The antisymmetrizer image in its own tabloid module is one-dimensional

The coefficient of $\{t\}$ in $e_t$ is $1$ by [F2], so $e_t\mathrm{\ne }0$.

Let $\{s\}$ be any basis tabloid. If ${\kappa }_t\cdot \{s\}=0$, its image already lies in $\mathbb{C}{e}_t$. Otherwise, [F5] implies that each row of $s$ meets each column of $t$ in at most one entry, and [F6] gives $\rho \in R_s$ and $\gamma \in C_t$ with $\rho \cdot s=\gamma \cdot t$; since $\rho$ stabilizes $\{s\}$ by [F4], this yields $\{s\}=\gamma \cdot \{t\}$.

For any $\gamma \in C_t$, reindex the defining sum by $d=c\gamma$ to obtain ${\kappa }_t\gamma ={\sum }_{c\in C_t}\mathrm{sgn}(c)c\gamma ={\sum }_{d\in C_t}\mathrm{sgn}(d{\gamma }^{-1})d=\mathrm{sgn}(\gamma ){\kappa }_t$, using [F1,F7] and $\mathrm{sgn}({\gamma }^{-1})=\mathrm{sgn}(\gamma )$; applying this to the tabloid equality of step 1.2 gives ${\kappa }_t\cdot \{s\}=\mathrm{sgn}(\gamma )e_t$ in its nonzero case. Thus every basis tabloid maps into $\mathbb{C}{e}_t$, and linearity with [F3] gives ${\kappa }_t{M}^{\lambda }\subseteq \mathbb{C}{e}_t$.

Since $e_t={\kappa }_t\cdot \{t\}$ by [F8], the vector $e_t$ belongs to the image ${\kappa }_t{M}^{\lambda }$; step 1.1 makes its span nonzero, so step 2.1 gives ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$ and proves the statement.