# Universal Property of Free Abelian Group on Set

## Theorem

Let $S$ be a set.

Let $\struct {\Z^{\paren S}, \iota}$ be the free abelian group on $S$.

Let $\struct {G, +}$ be an abelian group.

Let $f: S \to G$ be a mapping.

Then there exists a unique group homomorphism $g : \Z^{\paren S} \to G$ with $g \circ \iota = f$:

$\xymatrix{ S \ar[d]_\iota \ar[r]^{\forall f} & G\\ \Z^{\paren S} \ar[ru]_{\exists ! g} }$

## Proof

For $x \in S$, the characteristic map $\phi_x$ of $x$ is defined as:

$\map {\phi_x} s = \begin{cases} 1 & \quad \text{ if } s = x \\ 0 & \quad \text{ if } s \ne x \end{cases}$

The maps $\set {\phi_x \ : \ x \in S}$ form a basis for $\Z^{\paren S}$.

If $\phi \in \Z^{\paren S}$, then it can be expressed uniquely as a finite sum

$\ds \phi = \sum_i n_i \phi_{x_i}$

where $\phi$ maps each $x_i$ to $n_i \in \Z$, and maps each element of $S$ not present in this sum to $0$.

$\iota$ maps $x \in S$ to $\phi_x \in \Z^{\paren S}$.

So, in order to satisfy $g \circ \iota = f$, we must have $\map g {\phi_x} = \map f x$ for every $x \in S$.

Since $g$ has to be a homomorphism, it must satisfy $\map g {\phi + \phi'} = \map g \phi + \map g {\phi'}$.

So, for $\phi \in \Z^{\paren S}$:

$\ds \map g \phi = \map g {\sum_i n_i \phi_{x_i} } = \sum_i \map g {n_i \phi_{x_i} } = \sum_i n_i \map g {\phi_{x_i} } = \sum_i n_i \map f {x_i}$

Therefore, the value of $\map g \phi$ is unique, for every $\phi \in \Z^{\paren S}$.

So, the homomorphism $g$ that satisfies $g \circ \iota = f$ is unique.