Semistandard maps are independent and respect dominance

[construct] The rule $t_0(r,c):={\lambda }_1+\cdots +{\lambda }_{r-1}+c$ defines a bijection $[\lambda ]\to \{1,\dots ,n\}$, because the row blocks $B_r=\{{\lambda }_1+\cdots +{\lambda }_{r-1}+1,\dots ,{\lambda }_1+\cdots +{\lambda }_r\}$ are disjoint intervals of sizes ${\lambda }_r$ covering $\{1,\dots ,n\}$ and increase along each row, and $t_0(r,c)={\lambda }_1+\cdots +{\lambda }_{r-1}+c<t_0(r+1,c)={\lambda }_1+\cdots +{\lambda }_r+c$, so the entries strictly increase down every column. Hence $t_0$ is a standard $\lambda$-tableau.

The stabilizer $R_{t_0}$ of $\{t_0\}$ consists of the permutations that preserve each row set of $t_0$, and such a $\rho$ acts on a filling $f$ by $(\rho \cdot f)(x)=f(x^{\mathrm{\prime }\mathrm{\prime }})$ with $t_0(x^{\mathrm{\prime }\mathrm{\prime }})={\rho }^{-1}(t_0(x))$ by [F1]; since $t_0$ carries the labels ${\lambda }_1+\cdots +{\lambda }_{r-1}+1,\dots ,{\lambda }_1+\cdots +{\lambda }_r$ in row $r$, the box $x^{\mathrm{\prime }\mathrm{\prime }}$ lies in the same row of $[\lambda ]$ as $x$. So the row orbit $R_{t_0}\cdot f$ consists exactly of the fillings obtained from $f$ by permuting the entries within each row of $[\lambda ]$, and two fillings in the same row orbit have the same multiset of entries in every row.

For $f\in T_{\lambda ,\mu }$ define $N_f(i,j):=\mathrm{\#}\{(r,c):c\le j,f(r,c)\le i\}$ for $1\le i\le {\mu }_1^{\mathrm{\prime }}$ and $1\le j\le {\lambda }_1$, and set $N_f(0,j)=N_f(i,0)=0$ on the boundary. Because the number of entries equal to $i$ in column $c$ is $(N_f(i,c)-N_f(i-1,c))-(N_f(i,c-1)-N_f(i-1,c-1))$, the vector $N_f$ determines, and is determined by, the ordered tuple of the multisets of entries in the columns of $[\lambda ]$; thus $N_f=N_g$ holds exactly when every column of $g$ is a rearrangement of the corresponding column of $f$, and the relation $f⪯g$ defined by $N_f(i,j)\le N_g(i,j)$ for all $i,j$ is a preorder on the finite set $T_{\lambda ,\mu }$.

Let $T$ be a semistandard $\lambda$-tableau of content $\mu$ and let $i\ge 1$. The set $D_i:=\{(r,c)\in [\lambda ]:T(r,c)\le i\}$ of boxes with entries $\le i$ is closed to the left and upwards: if $(r,c)\in D_i$ and $c\ge 2$ then $T(r,c-1)\le T(r,c)\le i$ by weak increase along rows, and if $r\ge 2$ then $T(r-1,c)<T(r,c)\le i$ by strict increase down columns; hence $D_i$ is a Young diagram inside $[\lambda ]$ by [F6]. Moreover $(r,c)\in D_i$ implies $r\le i$, since the entries strictly increase down a column, so $1\le T(1,c)<\cdots <T(r,c)\le i$ gives $r\le T(r,c)\le i$. So $D_i$ lies in the first $i$ rows and has ${\mu }_1+\cdots +{\mu }_i$ boxes by the content condition [F2], whence ${\mu }_1+\cdots +{\mu }_i\le {\lambda }_1+\cdots +{\lambda }_i$ for all $i\ge 1$ (both prefixes equal $n$ for $i$ at least the number of parts of $\mu$), that is $\lambda ⊵\mu$: this proves claim 2. If moreover $\mu =\lambda$, then $\mathrm{\mid }{D}_i\mathrm{\mid }={\lambda }_1+\cdots +{\lambda }_i$ is the number of boxes in the first $i$ rows of $[\lambda ]$; a left- and upward-closed subdiagram with the same number of boxes as its ambient diagram equals it, so $D_i$ is exactly those first $i$ rows, a box in row $r$ carries an entry $\le r$ but not $\le r-1$, namely $r$, and $T$ is the filling whose $r$-th row is constant with entry $r$; conversely that filling is semistandard of shape and content $\lambda$, so $K_{\lambda ,\lambda }=1$, proving claim 3.

For $\gamma \in C_{t_0}$ the box $x^{\mathrm{\prime }}$ with $t_0(x^{\mathrm{\prime }})={\gamma }^{-1}(t_0(x))$ lies in the same column of $[\lambda ]$ as $x$ by step 1.1, since $\gamma$ preserves the column sets of $t_0$ by [F3]; hence $\gamma$ acts on fillings by permuting the entries within each column of $[\lambda ]$, and conversely every such columnwise permutation of labels lies in $C_{t_0}$. Therefore $N_{\gamma \cdot f}=N_f$ for all $\gamma \in C_{t_0}$ and $f\in T_{\lambda ,\mu }$ by step 1.3, and if $N_g=N_f$ then $g=\gamma \cdot f$ for some $\gamma \in C_{t_0}$.

Let $T\in T_{\lambda ,\mu }$ be semistandard and let $f$ be a filling obtained from $T$ by permuting the entries within rows. Fix $i$: in each row $r$ of $T$ the entries $\le i$ form an initial segment, since $T(r,1)\le T(r,2)\le \cdots$, so the number of entries $\le i$ in the first $j$ columns of row $r$ of $T$ is $\min (j,m_r(i))$ with $m_r(i):=\mathrm{\#}\{c:T(r,c)\le i\}$, while the same count for $f$ is at most $\min (j,m_r(i))$, because $f$ has the same entries in row $r$ as $T$ by step 1.2. Summing over rows gives $N_f(i,j)\le N_T(i,j)$ for all $i,j$, that is $f⪯T$. If moreover $N_f=N_T$, then for each row $r$ and all $i,j$ we have $\mathrm{\#}\{c\le j:f(r,c)\le i\}=\min (j,m_r(i))$, and induction on $j$ gives $f(r,j)=T(r,j)$: assuming $f(r,c)=T(r,c)$ for $c<j$, the difference of the identities for $j$ and $j-1$ yields $[f(r,j)\le i]=\min (j,m(i))-\min (j-1,m(i))=[m(i)\ge j]=[T(r,j)\le i]$ for every $i$, and a value is determined by the thresholds that dominate it. As $r$ was arbitrary, $f=T$; so among the fillings of the row orbit $R_{t_0}\cdot T$ the filling $T$ is the unique one with $N_f=N_T$.

Two distinct semistandard fillings $T,T^{\mathrm{\prime }}$ of content $\mu$ satisfy $N_T\mathrm{\ne }{N}_{T^{\mathrm{\prime }}}$: if $N_T=N_{T^{\mathrm{\prime }}}$ then by step 1.3 each column of $T^{\mathrm{\prime }}$ is a rearrangement of the corresponding column of $T$, and each column of a semistandard filling is strictly increasing, hence determined by its multiset, so $T=T^{\mathrm{\prime }}$.

For any finite $\mathbb{C}$-linear combination $F={\sum }_T{a}_T{\theta }_T$ of the maps attached to semistandard fillings, $S_n$-linearity of the ${\theta }_T$, the identity $e_{t_0}={\kappa }_{t_0}\cdot \{t_0\}$ of [F3] and the defining value of ${\theta }_T$ in [F1] give the identity $F(e_{t_0})={\kappa }_{t_0}\cdot \left({\sum }_T{a}_T{\sum }_{f\in R_{t_0}\cdot T}f\right)={\sum }_T{a}_T{\sum }_{f\in R_{t_0}\cdot T}{\kappa }_{t_0}\cdot f$ in $M^{\mu }$, the outer sums being finite because $T_{\lambda ,\mu }$ is finite by [F2].

For $f\in T_{\lambda ,\mu }$ the expansion of ${\kappa }_{t_0}\cdot f={\sum }_{\gamma \in C_{t_0}}\mathrm{sgn}(\gamma )\gamma \cdot f$ in the filling basis involves, by step 2.1, only fillings $g$ with $N_g=N_f$; so the coefficient of a filling $g$ in ${\kappa }_{t_0}\cdot f$ is zero unless $N_g=N_f$, and the coefficient of $f$ itself is ${\sum }_{\gamma \cdot f=f}\mathrm{sgn}(\gamma )$. The latter is $1$ when $f=T$ is semistandard: a nontrivial $\gamma \in C_{t_0}$ permutes two entries of some column of $T$, whereas $T$ has distinct entries in every column, so only $\gamma =1$ fixes $T$ and $\mathrm{sgn}(1)=1$ by [F7].

Suppose that ${\sum }_T{a}_T{\theta }_T=0$ with not all $a_T=0$, and among the semistandard $T$ with $a_T\mathrm{\ne }0$ choose $T_{\ast }$ whose vector $N_{T_{\ast }}$ is maximal in the preorder of step 1.3: whenever $N_{T_{\ast }}(i,j)\le N_T(i,j)$ for all $i,j$ and $a_T\mathrm{\ne }0$, then $N_T=N_{T_{\ast }}$; such $T_{\ast }$ exists because the semistandard fillings of content $\mu$ form a finite set by [F2]. In the filling-basis expansion of $F(e_{t_0})={\sum }_T{a}_T{\sum }_{f\in R_{t_0}\cdot T}{\kappa }_{t_0}\cdot f$ from step 2.4, the coefficient of $T_{\ast }$ is exactly $a_{T_{\ast }}\mathrm{\ne }0$: a term ${\kappa }_{t_0}\cdot f$ with $f\in R_{t_0}\cdot T$ can contribute only if $N_f=N_{T_{\ast }}$ by step 3.1, while $N_f⪯N_T$ by step 2.2, so maximality gives $N_T=N_{T_{\ast }}$ and then $T=T_{\ast }$ by step 2.3; within the orbit $R_{t_0}\cdot T_{\ast }$ the condition $N_f=N_{T_{\ast }}$ forces $f=T_{\ast }$ by step 2.2, and the coefficient of $T_{\ast }$ in ${\kappa }_{t_0}\cdot T_{\ast }$ is $1$ by step 3.1.

Hence every nonzero combination $F={\sum }_T{a}_T{\theta }_T$ satisfies $F(e_{t_0})\mathrm{\ne }0$ by steps 2.4 and 4.1, so the restrictions ${\theta }_T{\mathrm{\mid }}_{S^{\lambda }}$ for distinct semistandard $T$ are linearly independent; taking a combination with a single nonzero coefficient shows that each ${\theta }_T{\mathrm{\mid }}_{S^{\lambda }}$ is nonzero. Since these restrictions lie in ${\mathrm{Hom}}_{S_n}(S^{\lambda },M^{\mu })$ by [F4], that space has dimension at least the number of semistandard fillings, which is $K_{\lambda ,\mu }$ by [F2]. This proves claim 1, and claims 2 and 3 are step 1.4.

Claims 1, 2 and 3 are steps 5.1 and 1.4. No division and no choice principle is used: the order of step 1.3 compares finitely many integer vectors attached to the finitely many fillings of content $\mu$, and $T_{\ast }$ is a maximal element of a finite set.