Universal Property of Free Module on Set
Theorem
Let $R$ be a ring with unity.
Let $R^{\left({I}\right)}$ be the free $R$-module on $I$.
Let $M$ be an $R$-module.
Let $\left\langle{m_i}\right\rangle_{i \mathop \in I}$ be a family of elements of $M$.
Then there exists a unique $R$-module morphism:
- $\Psi: R^{\left({I}\right)}\to M$
that sends the $i$th canonical basis element to $m_i$, for all $i\in I$.
Moreover:
- $\displaystyle \Psi((r_i)_{i \mathop \in I}) = \sum_{i \mathop \in I} r_i m_i$
Proof
Existence
By Morphism from Ring with Unity to Module, for all $i$ there exists a morphism $\psi_i:R\to M$ with $\psi_i(1)=m_i$.
By Universal Property of Direct Sum of Modules, there exists a morphism $\Psi:R^{(I)}\to M$ such that $\Psi\circ\iota_i=\psi_i$ for all $i$.
Thus $\Psi(e_i)=\Psi(\iota_i(1))=\psi_i(1)=m_i$ for all $i$.
We have $(r_i)_{i\in I}=\sum_{i\in I}r_ie_i$, so the expression for $\Psi$ follows by linearity.
Uniqueness
Let $\Psi$ be such a morphism.
Then $\Psi\circ\iota_i$ sends $1$ to $m_i$. By Morphism from Ring with Unity to Module, $\Psi\circ\iota_i=\psi_i$, with $\psi_i$ as above.
By Universal Property of Direct Sum of Modules, $\Psi$ is determined by $\Psi\circ\iota_i$.
Thus $\Psi$ is unique.
$\blacksquare$