Kernel of $\phi \colon $Hom$_R(\prod_{i \in I}A_i, B) \to \prod_{i \in I}\text{Hom}_R(A_i, B)$ is isomorphic to Hom$_R(\prod A_i/\sum^\oplus A_i, B)$.

Question

Let $R$ be a ring, $\{A_i\}, B$ be $R$-modules. Let $\iota_i \colon A_i \to \prod_{i \in I} A_i$ be the canonical injection. Let $\phi \colon \text{Hom}_R(\prod_{i \in I}A_i, B) \to \prod_{i \in I}\text{Hom}_R(A_i, B)$ be given by $\phi(f) = (f_i)_{i \in I}$, where $f_i = f \circ \iota_i$. Prove that $\text{Ker}\phi \cong \text{Hom}_R(\prod A_i/\sum^\oplus A_i, B)$. I know that in general, the contravariant Hom-functor does not preserve the direct products. So, $\phi$ is not an isomorphism. I was trying to find an isomorphism between $\text{Ker}\phi$ and $\text{Hom}_R(\prod A_i/\sum^\oplus A_i, B)$, but it seems not easy to find one. I am guessing that I need to use the universal property of direct products??? If someone can direct me how to prove this, I will be appreciated.

Answer 1

Hint: If $M$ and $P$ are $R$-modules, and $N$ is a submodule of $M$, homomorphisms $M/N \to P$ are in a one-to-one corresponde with homomorphisms $M \to P$ whose kernel contains $N$, that is, if $\pi : M \to M/N$ is the canonical projection, then the map $$\operatorname{Hom}_R(M/N,P) \to \{\alpha \in \operatorname{Hom}_R(M,P) : N \subseteq \ker \alpha\}; \quad \phi \mapsto \phi \circ \pi$$ is a bijection, in fact, an isomorphism of $\mathbb Z$-modules ($R$-modules if $R$ is commutative).