TITLE: Every kernel is monic
QUESTION [6 upvotes]: Before I begin: I'm new to category theory.
I'm trying to show that if $k$ is a kernel of some morphism of some category, then it is monic.
Here is my reasoning so far: 
Suppose the kernel of some morphism $f: A \longrightarrow B$ is $k: K \longrightarrow B$. I want to show that for any object $C$ and for any morphisms $g,h: C \longrightarrow K$, $kg=kh$ implies $g=h$. 
By the definition of the kernel, $fk = 0_{KB}$ where $0_{KB}$ is the unique zero morphism $K \longrightarrow B$. Next, note that $kg, kh : C \longrightarrow A$ are morphisms such that $fkg = 0_{CB} = fkh$ by uniqueness of $0_{CB}$. So $0_{KB} \ g = 0_{CB} = 0_{KB} \ h$. Here is where I am stuck: intuitively it seems obvious that $g=h$ after this. Why? 
I also would like to use the existence of unique morphisms $u,v: C \longrightarrow K$ such that $ku = kg$ and $kv = kh$, respectively; however, I don't know where to apply them.
Thanks for any help.

REPLY [8 votes]: The universal property says that whenever $l:C\to A$ is a morphism such that $fl=0$, then there is a unique morphism $l':C\to K$ such that $kl'=l$.
Now $kg=kh$ is a morphism $C\to A$ such that $f(kg)=0$, so there is a unique morphism $l':C\to K$ such that $kl'=kg=kh$. We conclude that $l'=g=h$.
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