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.)