TITLE: Universal property of kernel of a homomorphism
QUESTION [5 upvotes]: I am in the beginning of Category Theory.
Let $\theta\colon G\rightarrow K$ be a group homomorphism, and $\ker\theta$ denote the kernel of this homomorphism. Let $\varepsilon\colon G\rightarrow K$ denote the trivial homomorphism and $i\colon \ker\theta\rightarrow G$ denote the inclusion, i.e. $i(x)=x$. Then the pair $(\ker\theta,i)$ satisfies the property that $$\varepsilon\circ i=\theta\circ i.$$
Question(Exercise) What is the universal property of $(\ker\theta,i)$?
I didn't get any direction towards the solution. But, I tried to move in the following direction: by definition, $\ker\theta$ is the largest subset (subgroup) of $G$ whose elements go to identity via $\theta$. This "largest" and  "..go to identity via $\theta$" I tried to put in categorical language. The answer I came up then is as follows:

Given a subset $H$ of $G$ and $j\colon H\rightarrow G$, the inclusion map, if $j$ and $\theta$ (or $\ker\theta$) satisfy
  $$\varepsilon\circ j=\theta\circ j,$$
  then there is a unique (inclusion?) map $f\colon H\rightarrow \ker\theta$ such that $j=i\circ f$. 

Is this answer (universal proerty of $(\ker\theta,i)$) correct?
(I don't know how to draw commutative diagram in the latex, here; one may edit to draw the diagram here.)

REPLY [5 votes]: You're close, but not quite there. As was mentioned already in the comments, "subset" is not a categorical notion.
Because we are dealing with the category $\sf Grp$ of groups and morphisms, all our objects should, in fact, be groups.
So we might contemplate replacing "subset" by "subgroup". However, the notion of "subgroup" is not preserved under group isomorphism because it relies on the set-theoretical nature of the groups.
However, we know that for every $j: G' \to G$, $\operatorname{im} j$ is a subgroup of $j$, and indeed the inclusion homomorphisms show that every subgroup $H$ of $G$ arises in this way.
We can thus phrase the UMP as follows:

Given a group homomorphism $j: H \to G$, if $j$ and $\theta$ satisfy $$\varepsilon j = \theta j,$$ then there is a unique $f: H \to \ker\theta$ such that $j = if$.

In categorical lingo, we say that $(\ker\theta, i)$ is the equalizer of $\varepsilon$ and $\theta$. See also ProofWiki and WikiPedia.