Universal Property of Free Module on Set

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Theorem

Let $R$ be a ring with unity.

Let $R^{\paren I}$ be the free $R$-module on $I$.

Let $M$ be an $R$-module.

Let $\family {m_i}_{i \mathop \in I}$ be a family of elements of $M$.


Then there exists a unique $R$-module morphism:

$\Psi: R^{\paren I} \to M$

that sends the $i$th canonical basis element to $m_i$, for all $i \in I$.


Moreover:

$\ds \map \Psi {\family {r_i}_{i \mathop \in I} } = \sum_{i \mathop \in I} r_i m_i$


Proof

Existence

By Morphism from Ring with Unity to Module, for all $i$ there exists a morphism $\psi_i: R \to M$ with $\map {\psi_i} 1 = m_i$.

By Universal Property of Direct Sum of Modules, there exists a morphism $\Psi: R^{\paren I}\to M$ such that $\Psi \circ \iota_i = \psi_i$ for all $i$.

Thus:

$\forall i \in I: \map \Psi {e_i} = \map \Psi {\map {\iota_i} 1} = \map {\psi_i} 1 = m_i$

We have $\ds \family {r_i}_{i \mathop \in I} = \sum_{i \mathop \in I} r_i e_i$, 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.

$\blacksquare$