91Ó°ÊÓ

First, we review the definitions needed to understand the bifunctors and the mappings. 1. A **group** \((G, \cdot)\) is a set \(G\) endowed with a binary operation \(\cdot\) that satisfies the following properties: closure, associativity, existence of an identity element, and existence of an inverse element. 2. A **subgroup** \(S\) of a group \(G\) is a group itself and is a subset of \(G\) such that it is closed under the operation of \(G\). 3. A **module** is an algebraic structure that generalizes the notion of vector spaces, where elements of the group (or ring) can be multiplied by elements of the module. Here, \(\operatorname{Mod}(G)\) and \(\operatorname{Mod}(S)\) represent the category of \(G\)-modules and \(S\)-modules, respectively. The elements are sets with additional structure and operations. 4. A **bifunctor** is a functor that has two arguments; here, our bifunctors map pairs of modules to another category of modules, \(\operatorname{Mod}(\mathbf{Z})\), where \(\mathbf{Z}\) represents the ring of integers. 5. In the given bifunctors, we have the notation \(\operatorname{Hom}_{G}(A, M_{G}^{S}(B))\) and \(\operatorname{Hom}_{S}(A, B)\). The \(\operatorname{Hom}\) notation refers to the set of all homomorphisms between the modules, the first subscript indicates the group that the homomorphisms respect. 6. The induced representation \(M_{G}^{S}(B)\) is a module of \(G\) that is induced from the module \(B\) of \(S\). This is done based on Frobenius' Reciprocity Theorem.

Given mappings are defined as follows: 1. \(\varphi \mapsto\left(a \mapsto g_{a}\right)\), for \(\varphi \in \mathrm{Hom}_{S}(A, B)\), where \(g_{a}(\sigma)=\varphi(\sigma a), g_{a} \in M_{G}^{S}(B) .\) 2. \(f \mapsto f(1)\) with \(f \in \operatorname{Hom}_{G}\left(A, M_{G}^{S}(B)\right)\).

To show that the bifunctors are isomorphic, we need to show that the given mappings are inverses of each other and that they are module homomorphisms. We will start with the first mapping. 1. Let \(\varphi \in \mathrm{Hom}_{S}(A, B)\). Then the first mapping sends \(\varphi\) to \(a \mapsto g_{a}\), where \(g_{a}(\sigma)=\varphi(\sigma a), g_{a} \in M_{G}^{S}(B) .\) 2. Now let's apply the second mapping to this result: \(f \mapsto f(1)\) with \(f=g_{a}.\) We have \(g_{a}(1)=\varphi( 1 \cdot a)=\varphi(a)\). Thus, the second mapping sends \(a \mapsto \varphi(a)\). 3. So, by applying the second mapping on the result of the first mapping, we get \(\varphi \mapsto a \mapsto \varphi(a)\). This shows that the two mappings are indeed inverses of each other. Now, we need to check if the mappings are module homomorphisms, meaning they preserve the module structure: 1. For the first mapping: Let \(\varphi_1, \varphi_2 \in \mathrm{Hom}_{S}(A, B)\). Then, \((\varphi_1+\varphi_2)(a) = \varphi_1(a)+\varphi_2(a)\). Apply the first mapping to the sum, we get: \(g_{a}(\sigma)=(\varphi_1+\varphi_2)(\sigma a) = \varphi_1(\sigma a)+\varphi_2(\sigma a)\). This is equal to the sum of the individual images under the first mapping. 2. For the second mapping: Let \(f_1, f_2 \in \operatorname{Hom}_{G}\left(A, M_{G}^{S}(B)\right)\). Then, \((f_1+f_2)(1) = f_1(1)+f_2(1)\). Apply the second mapping to the sum, we get: \((f_1(1)+f_2(1))(a) = f_1(1)(a)+f_2(1)(a)\). This is equal to the sum of the individual images under the second mapping. Since the mappings are inverses of each other and are both module homomorphisms, we conclude that the bifunctors \((A, B) \mapsto \operatorname{Hom}_{G}\left(A, M_{G}^{S}(B)\right)\) and \((A, B) \mapsto \operatorname{Hom}_{S}(A, B)\) are isomorphic.