The Schur functor sends Weyl and simple modules to Specht and Hecke simple modules

Under [F1], $\rho (h)\rho (k)=\rho (hk)$ because both act on $H$ by left multiplication, so the identification uses $H$, without an opposite algebra. The functor $X\mapsto Xe$ is exact: for a short exact sequence, a lift $x$ of an element $y\in X^{\mathrm{\prime }\mathrm{\prime }}e$ yields the lift $xe$, and an element in the kernel lies in $X^{\mathrm{\prime }}$ and is fixed by $e$, hence lies in $X^{\mathrm{\prime }}e$. The tensor description uses the $(S,eSe)$-bimodule $Se$, which is a direct summand as a left $S$-module.

The support in [F2] gives $e{\phi }_{ST}e={\delta }_{\mathrm{type}(S),\omega }{\delta }_{\mathrm{type}(T),\omega }{\phi }_{ST}$ and $w_{T}e={\delta }_{\mathrm{type}(T),\omega }{w}_T$. Semistandard tableaux of type $\omega$ have each label once and are exactly standard tableaux. For standard $s,t$, their defining sum $m_{st}$ has one term, so ${\phi }_{st}{\mathrm{\mid }}_H(a)=m_{st}a$ and ${\rho }^{-1}({\phi }_{st})=m_{st}$. Thus the higher Schur ideal cuts out exactly $I^{⊳\lambda }$ in the corner. Read the Schur right-cell law with first index any standard tableau of type $\omega$; coefficient independence permits this choice. Restricting the multiplication by $\rho (h)$ to $H$ gives $m_{st}h$, so its coefficients are the Murphy right-cell coefficients. Consequently ${\mathrm{\Psi }}_{\lambda }:W^{\lambda }e\to C_{\mathrm{M}}(\lambda )$, $w_t\mapsto c_t$, is an $H$-isomorphism.

By the Schur cell product, ${\phi }_{as}{\phi }_{tb}\equiv B_{\lambda }(w_s,w_t){\phi }_{ab}$ modulo higher Schur shapes, with coefficient independent of the outer indices. Choose all four indices standard of shape $\lambda$ and type $\omega$. They lie in $eSe$; applying ${\rho }^{-1}$, or evaluating their endomorphisms at $1\in H$, gives $m_{as}{m}_{tb}\equiv B_{\lambda }(w_s,w_t)m_{ab}$ modulo $I^{⊳\lambda }$. The unique coefficient in [F3] is $G_{\lambda }(c_s,c_t)$, proving $B_{\lambda }{\mathrm{\mid }}_{W^{\lambda }e}={\mathrm{\Psi }}_{\lambda }^{\ast }{G}_{\lambda }$ exactly, with factor $1$.

Since $e^{\ast }=e$, contravariance gives $B_{\lambda }(xe,y(1-e))=B_{\lambda }(x,y(1-e)e)=0$. The decomposition $W^{\lambda }=W^{\lambda }e\oplus W^{\lambda }(1-e)$ is therefore orthogonal. A vector in $W^{\lambda }e$ pairs trivially with its complement, so it is in the restricted form radical exactly when it lies in ${\mathrm{rad}}_{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{W}^{\lambda }$. Since the latter is an $S$-submodule, ${\mathrm{rad}}_{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}(W^{\lambda }e)=({\mathrm{rad}}_{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{W}^{\lambda })e$. This holds even if the corner form is identically zero.

For the canonical semistandard tableau $T^{\lambda }$ of shape and type $\lambda$, the map ${\phi }_{T^{\lambda },T^{\lambda }}$ is the actual projection onto $M^{\lambda }$: it sends $m_{\lambda }h$ to itself and is zero on every other summand. Its square equals itself, so the Schur cell product gives $B_{\lambda }(w_{T^{\lambda }},w_{T^{\lambda }})=1$. Hence the Weyl form quotient is nonzero and is its simple head $L^{\lambda }$ by [F4]. Exactness, step 4.1 and step 3.1 give $L^{\lambda }e\cong W^{\lambda }e\mathrm{/}({\mathrm{rad}}_{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{W}^{\lambda })e\cong C_{\mathrm{M}}(\lambda )\mathrm{/}{\mathrm{rad}}_{\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{m}}{C}_{\mathrm{M}}(\lambda )=D_{\mathrm{M}}(\lambda )$, including when this corner quotient is zero.

By [F6] and step 5.1, $L^{\lambda }e\mathrm{\ne }0$ precisely for $e_q$-restricted $\lambda$. Apply the exact functor to a finite composition series of $W^{\lambda }$: every factor $L^{\nu }$ gives $D_{\mathrm{M}}(\nu )$ or zero. Delete repeated terms. The remaining factors are the distinct nonzero Murphy simples in [F4], giving $[W^{\lambda }:L^{\nu }]=[C_{\mathrm{M}}(\lambda ):D_{\mathrm{M}}(\nu )]$ for each $e_q$-restricted $\nu$. No multiplicity preservation is asserted for a killed simple label.

Apply [F5] and the involution $\theta$ to obtain $C_{\mathrm{M}}({\lambda }^{\mathrm{\prime }})\cong (S^{\lambda }{)}^{\theta }$ and $D_{\mathrm{M}}({\lambda }^{\mathrm{\prime }})\cong (D^{\lambda }{)}^{\theta }$. Replacing the Weyl label by ${\lambda }^{\mathrm{\prime }}$ in steps 2.1 and 5.1 gives the ordinary dictionary, including zero quotients. Conjugation makes ${\lambda }^{\mathrm{\prime }}$ restricted precisely when $\lambda$ is regular. Twisting both modules by the same exact equivalence preserves composition multiplicities, so step 6.1 gives $[W^{{\lambda }^{\mathrm{\prime }}}:L^{{\nu }^{\mathrm{\prime }}}]=[S^{\lambda }:D^{\nu }]$ for regular $\nu$. The genuine unit-scaled ordinary form bridge in [F5] also identifies these with the historical DJ polytabloid modules and quotients. At $n=0$, $e=1$ and each form has value $1$, so the same proof includes the empty partition.