Difference between $F(S)$ and $G$ when $S$ generates $G$ Let $G$ be a group generated by a set $S$.
It is known that the inclusion map $i: S \to G$ can be extended to an epimorphism $I: F(S) \to G$.
Is there a difference between $F(S)$ and $G$? Clearly $G \subset F(S)$, but also it seems that every element in $F(S)$ is generated by elements from $G$ - so they're equal.
If so, what's the idea behind this map $I$? Why concentrate on an epimorphism? 
Lastly, perhaps this notion of equality isn't correct, as $G$ may not be free?
 A: 
Clearly $G \subset F(S)$,

That's "clearly" wrong. For starters the literal inclusion has no chance of happening. The construction of $F(S)$ (as a quotient of the space of all words) just doesn't allow that.
Secondly there is no embedding of $G$ to $F(S)$ in general. For example take $G=\mathbb{Z}_2$ and $S=\{1\}$. Then $F(S)\simeq\mathbb{Z}$ and it is impossible to embed $\mathbb{Z}_2$ into $\mathbb{Z}$. The situation when $G$ can be embedded into $F(S)$ is very rare and actually implies that $G$ is free.
So it's the other way around: there's a surjective group homomorphism $F(S)\to G$, that's your $I$. The reason why this is important is because that implies that $G$ is isomorphic to some quotient $F(S)/N$ for some normal subgroup $N\subseteq F(S)$. And so you can describe the group $G$ in terms of free generators (the $S$ subset) and relations (the $N$ subgroup).
For example 
$$\mathbb{Z}_2\oplus\mathbb{Z}_3\simeq \langle x, y\ | \ x^2=y^3=1;\ xy=yx\rangle$$
means that $\mathbb{Z}_2\oplus\mathbb{Z}_3$ is isomorphic to $F(\{x,y\})$ divided by the normal subgroup of $F(\{x,y\})$ generated by $\langle\{x^2, y^3, xyx^{-1}y^{-1}\}\rangle$.
There are multiple other applications.
Perhaps the confusion comes from the fact that in the set theory if there is a surjection $X\to Y$ then there is an injection $Y\to X$. But that doesn't work in the realm of groups and group homomorphisms.
