Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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 at 16:06

1 Answer 1

up vote 3 down vote accepted

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 $l'=g=h$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.