# Can any $\theta \in \text{Hom}(S^\lambda,M^\mu)$ be written as $\theta = \kappa_t$?

I am currently working on Bruce Sagan's The Symmetric Group.

I am struggling to understand why the following proposition should be true.

Suppose that the field of scalars is $\mathbb{C}$ and $\theta \in \text{Hom}(S^\lambda,M^\mu)$ is nonzero. Thus $\lambda\trianglerighteq\mu$, and if $\lambda = \mu$, then $\theta$ is multiplication by a scalar.

In the case $\lambda = \mu$, Corollary 2.4.3 yields $\theta(\mathbf{e}_t) = c\mathbf{e}_t$ for some constant $c$. So for any permutation $\pi$, $\theta(\mathbf{e}_{\pi t}) = \theta(\pi\mathbf{e}_t) = \pi\theta(\mathbf{e}_t) = \pi(c\mathbf{e}_t) = c\mathbf{e}_{\pi t}$. Thus $\theta$ is multiplication by $c$.

Corollary 2.4.3: If $\mathbf{u}\in M^\mu$ and sh $t = \mu$, then $\kappa_t\mathbf{u}$ is a multiple of $\mathbf{e}_t$.

Why can we apply this corollary here? The corollary is concerned with $\kappa_t = \sum \limits_{\pi \in C_t} \text{sgn}(\pi)\pi$ (where $C_t$ is the column-stabiliser of a tableau $t$), whereas for this proposition we are considering an arbitrary $\theta \in \text{Hom}(S^\lambda,M^\mu)$.

Thank you very much for your help!

Sagan has extended the map $\theta$ to $M^\lambda$. In $M^\lambda$, we have $\mathbf{e}_t = \kappa_t \mathbf{\left\{t\right\}}$, so that $\theta\left(\mathbf{e}_t\right) = \theta\left(\kappa_t \mathbf{\left\{t\right\}}\right) = \kappa_t \theta\left(\mathbf{\left\{t\right\}}\right)$ (since the extended map $\theta$ is a $\mathbb C\left[\mathcal{S}_n\right]$-module homomorphism). Now, apply Corollary 2.4.3 with $\mathbf{u} = \theta\left(\mathbf{\left\{t\right\}}\right)$.