The restriction coproduct is Schur skewing

For each partition $\lambda$, the set of subpartitions $\mu \subseteq \lambda$ is finite, so the displayed sum defining ${\mathrm{\Delta }}_{\mathrm{\Lambda }}(s_{\lambda })$ is an element of $\mathrm{\Lambda }\otimes \mathrm{\Lambda }$. The degreewise Schur basis [F10] and the direct-sum grading [F11] give a unique $\mathbb{Z}$-linear extension to all of $\mathrm{\Lambda }$. By the skew expansion [F9], the support condition [F8], and the definition of the skew Schur function [F13], this extension has the finite coefficient form ${\mathrm{\Delta }}_{\mathrm{\Lambda }}(s_{\lambda })={\sum }_{a+b=\mathrm{\mid }\lambda \mathrm{\mid }}{\sum }_{\mu ⊢a,\nu ⊢b}{c}_{\mu \nu }^{\lambda }{s}_{\mu }\otimes s_{\nu }$.

Fix $\lambda ⊢n$ and $a+b=n$, and let $\theta ={\iota }_{a,b}^{\ast }({\mathrm{Res}}_{H_{a,b}}^{S_n}{\chi }^{\lambda })\in R(S_a\times S_b)$, as in [F2]. By [F3], this honest character is a nonnegative integral sum of the orthonormal external-product basis characters. Thus the coefficient of ${\chi }^{\mu }⊠{\chi }^{\nu }$ in $\theta$ is $⟨{\chi }^{\mu }⊠{\chi }^{\nu },\theta {⟩}_{S_a\times S_b}$, because that coefficient is an integer and the basis is orthonormal.

Frobenius reciprocity [F4] identifies that coefficient with $⟨{\mathrm{Ind}}_{H_{a,b}}^{S_n}(({\chi }^{\mu }⊠{\chi }^{\nu })\circ {\iota }_{a,b}^{-1}),{\chi }^{\lambda }{⟩}_{S_n}$. Under the explicit ordered-block identification in [F2], this induced character is the outer product from [F5]; the zero-based to one-based relabeling required by its definition is checked in the proof of [F5]. By [F5] and orthonormality [F6], the inner product is exactly $c_{\mu \nu }^{\lambda }$.

Since the external products form a basis by [F3], the coefficient calculation in step 2.1 gives ${\iota }_{a,b}^{\ast }({\mathrm{Res}}_{H_{a,b}}^{S_n}{\chi }^{\lambda })={\sum }_{\mu ⊢a,\nu ⊢b}{c}_{\mu \nu }^{\lambda }{\chi }^{\mu }⊠{\chi }^{\nu }$. Applying ${\mathrm{\Phi }}_{a,b}^{-1}$ as in [F2] yields ${\mathrm{\Delta }}_{a,b}({\chi }^{\lambda })={\sum }_{\mu ,\nu }{c}_{\mu \nu }^{\lambda }{\chi }^{\mu }\otimes {\chi }^{\nu }$.

Applying $\mathrm{ch}\otimes \mathrm{ch}$ to the component formula in step 3.1 and using [F7] gives ${\sum }_{a+b=n}{\sum }_{\mu ⊢a,\nu ⊢b}{c}_{\mu \nu }^{\lambda }{s}_{\mu }\otimes s_{\nu }$. By [F9] and the definition [F13], this is ${\sum }_{\mu \subseteq \lambda }{s}_{\mu }\otimes s_{\lambda \mathrm{/}\mu }={\mathrm{\Delta }}_{\mathrm{\Lambda }}(s_{\lambda })$, using step 1.1. Thus the characteristic identity holds for each ${\chi }^{\lambda }$.

Conversely, [F1], [F7], [F10], [F11], and [F12] show that $\mathrm{ch}\otimes \mathrm{ch}:R_S\otimes R_S\to \mathrm{\Lambda }\otimes \mathrm{\Lambda }$ is an isomorphism: it sends the basis tensors ${\chi }^{\mu }\otimes {\chi }^{\nu }$ bijectively to $s_{\mu }\otimes s_{\nu }$. Hence the characteristic identity for ${\chi }^{\lambda }$ determines each bidegree component uniquely. Applying the inverse tensor basis map and then ${\mathrm{\Phi }}_{a,b}$ from [F2] recovers the restriction formula in the Statement. This proves the reverse implication in the stated equivalence.

Every element of $R_S$ is a finite integral linear combination of the basis characters by [F1] and [F7], and both coproducts and the characteristic map are $\mathbb{Z}$-linear, so the identity extends to every $f\in R_S$. For $\lambda =\mathrm{\varnothing }$ the only term is $1\otimes 1$; when $a=0$ or $b=0$, the empty-factor coefficient in [F8] is one and [F2] gives the endpoint identity. Coefficients outside $\mu \subseteq \lambda$ or the required sizes vanish by [F8]. All sums and basis expansions are finite, so no choice principle is used.