\lambda )\!\uparrow^P_{P_{\nu}}) = (e - 1) \lambda\!\uparrow_{G_{\nu}}^{G} -\sum_{i = 2}^{e}\varepsilon_i (({{\chi}_i \downarrow_{G_{\nu}})\lambda})\!\uparrow_{G_{\nu}}^{G} + \ \varepsilon ({{\chi}_{\nu}
\lambda})\!\uparrow_{G_{\nu}}^{G}$$ for any $\lambda \in $[Irr]{}$(P_{\nu}/\tilde{P}) = $[Irr]{}$(G_{\nu}/\tilde{G})$.

It suffices to show $1_{G} * ((\eta_{\hat{\nu}} \lambda )\!\uparrow^P_{P_{\nu}}) = (1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda)) \uparrow^G_{G_{\nu}}$. Let $P = \bigcup_{j = 1}^{l} v_j P_{\nu}$ (disjoint) and let $u \in P_{\nu}$. Since $G = \bigcup_{j = 1}^{l}v_j  G_{\nu}$ (disjoint) and $P_{\nu}$ is normal in $P$, for $\rho \in C_{G}(u)_{p'}$, $$(1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda))\uparrow^G_{G_{\nu}}\!(u\rho) = \sum_{j = 1}^{l} (1_{G_{\nu}} * (\eta_{\hat{\nu}} \lambda))((u\rho)^{v_j})$$ $$= \sum_{j = 1}^{l}( \eta_{\hat{\nu}} \lambda)(u^{v_j}) =  
((\eta_{\hat{\nu}} \lambda )\!\uparrow^P_{P_{\nu}})(u) = (1_{G} * ((\eta_{\hat{\nu}} \lambda)\!\uparrow^P_{P_{\nu}}))(u\rho).$$ This completes the proof. $\blacksquare$

\
For $\nu,$ $ \nu' \in \tilde{{\cal X}}$, $\tilde{\chi}_{\nu}$ and $\tilde{\chi}_{\nu'}$ are $G$-conjugate if and only if those are $P$-conjugate if and only if $\nu $ and $\nu'$ are $(P\rtimes E)$-conjugate. Since $b(G)$ is the unique block of $G$ which covers $b(\tilde{G})$, we have the following.

Let ${\cal X} (\subseteq \tilde{{\cal X}})$ be a set of representatives for the $(P \rtimes E)$-conjugacy classes of [Irr]{} $(\tilde{P}) \backslash \{ 1_{\tilde{P}} \}$. $${\rm Irr}(b(G)) = \bigcup_{i = 1}^e \Big\{ \chi_i \lambda \ | \  \lambda \in {\rm Irr}(G/\tilde{G}) \Big\}
\bigcup_{\nu \in {\cal X}} \Big\{ (\chi_{\nu} \lambda)\!\uparrow_{G_{\nu}}^{G} \ | \ 