Elementary ribbons have explicitly determined endpoints

The external Example 3.14 identifies $r(I,a)$ with ${\nu }_I(a)$ and states that these functions give a nu-structure; Definition 3.12 makes their values atoms of the positive ribbon category. In particular $r(I,a)$ is positive and conjugates the indicated positive parabolic monoids.

By [F3], conjugation by ${\mathrm{\Delta }}_I$ permutes $I$ and conjugation by ${\mathrm{\Delta }}_K$ sends it to ${\iota }_K(I)$. Therefore $r(I,a{)}^{-1}{A}_I^{+}r(I,a)={\mathrm{\Delta }}_K^{-1}{\mathrm{\Delta }}_I{A}_I^{+}{\mathrm{\Delta }}_I^{-1}{\mathrm{\Delta }}_K=A_J^{+}$. The same calculation gives the subgroup equality. This fixes the endpoint independently of any source/target naming.

If $r(I,a)$ belonged to $A_I^{+}$, the positive identity ${\mathrm{\Delta }}_K={\mathrm{\Delta }}_{I}r(I,a)$ would have support contained in $I$ by [F2]. But its support is $K$, containing $a\mathrm{\notin }I$. This contradiction proves it is outside $A_I^{+}$.

For $X\subseteq I$, apply [F1] to the restricted spherical system from [F7]. Its Artin group maps onto $A_X$ by sending each restricted generator to the ambient generator; this is a homomorphism by [F6]. In that restricted group every element is $u{v}^{-1}$ with $u,v$ positive: such expressions include each generator and its inverse, and are closed under multiplication, since a common right multiple $bp=cq$ of positive $b,c$ gives $(a{b}^{-1})(c{d}^{-1})=(ap)(dq{)}^{-1}$. Such common right multiples exist by [F1]. Mapping these expressions into $A$ proves that every element of $A_X$ is $u{v}^{-1}$ with $u,v\in A_X^{+}$. No injectivity of the restricted-group map is assumed here.

Let $C_1,\dots ,C_k$ be the diagram components of $I$. Distinct components have $m(u,v)=2$ between their generators, so the Artin relations make the subgroups $A_{C_j}$ commute elementwise; they generate $A_I$. To prove uniqueness, suppose $g_1\cdots g_k=1$ with $g_j\in A_{C_j}$, and use step 1.4 to write $g_j=u_j{v}_j^{-1}$ with $u_j,v_j\in A_{C_j}^{+}$. Commuting distinct factors gives $u_1\cdots u_k=v_1\cdots v_k$ in $A$. By [F1] this is an equality in the presented positive monoid, hence a finite sequence of positive braid replacements connects the two words. By [F2], all intermediate words use only $I$. For a fixed $C_j$, erase all letters outside $C_j$ throughout this sequence. A braid relation within $C_j$ remains that relation, one outside it becomes empty on both sides, and one crossing components is a commuting relation and becomes the same word on both sides. Thus $u_j=v_j$ and $g_j=1$ for every $j$. Each factor is normal, since it is preserved by conjugation from itself and commutes with the other generating factors. The factors therefore form an internal direct product in [F9]. When $I=\mathrm{\varnothing }$ this is the empty product.

Conversely, suppose $A_X$ is an internal direct product of two proper standard parabolic subgroups $A_U,A_V$, where $U,V\subseteq X\subseteq I$. The generators are nonidentity by [F6] and [F7], so trivial intersection forces $U\cap V=\mathrm{\varnothing }$. Projecting the equality $A_X=A_U{A}_V$ to $W$ gives $W_X\subseteq W_{U\cup V}$, and hence $X=U\cup V$ by $W_{U\cup V}\cap S=U\cup V$ in [F7]. Normality of both factors and trivial intersection make them commute: each commutator belongs to both factors and is therefore the identity. For $u\in U,v\in V$, their Coxeter images commute, so $(uv{)}^2=1$; the order convention [F8] forces $m(u,v)=2$. Thus the nonempty parts $U,V$ have no diagram edge between them. In particular a connected $X$ admits no such decomposition. Together with step 2.1, this proves that the indecomposable standard parabolic factors are exactly the diagram components, and Godelle's $L_s$ equals $I_s$.

For $s\in I$, apply Example 3.14 only after the identification in step 3.1: it gives ${\nu }_I(s)={\mathrm{\Delta }}_{L_s}={\mathrm{\Delta }}_{I_s}$. By [F2] its support is $I_s\subseteq I$, so it belongs to $A_I^{+}$. Combined with steps 1.1–1.3, this proves all the assertions.