Universal Property of Free Module on Set

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Theorem

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:

$\displaystyle \Psi((r_i)_{i \mathop \in I}) = \sum_{i \mathop \in I} r_i m_i$

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.

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.

$\blacksquare$