[proofplan] We write $s_\lambda$ using the Jacobi-Trudi determinant in the complete homogeneous symmetric functions. Since $\omega$ is a ring homomorphism and sends each $h_r$ to $e_r$, applying $\omega$ to that determinant replaces every entry by the corresponding elementary symmetric function. The resulting determinant is exactly the dual Jacobi-Trudi determinant for $s_{\lambda'}$, with the standard conventions $h_0=e_0=1$ and $h_r=e_r=0$ for $r<0$ handling all boundary entries. [/proofplan] [step:Choose a Jacobi-Trudi determinant for $s_\lambda$] Let $\lambda=(\lambda_1,\dots,\lambda_\ell)$ be a partition, where $\ell=\ell(\lambda)$ denotes its number of nonzero parts. We use the standard conventions \begin{align*} h_0=e_0=1, \qquad h_r=e_r=0 \quad \text{for every } r<0. \end{align*} By the [Jacobi-Trudi identity](/theorems/5181) (citing a result not yet in the wiki: Jacobi-Trudi Identity), the Schur function $s_\lambda \in \Lambda$ is given by \begin{align*} s_\lambda=\det\bigl(h_{\lambda_i-i+j}\bigr)_{1\leq i,j\leq \ell}. \end{align*} [/step] [step:Apply $\omega$ entrywise to the determinant] Define the $\ell \times \ell$ matrix $H_\lambda$ over $\Lambda$ by \begin{align*} (H_\lambda)_{ij}=h_{\lambda_i-i+j} \end{align*} for $1\leq i,j\leq \ell$. Since $\omega:\Lambda\to\Lambda$ is a ring homomorphism, it commutes with the determinant polynomial: \begin{align*} \omega(\det H_\lambda) &= \omega\left(\sum_{\sigma\in S_\ell}\operatorname{sgn}(\sigma) \prod_{i=1}^{\ell} h_{\lambda_i-i+\sigma(i)}\right) \\ &= \sum_{\sigma\in S_\ell}\operatorname{sgn}(\sigma) \prod_{i=1}^{\ell} \omega\left(h_{\lambda_i-i+\sigma(i)}\right). \end{align*} Using $\omega(h_r)=e_r$ for $r\geq 0$ and the shared convention $h_r=e_r=0$ for $r<0$, this becomes \begin{align*} \omega(s_\lambda) = \det\bigl(e_{\lambda_i-i+j}\bigr)_{1\leq i,j\leq \ell}. \end{align*} [guided] The determinant is a polynomial expression in its entries, with integer coefficients. This matters because $\omega$ is a ring homomorphism, so it preserves addition, multiplication, and integer scalar multiplication. Explicitly, if $S_\ell$ denotes the symmetric group on $\{1,\dots,\ell\}$, then \begin{align*} \det H_\lambda = \sum_{\sigma\in S_\ell}\operatorname{sgn}(\sigma) \prod_{i=1}^{\ell} h_{\lambda_i-i+\sigma(i)}. \end{align*} Applying $\omega$ gives \begin{align*} \omega(\det H_\lambda) &= \sum_{\sigma\in S_\ell}\operatorname{sgn}(\sigma) \prod_{i=1}^{\ell} \omega\left(h_{\lambda_i-i+\sigma(i)}\right). \end{align*} For every index $r=\lambda_i-i+\sigma(i)$ with $r\geq 0$, the defining property of the omega involution gives $\omega(h_r)=e_r$. If $r<0$, then the determinant convention sets $h_r=0$ and $e_r=0$, so the same replacement remains valid. Therefore every entry $h_{\lambda_i-i+j}$ is replaced by $e_{\lambda_i-i+j}$, and hence \begin{align*} \omega(s_\lambda) = \det\bigl(e_{\lambda_i-i+j}\bigr)_{1\leq i,j\leq \ell}. \end{align*} [/guided] [/step] [step:Identify the elementary determinant with $s_{\lambda'}$] By the [dual Jacobi-Trudi identity](/theorems/5199) (citing a result not yet in the wiki: Dual Jacobi-Trudi Identity), applied to the partition $\lambda$, the Schur function of the conjugate partition $\lambda'$ satisfies \begin{align*} s_{\lambda'} = \det\bigl(e_{\lambda_i-i+j}\bigr)_{1\leq i,j\leq \ell}. \end{align*} Comparing this identity with the determinant obtained above gives \begin{align*} \omega(s_\lambda)=s_{\lambda'}. \end{align*} This proves the theorem for every partition $\lambda$. [/step]