The free group is often defined by its universal property. A group $F$ is said to be free on a subset $S$ with inclusion map $\iota : S \rightarrow F$ if for every group $G$ and set map $\phi:S \rightarrow G$ there exists a unique homomorphism $\overline{\phi}:F \rightarrow G$ such that $\overline{\phi} \circ \iota (s) = \phi (s)$ $ \forall s \in S$.
It is said that the existence of $\overline{\phi}$ is what determines there are no relations. My question is (without defining the (free) group of reduced words) can you show there are no relations just by picking groups G to map into? That is, can you show no reduced words on $S^{\pm1}$ (excluding the empty word) are equal to the identity element in the free group?
Example attempt:
Suppose the reduced word $w$ is not the empty word, so it contains some letter $a$. Suppose adding all the powers of $a$ in the word $w$ gives the integer $k$. Define the set map $\phi : S \rightarrow \mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z} $ by
$\phi (s) = \left\{
\begin{array}{@{}l@{\thinspace}l}
0 & \text{ if } s \neq a\\
1 & \text{ if } s = a\\
\end{array}
\right.$
then the homomorphism from $F$ to $\mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z}$ extending the set map $\phi$ sends $w$ to $k$. As homomorphisms preserve group identities, and k is not the identity in $\mathbb{Z}/( \lvert k \rvert +1)\mathbb{Z}$, then $w$ is not the identity in $F$.
This was my first idea, but fails because it misses the case where $k=0$.
Best Answer
Let $u$ be a reduced word, for instance $u = ab\bar a$. Now, spell your word in the Cayley graph of the free group, for instance $$ 1 \xrightarrow{a} 2 \xrightarrow{b} 3 \xleftarrow{a} 4 $$ and complete $a$ and $b$ into permutations, for instance $a = (1,2)(3,4)$ and $b = (1)(2,3)(4)$. In the permutation group generated by $a$ and $b$, $u$ will not be the identity, since it maps $1$ to $4$.
This is basically the idea behind the Stallings automaton.
EDIT. Another example. Let $u = a^2b^{-3}cb^2$. Spelling $u$ yields the following diagram: $$ 1 \xrightarrow{a} 2 \xrightarrow{a} 3 \xleftarrow{b} 4 \xleftarrow{b} 5 \xleftarrow{b} 6 \xrightarrow{c} 7 \xrightarrow{b} 8 \xrightarrow{b} 9 $$ Now complete $a$, $b$ and $c$ into permutations, for instance $a = (1, 2, 3)$, $b = (6, 5, 4, 3)(7,8,9)$ and $c = (6,7)$.