Evaluated double leaves form bases of type-A Soergel bimodule homs

Unit evaluation is an isomorphism: [F1] gives a basis of ${\mathrm{Hom}}_D(\underset{‾}{u},\mathrm{\varnothing })$, and [F2] says its evaluated images form a basis of ${\mathrm{Hom}}_{R\text{-}R}(B_{\underset{‾}{u}},R)$. Evaluation is left $R$-linear because a polynomial in the leftmost region acts by left multiplication on the output. Thus the map $E_{\underset{‾}{u},\mathrm{\varnothing }}$ on these Hom spaces is an isomorphism of graded left $R$-modules.

Right bending: write $X=\underset{‾}{x}$, $Y=\underset{‾}{y}$ and $Y^{\ast }=\mathrm{rev}(\underset{‾}{y})$. The cups and caps of [F3], nested in reverse order, give ${\epsilon }_Y:Y\otimes Y^{\ast }\to \mathbf{1}$ and ${\eta }_Y:\mathbf{1}\to Y^{\ast }\otimes Y$. In either category the maps $\beta (f)={\epsilon }_Y\circ (f\otimes {\mathrm{id}}_{Y^{\ast }}),{\beta }^{-1}(g)=(g\otimes {\mathrm{id}}_Y)\circ ({\mathrm{id}}_X\otimes {\eta }_Y)$ are inverse degree-zero bijections between $\mathrm{Hom}(X,Y)$ and $\mathrm{Hom}(X\otimes Y^{\ast },\mathbf{1})$ by the two triangle identities. They are left $R$-linear: bending takes place on the right and leaves the leftmost polynomial region fixed; the same assertion for bimodules follows from left linearity of the evaluation map.

Compatibility: because $\mathcal{F}$ preserves composition, tensor products and the specified cups and caps, the two bending maps satisfy ${\beta }_{\mathrm{b}\mathrm{i}\mathrm{m}}\circ E_{\underset{‾}{x},\underset{‾}{y}}=E_{\underset{‾}{u},\mathrm{\varnothing }}\circ {\beta }_D\mathrm{.}$ Both bending maps are isomorphisms by step 1.2 and the unit evaluation map is an isomorphism by step 1.1. Hence $E_{\underset{‾}{x},\underset{‾}{y}}={\beta }_{\mathrm{b}\mathrm{i}\mathrm{m}}^{-1}{E}_{\underset{‾}{u},\mathrm{\varnothing }}{\beta }_D$ is an isomorphism of graded left $R$-modules. This conclusion uses no identification of the bending of an individual double leaf with an individual unit-target light leaf.

The double leaves of [F1] are a homogeneous basis of the source of $E_{\underset{‾}{x},\underset{‾}{y}}$, so their images are a homogeneous basis of its target by step 2.1. A graded functor preserves their degrees $d(e)+d(f)$. There are finitely many indexing pairs, since each word has finitely many subexpressions. Thus the graded rank is exactly ${\sum }_w{\sum }_{e,f:w_e=w_f=w}{v}^{d(e)+d(f)}$, for this pair of words.

For finite direct sums a bimodule map is uniquely a matrix of maps between the summands. A degree-$d$ map $B_{{\underset{‾}{x}}_a}\to B_{{\underset{‾}{y}}_b}$ has degree $d+r_a-s_b$ when viewed from $B_{{\underset{‾}{x}}_a}(r_a)$ to $B_{{\underset{‾}{y}}_b}(s_b)$, because these shifts lower element degrees by $r_a$ and $s_b$. Thus the individual evaluated bases, placed in one matrix entry at a time, form a free homogeneous basis of the total Hom module, with the displayed sum of shifted ranks. Empty sums give the zero module and rank zero; for $M=R\oplus R$, $N=R$ there are two degree-zero basis entries and rank two. Taking degree zero recovers categorical morphisms, without asserting they form an $R$-submodule. ∎