Let $R$ be a ring and let $A_1, \ldots, A_n$ be $R$-modules.
I have to show that there are $R$-homomorphisms $e_i : \prod\limits_{i =1}^n A_i \rightarrow \prod\limits_{i =1}^n A_i$ for $i = 1, \ldots, n$ satisfying the following conditions :
(I) $\quad$ $e_i \circ e_i = e_i$ for all $i = 1, \ldots ,n$
(II) $\ \ \ \ e_i \circ e_j = 0$ for all $1 \leq i, j \leq n$ and $i \neq j$
(III) $\ \ \ id_A = e_1 + \cdots + e_n$
My problem is that I have no idea how to construct such a map $e_i$.
Furthermore, I have to prove that $\varphi : A \rightarrow \prod\limits_{i =1}^n A_i$ is an $R$-isomorphism, where $A$ is an $R$-module, $e_1, \ldots, e_n \in Hom_R(A, A)$ a complete set of orthogonal idempotents of $M$ and $A_i = e_i(A)$.
It is clear to me how to prove that a map between modules is an $R$-isomorphism, but in this case I may need some hints how $\varphi$ is defined.
Thanks for your help.
Since the first part is obvious, I answer the second part of the question.
By the universal property of $\prod\limits_{i = 1}^n A_i$, there exists a unique homomorphism $\phi : A \rightarrow \prod\limits_{i = 1}^n A_i$ such that $\pi_i \circ \phi = e_i$ for all $i \in \{1, \ldots ,n\}$. By condition III we have $(e_1 + \cdots + e_n) \circ \phi = id_A$.
It remains to show that $\phi$ is a bijection.
Let $x \in ker(\phi)$. Then $(\pi_1 + \cdots + \pi_n) \circ \phi(x) = x$ implies that $x = 0$. Hence, $\phi$ is injective.
We have $A_i = e_i(A)$. Since $\pi_i$ is surjective, $\pi_i(\prod\limits_{i = 1}^n A_i) = A_i$. Moreover, $(\pi_i \circ \phi)(A) = e_i(A)$. It follows that $\pi_i(\prod\limits_{i = 1}^n A_i) = \pi_i(\phi(A))$ for all $i = 1, \ldots,n$.
Finally, $\prod\limits_{i = 1}^n A_i = \phi(A) \Rightarrow \phi$ is surjective.