TITLE: Definition of the quotient $G/H$
QUESTION [5 upvotes]: my question concerns the definition of the quotient $G/H$ in the sense of category theory. The following definition was obtained from the lecture notes linked to at the end of the question(p.2)

Def. Let G be a group, and let S be a subset. The quotient of G by S is a group homomorphism $$q: G \rightarrow Q$$ such that $S \subset \mathrm{Ker}(q)$ which is universal amongst all such group homomorphism. That is, suppose that $$\phi : G \rightarrow G'$$ is a group homomorphism, such that $S \subset \mathrm{Ker}(\phi)$. Then there is a unique group homomorphism $\Psi: Q \rightarrow G'$ which makes the following diagram commute.


I am having trouble understanding what role $S$ is playing and why this definition makes sense. Thanks in advance.
http://www.math.ucsb.edu/~mckernan/Teaching/05-06/Autumn/220A/l_4.pdf

REPLY [2 votes]: To make sense of the definition from category theory, you need to check that the definition of quotient from set theory is universal.
It is better to work with a normal subgroup $N$ instead of a subset $S$. Let $q$ be the quotient map of $G/N$, then $q$ is surjective, $ker(q)=N$.
Take $\phi$ such that $ker(q)\subseteq ker(\phi)$. Make a function $\psi$, for every $x\in Q$ choose $\psi(x)$ to be an image along $\phi$ of a preimage along $q$ of $x$ (which exists because $q$ is surjective).
$ker(q)\subseteq ker(\psi)$, then $\forall(g_0\in G)(g_1\in G). q(g_0)=q(g_1)\to \phi(g_0)=\phi(g_1)$, i.e. $\psi$ is well-defined.
Using the above and routine work prove that:

your diagram commutes;
$\forall\psi'. \psi'\circ q=\phi \to \psi'=\psi$ , i.e. $\psi$ is unique such that your diagram commutes;
$\psi$ is a homomorphism.

In summary, $\forall\phi. ker(q)\subseteq ker(\phi) \to \exists!\psi. \psi\circ q=\phi$, i.e. $q$ is universal.
To prove the converse, i.e. that every universal object is isomorphic to $q$, use the theorem of pure category theory that says that every 2 universal objects are isomorphic by the unique isomorphism.
You can adapt this proof for other algebraic structures, instead of kernels take congruences.