then the composition \[M_I^{\rank M} \hookrightarrow M \hookrightarrow M_J^{\rank M}\] surjects at vertex \(j\) and so it is a \(\mathbb{C}[[t]]\)-generator for \(\Hom(M_I,M_J)\) and \(M \in \rij(I,J).\) ◻

Proof. First observe that since \[M_I \hookrightarrow M_J\] surjects at vertex \(j,\) \[\calrij(I,J) \subseteq \su(I,j)\] by Lemma 49. By Theorem 43 we know that \[\pi_I: \su(I,j) \rightarrow \emb {j}(\upcat {v_I})\] is a quotient by \(M_I\) and induces an equivalence of extension groups. Using this we see that \(\calrij(I,J)\) is sent to \[\{U \in \sub(\pi_I(M_J)): \ext^{1}(U,\pi_I(M_J)) = 0\} \subseteq \upcat {v_I}.\] It therefore suffices to prove that \[\{U \in \sub(\pi_I(M_J)): \ext^{1}(U,\pi_I(M_J)) = 0\} = \lec {v_I} {w_J}.\] Since \((\downcat w, \upcat w)\) is a torsion pair we can make identifications: \[\begin{align} \lec {v_I} {w_J} &= \upcat {v_I} \cap \downcat {w_J} \\ &= \{U \in \upcat {v_I}: \Hom(U, \upcat {w_J}) = 0\} \\ &= \{U \in \sub(\pi_I(\cmcpro{j-k})) : \Hom(U,\pi_J(\cmcpro{j-k})) = 0\}. \end{align}\] We will now show \[\{U \in \sub(\pi_I(M_J)): \ext^{1}(U,\pi_I(M_J)) = 0\} = \{U \in \sub(\pi_I(\cmcpro{j-k})) : \Hom(U,\pi_J(\cmcpro{j-k})) = 0\}.\] First, observe that by taking the exact sequence \[M_J \hookrightarrow \cmcpro{j-k} \twoheadrightarrow \pi_J(\cmcpro{j-k})\] and pushing out via the morphism \[M_J \twoheadrightarrow \pi_I(M_J)\] we get a commuting diagram \[\begin{tikzcd} {M_I} && {M_I} \\ \\ {M_J} && {\cmcpro{j-k}} && {\pi_J(\cmcpro{j-k})} \\ \\ {\pi_I(M_J)} && {\pi_I(\cmcpro{j-k})} && {\pi_J(\cmcpro{j-k}).} \arrow[shift right, no head, from=1-1, to=1-3] \arrow[shift left, no head, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-1, to=5-1] \arrow[two heads, from=3-3, to=3-5] \arrow[two heads, from=3-3, to=5-3] \arrow[shift right, no head, from=3-5, to=5-5] \arrow[shift left, no head, from=3-5, to=5-5] \arrow[hook, from=5-1, to=5-3] \arrow[two heads, from=5-3, to=5-5] \end{tikzcd}\] This gives us the exact sequence \[\pi_I(M_J) \hookrightarrow \pi_I(\cmcpro{j-k}) \twoheadrightarrow \pi_J(\cmcpro{j-k}).\] Take \(U \in \{U \in \sub(\pi_I(M_J)): \ext^{1}(U,\pi_I(M_J)) = 0\}.\) By applying \(\Hom(U,-)\) to this exact sequence we get \[\Hom(U,\pi_I(M_J)) \hookrightarrow \Hom(U,\pi_I(\cmcpro{j-k})) \twoheadrightarrow \Hom(U,\pi_J(\cmcpro{j-k})).\] Since for \(M \in \rij(I,J)\) we know that any morphism \[M \rightarrow \cmcpro{j-k}^{\rank M}\] factors via \(M_J^{\rank M}\) it follows that any morphism \[\pi_I(M) \rightarrow \pi_I(\cmcpro{j-k}^{\rank M})\] factors via the embedding \[\pi_I(M_J^{\rank M}) \hookrightarrow \pi_I(\cmcpro{j-k}^{\rank M}).\] When applied to the exact sequence \[\Hom(U,\pi_I(M_J)^{\rank M}) \hookrightarrow \Hom(U,\pi_I(\cmcpro{j-k})^{\rank M}) \twoheadrightarrow \Hom(U,\pi_J(\cmcpro{j-k})^{\rank M})\] this implies the first map is a surjection so \(\Hom(U,\pi_J(\cmcpro{j-k})) = 0.\) Conversely take \(U \in \{U \in \sub(\pi_I(\cmcpro{j-k})) : \Hom(U,\pi_J(\cmcpro{j-k})) = 0\}.\) Since \[\Hom(U,\pi_J(\cmcpro{j-k})) = 0\] we see from the universality of kernels applied to \[\begin{tikzcd} && U \\ \\ {\pi_I(M_J)^r} && {\pi_I(\cmcpro{j-k})^r} && {\pi_J(\cmcpro{j-k})^r} \arrow[dotted, hook, from=1-3, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[dashed, from=1-3, to=3-5] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-3, to=3-5] \end{tikzcd}\] that \(U \in \sub(\pi_I(M_J)).\) Again applying \(\Hom(U,-)\) to the same exact sequence before we get \[\ext^{1}(U,\pi_I(M_J)) \hookrightarrow \ext^{1}(U,\pi_I(\cmcpro{j-k})).\] Since \(\pi_I(\cmcpro{j-k})\) is injective in \(\upcat {v_I}\) then \(\ext^{1}(U,\pi_I(M_J)) = 0.\) ◻

Proof. Take an injective envelope \[\pi_I(M) \hookrightarrow A \twoheadrightarrow N\] in \(\lec {v_I} {w_J}\) and lift this via Theorem 41 to a sequence \[M \hookrightarrow \widetilde{A} \twoheadrightarrow \widetilde{N}\] in \(\Mod C.\) All of these terms are in \(\calrij(I,J)\) since \(\calrij(I,J)\) is extension closed and since Theorem 41 gives us a triangle equivalence we know \(\ext^1_C(A,-)\) and \(\ext^1_C(-,A)\) vanish on \(\calrij(I,J)\) so \(A\) is projective-injective. A dual argument shows there are enough projectives. ◻

Proof. The proof of Lemma 13 tells us that for \(I = v^{-1}[j]\) then \(\pi_I(\cmcpro{n-j}) = \soc {v^{-1}} Q_j.\) This was stated for \(v \in \quotmax\) in the proof but the proof holds more generally for any \(v \in W\) by using \(v = id\) as the initial step of the induction. In particular, we apply this in the case \(v = (w_0r)^{-1}\) and we have the exact sequence \[M_{rw_0[j]} \hookrightarrow \cmcpro{n-j} \twoheadrightarrow \soc {rw_0} Q_j.\] We pushout to get the diagram \[\begin{tikzcd} {\cmcpro{n}} & {\cmcpro{n}} \\ {M_{rw_0[j]}} & {\cmcpro{n-j}} & {\soc {rw_0} Q_j} \\ {\pi_n(M_{rw_0[j]})} & {Q_j} & {\soc {rw_0} Q_j.} \arrow[shift right, no head, from=1-1, to=1-2] \arrow[shift left, no head, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[hook, from=1-2, to=2-2] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-2, to=2-3] \arrow[two heads, from=2-2, to=3-2] \arrow[two heads, from=2-3, to=3-3] \arrow[hook, from=3-1, to=3-2] \arrow[two heads, from=3-2, to=3-3] \end{tikzcd}\] By [5] we have \(\pi_n(M_{rw_0[j]}) = \head {r w_0 w_0} Q_j = \head {r} Q_j\) and since \(w_0[j] = [n+1-j,n]\) this completes the proof. ◻

Proof. All three of these are naturally submodules of \(Z_i.\) In order to show that they agree we will show \(t_v(Z_i) \subseteq Z_i\cap Y \subseteq \ker(Z_i \twoheadrightarrow \soc {v^{-1}} Z_i) \subseteq t_v(Z_i).\) Since these are finite dimensional modules the result will then follow. The proof that \(t_v(Z_i) \subseteq Z_i\cap Y\) is the same as in the proof of Lemma 33. Observe that \(Z_i \cap Y \subseteq \ker(Z_i \twoheadrightarrow \soc {v^{-1}} Z_i)\) since \(\soc {v^{-1}} Y = 0\) so \(Z_i\cap Y \subseteq \ker(Z_i \twoheadrightarrow \soc {v^{-1}} Z_i).\) It remains to show \(\ker(Z_i \twoheadrightarrow \soc {v^{-1}} Z_i) \subseteq t_v(Z_i).\) Since \(v \leq w\) then \(\head {v^{-1}w_0} Q_i \subseteq \head {w^{-1}w_0} Q_i.\) Since \((\downcat v, \upcat v)\) is a torsion pair the composition \[\head {v^{-1}w_0} Q_i \hookrightarrow \head {w^{-1}w_0} Q_i \twoheadrightarrow \soc {v^{-1}} \head {w^{-1}w_0} Q_i\] is zero. It is exact at the middle term since \(\ker(\head {w^{-1}w_0} Q_i \twoheadrightarrow \soc {v^{-1}} \head {w^{-1}w_0} Q_i) \subseteq \ker(Q_i \twoheadrightarrow \soc {v^{-1}} Q_i) = \head {v^{-1}w_0} Q_i.\)

Proposition 5.4 of [17] says \(t_{w^k_0}(Q_i) = \head {w^k_0w_0} Q_i,\) and since \(w^k_0 \leq v\) and we have inclusions \(t_{w^k_0}(Q_i) = \head {w^k_0w_0} Q_i \subseteq \head {v^{-1}w_0} Q_i \subseteq \head {w^{-1}w_0} Q_i.\) This implies that \[\begin{tikzcd} {t_{w_0^k}(\head{v^{-1}w_0} Q_i)} & {t_{w_0^k}(\head {w^{-1}w_0} Q_i)} & 0 \\ {\head {v^{-1}w_0} Q_i} & {\head {w^{-1}w_0} Q_i} & {\soc {v^{-1}} \head {w^{-1}w_0} Q_i} \arrow[from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[from=1-2, to=1-3] \arrow[hook, from=1-2, to=2-2] \arrow[from=1-3, to=2-3] \arrow[from=2-1, to=2-2] \arrow[from=2-2, to=2-3] \end{tikzcd}\] commutes. The snake lemma along with [17] implies that \(\ker(\soc {w_0^k} \head {w^{-1}w_0} Q_i \twoheadrightarrow \soc {v^{-1}} \head {w^{-1}w_0} Q_i) = \soc {w_0^k} \head {v^{-1}w_0} Q_i.\) Since \(\soc {w_0^k} \head {v^{-1}w_0} Q_i \in \downcat v\) then this implies \(\ker(Z_i \twoheadrightarrow \soc {v^{-1}} Z_i) \subseteq t_v(Z_i).\) ◻

Proof. Recall that by [3] then if \[M_{J_1} \hookrightarrow M_{J_2}\oplus M_{J_3} \twoheadrightarrow M_{J_4}\] we have \(e(J_4) = e(J_2) + e(J_3) - e(J_1).\) We know that \[M_R \hookrightarrow M_I \oplus M_J \twoheadrightarrow M_{L_i}.\] We have shown that \(J = \mink{w^{-1}[i]}\) when \(i \geq k\) and \([k-i]\cup w^{-1}[i]\) otherwise. In the proof of Lemma 55 we also showed that \(t_v(Z_i) = \soc {w_0^k} \head {v^{-1}w_0} Q_i.\) Therefore \(R = \mink{v^{-1}[i]}\) when \(i \geq k\) and \(R = [k-i] \cup v^{-1}[i]\) otherwise. The result then follows. ◻

Proof. By applying \(\Hom_C(-,M_I)\) to the sequence \[M_R \hookrightarrow M_J\oplus M_I \twoheadrightarrow M_{L_i}\] we use the same argument as in Proposition 36 to see that \[\ext^1_C(M_{L_i},M_I) \hookrightarrow \ext^1_C(M_J,M_I) \rightarrow \ext^1_C(M_R,M_I).\] Since \(\ext^1_C(M_R,M_I) = \ext^1_C(\soc {w_0^k} \head {v^{-1}w_0} Q_i, \head {v^{-1}w_0} Q_k) = 0\) then \(\ext^1_C(M_{L_i},M_I) = \ext^1_C(M_J,M_I) = \ext^1_{\propi}(Z_i, \head {v^{-1}w_0} Q_k).\) We know \(\ext^1_{\propi}(X,Y) = \underline{\Hom}_{\propi}(X,\Omega^{-1}(Y))\) where \(\Omega^{-1}(Y)\) is the cokernel of the injective envelope of \(Y.\) For \(\head{v^{-1}w_0} Q_k\) we have \[\Omega^{-1}(\head{v^{-1}w_0} Q_k) = \soc {v^{-1}} Q_k\] so \[\ext^1_C(M_{L_i},M_I) = \underline{\Hom}_{\propi}(Z_i, \soc {v^{-1}} Q_k).\]

Finally consider the triangle \[t_{w_0^k} Q_i \rightarrow \head{w^{-1}w_0} Q_i \rightarrow Z_i \rightarrow \Omega^{-1}(t_{w_0^k} Q_i)\] in \(\underline{\Mod} \propi\) and apply \(\underline{\Hom}(-,\soc {v^{-1}} Q_k)\) to get the sequence \[\underline{\Hom}(\Omega^{-1}(t_{w_0^k} Q_i), \soc {v^{-1}} Q_k) \rightarrow \underline{\Hom}(Z_i,\soc {v^{-1}} Q_k) \rightarrow \underline{\Hom}(\head{w^{-1}w_0} Q_i, \soc {v^{-1}} Q_k) \rightarrow \underline{\Hom}(t_{w_0^k} Q_i,\soc {v^{-1}} Q_k).\] We have \(\underline{\Hom}_{\propi}(\Omega^{-1}(t_{w_0^k} Q_i), \soc {v^{-1}} Q_k) = \underline{\Hom}_{\propi}(t_{w_0^k} Q_i, \head {v^{-1}w_0} Q_k) = 0\) since \(\head {v^{-1}w_0} Q_k \in \upcat {w_0^k}.\) Notice that for \[f \in \underline{\Hom}_{\propi}(\head{w^{-1}w_0} Q_i, \soc {v^{-1}} Q_k)\] we have \(\head {v^{-1}w_0} Q_i \subseteq \ker(f).\) But \(t_{w_0^k} Q_i = \head {w_0^kw_0} Q_i \subseteq \head {v^{-1}w_0} Q_i\) so the final map is the zero map. In particular \[\underline{\Hom}_{\propi}(Z_i,\soc {v^{-1}} Q_k) = \underline{\Hom}_{\propi}(\head{w^{-1}w_0} Q_i, \soc {v^{-1}} Q_k) = \ext^1_{\propi}(\head {w^{-1}w_0} Q_i, \head {v^{-1}w_0} Q_k).\] ◻

Proof. Given a surjection \(B_{v,w}^t \twoheadrightarrow M\) this surjects at vertex \(n.\) By construction \(B_{v,w} \in \su(I,n)\) so there is a morphism \(M_I^r \rightarrow B_{v,w}\) which surjects at vertex \(n.\) Composing these gives \(M \in \fa(I,n).\) It is then part of the definition that \(\ext^1(M_I,M) = 0.\) ◻

Proof. First we show \(\pi_I(\rich) \subseteq \lec v w.\) Given \(M \in \rich\) then we can form the diagram \[\begin{tikzcd} {M_I^r} & {M_I^s} \\ {B_{v,w}^s} & M \\ {Q_{v,w}^t} & {\pi_I(M).} \arrow[hook, from=1-1, to=2-1] \arrow[hook, from=1-2, to=2-2] \arrow[two heads, from=2-1, to=2-2] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-2, to=3-2] \end{tikzcd}\] Where we have denoted \(Q_{v,w} = \oplus_i \soc {v^{-1}} \head {w^{-1}w_0} Q_i.\) Since the kernel of the projection of \(M\) to \(\pi_I(M)\) is the images of all maps from \(M_I\) then it follows from universality of cokernels that there is a surjection \(Q_{v,w}^t \twoheadrightarrow \pi_I(M).\) Since \(Q_{v,w} = \oplus_i \soc {v^{-1}} \head {w^{-1}w_0} Q_i\) is a quotient of \(\oplus_i \head {w^{-1}w_0} Q_i\) then \(\pi_I(M) \in \downcat w.\) We know \(\pi_I(M) \in \upcat v\) so it follows \(\pi_I(M) \in \lec v w.\) Conversely suppose we have \(X \in \lec v w.\) Take its minimal lift \(\widetilde{X} \in \su(I,n)\) and the exact sequence \[M_I^t \hookrightarrow \widetilde{X} \twoheadrightarrow X.\] The category \(\lec v w\) has enough projectives so there is a surjection \(Q_{v,w}^r \twoheadrightarrow X.\) We extend this to get a surjection \(B_{v,w}^r \twoheadrightarrow X.\) By Theorem 41 we see that \(B_{v,w}\) is rigid and in particular \(\ext^1(B_{v,w},M_I) = 0.\) We can therefore repeat the proof used for Lemma 25 to see that there is a surjection \(B_{v,w}^l \twoheadrightarrow \widetilde{X}.\) Since \(\widetilde{X} \in \su(I,n)\) it follows \(\widetilde{X} \in \rich.\) ◻

Proof. Observe as a consequence the above proposition and Theorem 41 \[\rich = \{M \in \cm(C): B_{v,w}^t \twoheadrightarrow M, \, \ext^1(B_{v,w},M) = 0\}.\] This means \(\rich = \overline{S}(B_{v,w},[n]).\) In particular by Lemma 25 it is extension closed. Now consider \(M \in \rich.\) Let \(\pi_I(M) \hookrightarrow Q_{\pi_I(M)} \twoheadrightarrow \Omega^{-1}(\pi_I(M))\) be an injective envelope of \(\pi_I(M)\) in \(\lec v w.\) By Theorem 41 we see there is a sequence \[M \hookrightarrow Q \twoheadrightarrow Y\] with \(Y \in \rich\) and by Proposition 22 we have \(Q \in \mathrm{add}(B_{v,w}).\) This means the category has enough injectives. The same argument works for projectives. These objects coincide since \(\rich\) is extension closed in a Frobenius category. ◻

Proof. The \(\mathbb{C}^{\times}\) action gives a \(\mathbb{Z}\)-grading \[\mathbb{C}[\widetilde{C^I}] = \bigoplus_{r \in \mathbb{Z}} \mathbb{C}[\widetilde{C^I}]_r\] where \(\mathbb{C}[\widetilde{C^I}]_r\) is the homogeneous functions of degree r. The coordinate ring \(\mathbb{C}[C^I]\) is the \(\mathbb{C}^{\times}\)-invariant functions. That is to say \[\mathbb{C}[C^I] \cong \mathbb{C}[\widetilde{C^I}]_0.\] We then have an explicit isomorphism \[\begin{align} \mathbb{C}[\widetilde{C^I}&]_r \xrightarrow{\sim} \mathbb{C}[\widetilde{C^I}]_0 \cong \mathbb{C}[C^I]\\ &f \longmapsto \Delta_I^{-r}\cdot f.\end{align}\] ◻

Proof. We first start with the obvious isomorphism \[N^{'}(v) \cong vN^{'}(v).\] It is a classical result (see for example [30]) that the restriction of the quotient \[G \rightarrow B\backslash G\] gives an isomorphism \[vN^{'}(v) \cong C^v.\] We recall the projection \[B\backslash G \rightarrow P_k\backslash G \cong \gr(k,n).\] By an argument of Lusztig [15] (see [31] for a more detailed argument) this restricts to an isomorphism \[C^v \cong C^{v^{-1}[k]}.\] We stress that the variety on the left is the opposite Schubert cell in the flag variety whereas the variety on the right is the opposite Schubert cell associated to \(v^{-1}[k]\) in the Grassmannian. ◻

Proof. This follows from the equivalent statement for \(\varphi\) along with Proposition 22. ◻

Proof. We recall that Leclerc showed in [5] that \(\varphi(\upcat v)\) generates the algebra \(\mathbb{C}[N^{'}(v)].\) Combine this with the isomorphism \(C^{v^{-1}[k]} \cong N^{'}(v)\) to generate \(\mathbb{C}[C^{v^{-1}[k]}].\) We recall that \[\mathbb{C}[\widetilde{C^I}] = \bigoplus_{r \in \mathbb{Z}} \mathbb{C}[C^I]\cdot \Delta_I^r.\] The result then follows. ◻

Proof. The same argument as at the beginning of this section gives that \(\mathbb{C}[\widetilde{C}_{v,w}] = \oplus_i \mathbb{C}[C_{v,w}]\cdot \Delta_I^i.\) By [5] we know that the localisation of \(^{N(v)}\mathbb{C}[N_-]^{N^{'}(w)}\) at the subset \(\{\varphi_X: X \in \mathrm{add}(Q_{v,w})\}\) is isomorphic to \(\mathbb{C}[C_{v,w}].\) Since \(\clchar_{B_{v,w}} = \Delta_I^j\cdot \varphi_{Q_{v,w}}\) for some \(j \in \mathbb{N}\) it follows that localising at \(\{\clchar_X: X \in \mathrm{add}(B_{v,w})\}\) agrees with localising at \(\{\clchar_X: X \in \mathrm{add}(\Delta_I)\}\) and then \(\{\clchar_X: X \in \mathrm{add}(Q_{v,w})\}.\) First localising at \(\{\clchar_X: X \in \mathrm{add}(\Delta_I)\},\) [5] tells us \[\clalg v w [\Delta_I^{r}: r \in \mathbb{Z}] \cong ^{N(v)}\mathbb{C}[N_-]^{N^{'}(w)}[\Delta_I^r:r \in \mathbb{Z}].\] Here we have identified \(^{N(v)}\mathbb{C}[N_-]^{N^{'}(w)}\) with its image in \(\mathbb{C}[\widehat{C^I}].\) We now localise at \(\{\clchar_X: X \in \mathrm{add}(Q_{v,w})\}\) giving \[\clalg {v} {w}^{\circ} \cong \mathbb{C}[C_{v,w}][\Delta_I^r: r \in \mathbb{Z}] \cong \mathbb{C}[\widetilde{C}_{v,w}].\] ◻

Proof. For \(M_J \in \su(I,n)\) then by Corollary 6 and Proposition 31 we have \(\pi_I(M_J) = \soc {v^{-1}} \head {w^{-1}w_0} Q_k\) where \(v, w \in \quotmax\) and \(v^{-1}[k] = I,\) \(w^{-1}[k] = J.\) The results in [17] tell us how to express \(\varphi_{\pi_I(M_J)}\) as a generalised minor. We can combine this with [32] to say \(v\cdot\varphi_{\pi_I(M_J)} = \Delta_{\omega_k, w^{-1}\omega_k} = \Delta_J/\Delta_I.\) We should comment that the identification of \(\Delta_{\omega_k, w^{-1}\omega_k} = \Delta_J/\Delta_I \in \mathbb{C}[\widetilde{C^I}]\) is because we are considering \(\mathbb{C}[C^I]\) as the degree 0 homogeneous part of \(\mathbb{C}[\widetilde{C^I}].\) In particular, since \(\rank M_J = 1\) then \(\clchar_{M_J} = \Delta_I \cdot \Delta_J/\Delta_I = \Delta_J.\) ◻

Proof. This can either been shown directly using Leclerc’s tilting object \(U_{\mathbf{i}} \in \upcat v\) or using Proposition 40 and [33]. ◻

Proof. By Lemma 74 this result is true for rank one modules. By Lemma 75 there is a cluster-tilting object in \(\su(I,n)\) consisting of rank one indecomposable summands. In particular, \(\clchar\) and \(\mathrm{res}\cdot \Psi\) agree on this cluster tilting object. By [3] the map \(\mathrm{res}\cdot \Psi\) is a cluster character. Therefore the two maps agree on reachable cluster variables. See [5] for the definition of reachable. By the paragraph preceding Proposition 5.1 of [5] we know that the coordinate ring \(\mathbb{C}[C^I]\) is spanned by the cluster structure determined by \(\upcat v.\) Since \(\clchar_{M_I} = \mathrm{res}\cdot \Psi_{M_I}\) again by Lemma 74 then the two maps agree on the whole coordinate ring \(\mathbb{C}[\widetilde{C^I}].\) ◻

Proof. Let \[U \hookrightarrow Q_U\] be the injective envelope of \(U \in \upcat v.\) Form the pullback diagram \[\begin{tikzcd} {M_I^t} && {M_I^t} \\ X && {\widetilde{Q}_U} && W \\ U && {Q_U} && W \arrow[shift right, no head, from=1-1, to=1-3] \arrow[shift left, no head, from=1-1, to=1-3] \arrow[hook, from=1-1, to=2-1] \arrow[hook, from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-3] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-3, to=2-5] \arrow[two heads, from=2-3, to=3-3] \arrow[shift right, no head, from=2-5, to=3-5] \arrow[shift left, no head, from=2-5, to=3-5] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-3, to=3-5] \end{tikzcd}\] By Corollary 3 we know that \(t = \rank (\widetilde{Q}_U) = \udim {U}.\) Since rank is additive on exact sequences we see \(\rank (X) = \udim {U}.\) Recall from Remark 1 that we are considering \(\upcat v\) as a full subcategory of \(\Mod C\) with no support at vertex n. Since \(U\) is not supported at vertex \(n\) then we see \(X \in \fa(I,n).\) It remains to show \(X \in \su(I,n)\) and that \(M_I\) is not a summand of X. First we show \(\ext^{1}(M_I,X) = 0.\) By applying \(\Hom(M_I,-)\) to the middle row of the diagram above we get \[\Hom(M_I,W) \twoheadrightarrow \ext^{1}(M_I,X).\] Therefore it suffices to show \(\Hom(M_I,W) = 0.\) Note that since we took an injective envelope and \(\upcat v\) is Frobenius then \(W \in \upcat v.\) Suppose we have a non-zero morphism \[f: M_I \rightarrow W.\] Since \(W\) is not supported at vertex \(n\) then the submodule of \(M_I\) generated at vertex \(n\) is contained in the kernel of \(f.\) This means there is a non-zero map \[\overline{f}: \pi_n(M_I) \rightarrow W.\] But by Lemma 30 we know \(\pi_n(M_I) \in \downcat v.\) Since \((\downcat v, \upcat v)\) is a torsion pair then this can’t be the case so \(\Hom(M_I,W) = 0.\) We now have \(X \in \su(I,n).\) It remains to show \(X\) has no summands in \(\mathrm{add}(M_I).\) This follows from an argument found in the proof of Lemma 3.3 of in [36]. Suppose \(X = Y\oplus M_I.\) This implies that \(W\) contains a summand which is projective-injective in \(\upcat v.\) Applying this to the bottom row implies that \(Q_U\) contains a projective-injective summand onto which the embedding from \(U\) does not map. This contradicts the fact that \[U \hookrightarrow Q_U\] was a injective envelope. ◻

Proof. By a result of Zwara [37] it follows that there is an exact sequence

\[\begin{tikzcd} M && {Y \oplus M} && {X} \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \end{tikzcd}\] in \(\Mod \propi.\)

Now consider the pushout \[\begin{tikzcd} {t_v(M)} && {t_v(M)} \\ \\ M && {Y \oplus M} && {X} \\ \\ {M/t_v(M)} && U && {X} \arrow[shift right, no head, from=1-1, to=1-3] \arrow[shift left, no head, from=1-1, to=1-3] \arrow[hook, from=1-1, to=3-1] \arrow[hook, from=1-3, to=3-3] \arrow[hook, from=3-1, to=3-3] \arrow[two heads, from=3-1, to=5-1] \arrow[two heads, from=3-3, to=3-5] \arrow[two heads, from=3-3, to=5-3] \arrow[no head, from=3-5, to=5-5] \arrow[shift left=3, no head, from=3-5, to=5-5] \arrow[hook, from=5-1, to=5-3] \arrow[two heads, from=5-3, to=5-5] \end{tikzcd}\] and recall that \(\Hom(\downcat v, \upcat v) = 0.\) In particular we see \(U \cong M/t_v(M) \oplus Y.\) Setting \(N = M/t_v(M)\) this gives us an exact sequence

\[\begin{tikzcd} {N} && {Y\oplus N} && {X} \arrow[hook, from=1-1, to=1-3] \arrow[two heads, from=1-3, to=1-5] \end{tikzcd}\] in \(\upcat v.\) Consider \(Q_{X}\) an injective envelope of \(X\) and \(Q_N\) an injective envelope of \(N.\) The horseshoe lemma gives us

\[\begin{tikzcd} N & {N\oplus Y} & {X} \\ {Q_N} & {Q_N\oplus Q_{X}} & {Q_{X}} \\ A & B & D \arrow[hook, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[two heads, from=1-2, to=1-3] \arrow[hook, from=1-2, to=2-2] \arrow[hook, from=1-3, to=2-3] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-2, to=2-3] \arrow[two heads, from=2-2, to=3-2] \arrow[two heads, from=2-3, to=3-3] \arrow[hook, from=3-1, to=3-2] \arrow[two heads, from=3-2, to=3-3] \end{tikzcd}\] where the bottom row comes from the snake lemma. Since \[N \hookrightarrow Q_N\] and \[X \hookrightarrow Q_{X}\] are injective envelopes then \(A,D \in \upcat v.\) The category \(\upcat v\) is extension closed and therefore \(B \in \upcat v.\) This then allows us to apply the lift method of [36] as in the previous proof to produce the lift

\[\begin{tikzcd} {M_I^{t}} & {M_I^t} \\ Z & {\widetilde{Q}_N\oplus \widetilde{Q}_{X}} & B \\ {Y\oplus N} & {Q_N\oplus Q_{X}} & B \arrow[shift right, no head, from=1-1, to=1-2] \arrow[shift left, no head, from=1-1, to=1-2] \arrow[hook, from=1-1, to=2-1] \arrow[hook, from=1-2, to=2-2] \arrow[hook, from=2-1, to=2-2] \arrow[two heads, from=2-1, to=3-1] \arrow[two heads, from=2-2, to=2-3] \arrow[two heads, from=2-2, to=3-2] \arrow[shift right, no head, from=2-3, to=3-3] \arrow[shift left, no head, from=2-3, to=3-3] \arrow[hook, from=3-1, to=3-2] \arrow[two heads, from=3-2, to=3-3] \end{tikzcd}\] where \(t = \udim{Y} + \udim{N}\) and \(\ext^{1}(M_I,Z) = 0.\) In particular \(Z \in \su(I,n)\) and \(\pi_I(Z) = Y\oplus N.\) By Lemma 22 this implies \(Z = \widetilde{Y} \oplus\widetilde{N} \oplus M_I^l\) for some \(l \in \mathbb{N}.\) We know \(\rank Z = \rank \widetilde{X} + \rank \widetilde{N}.\) Cancelling out \(\rank \widetilde{N}\) we see \(\rank \widetilde{X} = \rank \widetilde{Y} + l.\) Since \(l \in \mathbb{N}\) we have \(\rank \widetilde{X} \geq \rank\widetilde{Y}.\) ◻

Proof. By Proposition 4.2 of [5] we know that the dual semicanonical basis induces a basis of \(S(\lec v {w_0} ).\) A function \(\varphi_U\) is in this dual semicanonical basis if and only if \(U\) is generic in an irreducible component of its representation variety.

Since functions of different homogeneous degrees are linearly independent then if \(\semican v w\) were not linearly independent then this must happen in a single homogeneous component. But under the isomorphism \[\mathbb{C}[\widetilde{C^I}]_r \xrightarrow{\sim} \mathbb{C}[C^I]\] this would give a linear dependence amongst the dual semicanonical basis elements of \(\mathbb{C}[C^I].\) In particular \(\semican v w\) must be linearly independent. ◻

Proof. Since \((\upcat v, \downcat v)\) is a torsion pair and \(\downcat v = \mathrm{Fac}(\head {v^{-1}w_0} Q)\) we have \(M \in \upcat v\) if and only if \[\Hom(\head {v^{-1}w_0} Q, M) = 0.\] Since \(\Hom(X,-)\) is upper semicontinuous [39] the set \(\Hom(\head {v^{-1}w_0} Q, -) = 0\) is open. ◻

Proof. Since \(\Lambda_\boldsymbol{d} \cap \upcat v\) is open \(\Lambda_\boldsymbol{d} \backslash \upcat v\) is closed. Since the closure of a \(\mathrm{G}_\boldsymbol{d} \cdot M\) is the intersection of all closed sets containing \(\mathrm{G}_\boldsymbol{d} \cdot M\) it follows that \[\overline{\mathrm{G}_\boldsymbol{d} \cdot M} \subseteq \Lambda_\boldsymbol{d} \backslash \upcat v.\] ◻

Proof. Under Conjecture 85 then \(f_{Z_t}(\pi_I(M)) \neq 0\) implies that \(\pi_I(M)\) is in the irreducible component in which \(U_t\) is generic. Since \(U_t\) is generic in \(Z_t\) and \(\Lambda_\boldsymbol{d} \cap \upcat v\) is open \(Z_t \cap \upcat v\) is dense in \(Z_t.\) This could also have been obtained from [12]. This means that \(\pi_I(M) \in \overline{Z_t \cap \upcat v}.\) Since \(\pi_I(M) \in \upcat v\) it can not be in the closure of a generic module not in \(\upcat v.\) It follows that there is a generic module \(U_t \in Z_t \cap \upcat v\) such that \(\pi_I(M) \in \overline{\mathrm{G}_\boldsymbol{d} \cdot U_t}.\) We now conclude by Proposition 79 that \(\kappa(U_t) \leq \rank M.\) ◻

Proof. The second result will follow directly from the first since \(\ext\) is additive. Clearly for \(M \in \bigcap_j \fa(I_j,j-1)\) then \(M \in \cm(B).\) We now need to show \(M \in \cm(B)\) implies \(M \in \bigcap_j \fa(I_j,j-1).\) For such a M we have a surjection \[B^t \twoheadrightarrow M.\] In particular this must also surject at vertex \(j-1\). In particular there is a map \[\bigoplus_j M_{I_j}^{t_j} \hookrightarrow M\] that surjects at vertex \(j-1\). But for a necklace we have \(I_j \leq_{j} I_i\) so by Lemma 29 we have an embedding \[M_{I_j}^r \hookrightarrow \bigoplus_j M_{I_j}^{t_j}\] that also surjects at vertex \(j-1\). The composition of these embeddings tells us that there is an embedding \[M_{I_j}^r \hookrightarrow M\] that surjects at vertex \(j-1\). Repeating this at each vertex tells us \(M \in \bigcap_j \fa(I_j,j-1).\) ◻