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.

Problem Suppose $\mathscr{C}$ is an arbitrary category with zero object. $A$ and $B$ are two objects of $\mathscr{C}$. Let $f\in Mor_\mathscr{C}(A,B)$. It's given that $f$ is monic. I need to show that $f$ has a kernel.

My claim Object $A$ with morhism $\iota$(will be def'd below) is the kernel of $f$. I did the followings:

1- Suppose $Z$ is zero object. I defined the morhisms:
$ \phi_{a} : Z \rightarrow A $ , $ \phi_{b} : Z \rightarrow B $
$ \psi_{a} : A \rightarrow Z $ , $ \psi_{a} : B \rightarrow Z $
Since Z is both initial and terminal these morphisms are in some sense "unique".

2- I defined $\iota=\phi_{a} \circ \psi_{a}$. I showed that $f \circ \iota = 0_{AB}$ ie. the zero morhism from $A$ to $B$.

3- Suppose there is an object $D$ with morphism $g: D \rightarrow A$ such that $f \circ g = 0_{DB}$. I showed that for every morphism $h: D \rightarrow A$ we have $g = \iota \circ h$.

4- I failed to prove uniqueness of h.

My question Could you please tell me that if i am in the right path or not? If i am, how can i fix the 4th step? Thanks

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Having a zero object in your category gives you a (unique) zero morphism $A\to B$ obtained as composition $A\to 0\to B$ for any two objects $A,B\in\cal C$.

This gives on its own right a practical definition for $\ker (f)$: it is the equalizer of the arrow $f$ and the zero map; you can have fun in finding that this is precisely the right definition for "what is sent to $0$ by $f$" (for example, what if $\cal C$ is the category of modules over a ring, or more simply vector spaces, or less trivially pointed sets?

Once you know this, and the fact that $\ker(f)$, whenever it exists, is unique up to a unique iso, your problem boils down to showing that the unique arrow $0\to A$ from zero object to the domain of $f$ serves as the kernel of any monic arrow $f\colon A\to B$.

share|improve this answer
    
Woops, Berci arrived first! :) nevermind, at least our answers are not the same. –  tetrapharmakon Dec 28 '13 at 16:39
    
equalizer concept was totally helpful. I really appreciate your help. –  iamvegan Dec 28 '13 at 17:22

Well, basically on the right direction, but not exactly the right track. You should start it over and simplify.

You miss the observation that the domain of the kernel is going to be the zero object itself, and not $A$.

Claim: The kernel of a monic $f:A\to B$ will be the unique arrow $Z\to A$.
Proof: (your turn)

share|improve this answer
    
thanks a lot for your answer. i completed the proof of my turn part. –  iamvegan Dec 28 '13 at 17:23

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.