# Every kernel is monic

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.

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This statement is a special case of more general abstract categorical statement: every equalizer is monic. The kernel of $f$ is the equalizer of $f$ and $0_{AB}$. – Oskar Jan 21 '14 at 16:06
i have a question, in your question, you define $k : K \to B$. i don't understand, by my decision, kernel is $k : K \to A$. – user207839 Jan 14 '15 at 14:10

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 $l'=g=h$.