Proposition 5

Let A be a ring, and let $\mathscr{S}$ be the set of classes of simple A-modules. Let M and N be A-modules.

a) Let $f: M\rightarrow N$ be a homomorphism. For every $\lambda \in \mathscr{S},f$ induces a homomorphism $f_{\lambda}$ from $M_{\lambda}$ to $N_{\lambda}$; if M is semisimple and $f$ surjective, then each of the homomorphisms $f_{\lambda}$ is surjective.

b) Suppose that M is semisimple. The mapping $f\mapsto (f_{\lambda})_{\lambda\in\mathscr{S}}$ is a group isomorphism from Hom$_A(M,N)$ to $\prod_{\lambda\in\mathscr{S}}$ Hom$_A(M_{\lambda},N_{\lambda})$. When M is equal to N, the mapping is a ring isomorphism from End$_A(M)$ to $\prod_{\lambda\in\mathscr{S}}$ End$_A(M_{\lambda})$.

For every $\lambda \in \mathscr{S}$, the submodule $f(M_{\lambda})$ of N is isomorphic to a quotient of an isotypical module of type $\lambda$; it is therefore isotypical of type $\lambda$ and, consequently, contained in $N_{\lambda}$.

Suppose that M is semisimple and $f$ surjective. Then $f$ induces an isomorphism from $M/$ Ker($f$) to N that sends $(M_{\lambda}$+Ker($f$))$/$ Ker($f$) to $f(M_{\lambda})$. By Proposition 4 of VIII, p. 65, we have $N_{\lambda}=f(M_{\lambda})$, which completes the proof of a).

The mapping considered in b) is clearly a group homomorphism, and it is a ring homomorphism when M is equal to N. Let $(f_{\lambda})_{\lambda\in\mathscr{S}}$ be an element of $\prod_{\lambda\in\mathscr{S}}$ Hom(M$_{\lambda},N_{\lambda}$). The unique element of its inverse image under the mapping in b) is the homomorphism $f: M\rightarrow N$ defined by

for every $(x_{\lambda})_{\lambda\in\mathscr{S}}\in \bigoplus_{\lambda}M_{\lambda}$.