Prove the uniqueness of a group homomorphism (extension from a generating set to original group)

Question

The following statement is from Lang, Algebra. p.10.

Let $G$ be a group, $S$ a set of generators for $G$, and $G'$ another group. Let $f: S \rightarrow G'$ be a map. If there exists a homomorphism $\overline{f}$ of $G$ into $G'$ whose restriction to $S$ is $f$, then there is only one.

Since $S$ is a set of generators for $G$, for any $g \in G$, there exist $x_1, x_2, \ldots, x_n \in S \cup S^{-1} \cup \{e\}$ such that $g = x_1 x_2 \cdots x_n$.

Thus if such $\overline{f}$ exists, then $\overline{f}(g) = \overline{f}(x_1 \cdots x_n) = \overline{f}(x_1) \cdots \overline{f}(x_n)$.

And if $x_i \in S$ then $\overline{f}(x_i) = f(x_i)$, while if $x_i \in S^{-1}$ then $\overline{f}(x_i) = \overline{f}((x_i^{-1})^{-1}) = (\overline{f}(x_i^{-1}))^{-1} = f(x_i^{-1})^{-1}$.

In this way, we can define $\overline{f}(g)$.

Question : For each $g \in G$, there may be many ways to make $g$ by multiplying some elements of $S \cup S^{-1} \cup \{e\}$. For example, it is possible that $g = x_1 \cdots x_n = y_1 \cdots y_m$ for some $x_i, y_j \in S \cup S^{-1} \cup \{e\}$. Then how can we show that $\overline{f}(x_1 \cdots x_n) = \overline{f}(y_1 \cdots y_m)$?

Answer 1

You seem to be misreading the question. It says

If there exists [such] a homomorphism $\overline f$ ... then there is only one.

This is a statement of uniqueness. Uniqueness statements can always be translated into a form which says "If you are given two of these things, then they are the same". So, when this statement is translated in that fashion, here's what it says:

If $\overline f_1 : G \to G'$ and $\overline f_2 : G \to G'$ are two homomorphisms, and if $\overline f_1(x) = \overline f_2(x)$ for all $x \in S$, then $\overline f_1(g) = \overline f_2(g)$ for all $g \in G$.