2
$\begingroup$

I am studying Vakil's notes, Section 1.6.3, on the concept of abelian categories.

A kernel of a morphism $f: B\rightarrow C$ is a map $i: A\rightarrow B$ such that $f\circ i=0$, and that is universal with respect to this property. Diagramatically:

enter image description here

A cokernel is defined dually by reversing the arrows.

The image of a morphism $f: A\rightarrow B$ is defined as $\text{im}(f)=\ker(\text{coker (f)})$.

It is mentioned as a fact that

For a morphism $f: A\rightarrow B$, $A\rightarrow \text{im}(f)$ is an epimorphism. I will show an equivalent statement: $f: A\rightarrow B$ is an epimorphism if $\text{im}(f)=B$.

Here is my effort so far to show it.

We have the following diagram: enter image description here

$e: B\rightarrow C$ is the cokernel of $f$, $i: B\rightarrow B$ is the kernel of the cokernel, hence the image of $f$. To show $f$ is an epimorphism, suppose that $\phi\circ f=0$ for some $\phi: B\rightarrow D$. We need to show that $\phi=0$.

We see that $\alpha\circ e\circ i=\phi\circ i=0$. My question is, when we say $\text{im}(f)=B$, does it implicitly imply that $\text{im}(f)$ is the identity map $\text{id}_B: B\rightarrow B$? If that is true, then I have $\phi=0$. Then the proof is done. If it is not, I don't know how to continue to show $\phi=0$.

Thank you for your help!

$\endgroup$
  • 1
    $\begingroup$ If you edit your diagram to denote the dotted $f$ by another letter, say $g$, it becomes easy to check that the statements are equivalent in case $\text{im } f = B$ is taken to mean $i = 1_B$. So for the purposes of the proof, you're fine to take it that way. It's hard to say there's a "correct" interpretation, since the statement constitutes an abuse of notation. $\endgroup$ – Mr. Chip Aug 8 '16 at 13:31
  • $\begingroup$ @Mr.Chip: Thank you for your comment. I put $f$ there since I think that map is unique, so it has to be $f$. But I see the abuse of notation there. Thanks for pointing it out. $\endgroup$ – KittyL Aug 8 '16 at 14:58
  • 1
    $\begingroup$ You can also show the way more useful statement: Every morphism factors as a coimage (cokernel of kernel) followed by an image (kernel of cokernel). Now remember that equalizers are mono and dually coequalizers are epi, hence kernels are mono and cokernels are epi. $\endgroup$ – Stefan Perko Aug 9 '16 at 21:19
  • $\begingroup$ @StefanPerko: Thank you! That is very enlightening. $\endgroup$ – KittyL Aug 10 '16 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.