 Solution

Let $V$ be an irreducible $G$-module with character $\chi \in \mathrm{Irr}(G\mid \theta )$, and let $W=V_{\theta }\ne 0$ be its $\theta$-isotypic component. It is invariant under the inertia group of a character $T$. The sum of its $G$-translates is a nonzero $G$-submodule, so irreducibility makes it all of $V$. Translates associated with different cosets $gT$ have distinct $N$-types and hence form a direct sum:To see that $W$ is irreducible over $T$, let $0\ne W_0\subseteq W$ be a $T$-submodule. Then ${\sum }_{g\in \mathcal{R}}g{W}_0$ is a nonzero $G$-submodule of $V$, hence equals $V$. Its $\theta$-isotypic part is exactly $W_0$, whereas the corresponding part of $V$ is $W$. Thus $W_0=W$.

Let $\xi$ be the irreducible character of $W$ as a $T$-module. It lies over $\theta$, and the natural map of induced representationsis an isomorphism: it identifies each summand with the corresponding $gW$, and these summands are direct and exhaust $V$. Consequently $\chi ={\xi }^G$, proving surjectivity of induction on the stated sets.

For injectivity, the construction in (a) shows that the $\theta$-isotypic component of ${\mathrm{Ind}}_T^{G}W$ is exactly $1\otimes W$, with its original $T$-action. Isomorphic induced $G$-modules therefore have isomorphic intrinsic $\theta$-components as $T$-modules. Since complex representations with the same character are isomorphic, ${\xi }_1^G={\xi }_2^G$ forces ${\xi }_1={\xi }_2$. Hence the Clifford correspondence is the bijectionwhose inverse takes the character of the $\theta$-isotypic component.