Epimorphisms Between Groups: When is a Homomorphism Onto?

In summary: No, but I can certainly exhibit an element of that cannot be reached by : one example is . In general, is even iff is even for .Ah. I think I've found the problem with my proposal. The problem with my example isn't surjectivity: it's whether it's well-defined. You could have and .
  • #1
TheBigBadBen
80
0
Interesting question I've happened upon:

If there is an epimorphism (i.e. onto homomorphism) , is there necessarily an epimorphism ? If not, under what conditions can we ascertain such an epimorphism given the existence of ?

I would think that this is true for abelian groups and by their respective decompositions, but I'm not sure how I'd show that to be the case. Also, I don't have a good guess for the general case. It seems clear that the isomorphism theorem should play a role at some point, but that doesn't lead to any obvious conclusions.
 
Physics news on Phys.org
  • #2
TheBigBadBen said:
Interesting question I've happened upon:

If there is an epimorphism (i.e. onto homomorphism) , is there necessarily an epimorphism ? If not, under what conditions can we ascertain such an epimorphism given the existence of ?

I would think that this is true for abelian groups and by their respective decompositions, but I'm not sure how I'd show that to be the case. Also, I don't have a good guess for the general case. It seems clear that the isomorphism theorem should play a role at some point, but that doesn't lead to any obvious conclusions.

Wouldn't there have to be? Let be arbitrary. Then there exist such that . Let be defined by the action of on only the first component. I think it'd be fairly straightforward to show that inherits all the required properties straight from . Onto seems clear: since is arbitrary, and , it follows that , and you have your element in that maps to . And the fact that is a homomorphism follows from the fact that is, right?
 
  • #3
Ackbach said:
Wouldn't there have to be? Let be arbitrary. Then there exist such that . Let be defined by the action of on only the first component. I think it'd be fairly straightforward to show that inherits all the required properties straight from . Onto seems clear: since is arbitrary, and , it follows that , and you have your element in that maps to . And the fact that is a homomorphism follows from the fact that is, right?

Your statement assumes that must act on the components separately. I thought this might be at first, but I don't believe this is necessarily the case.

A possible counter-example: given by is an epimorphism in which you can't separate the components in the manner you described.

EDIT: The counter-example is not onto. Still trying to think of something to prove my point.
 
Last edited:
  • #5
TheBigBadBen said:
Your statement assumes that must act on the components separately. I thought this might be at first, but I don't believe this is necessarily the case.

A possible counter-example: given by is an epimorphism in which you can't separate the components in the manner you described.

EDIT: The counter-example is not onto. Still trying to think of something to prove my point.

I'm not seeing how the counter-example fails to be onto. Can you exhibit an element of that cannot be reached by the quantity ?
 
  • #6
Ackbach said:
I'm not seeing how the counter-example fails to be onto. Can you exhibit an element of that cannot be reached by the quantity ?

No, but I can certainly exhibit an element of that cannot be reached by : one example is . In general, is even iff is even for .
 
Last edited:
  • #7
Ah. I think I've found the problem with my proposal. The problem with my example isn't surjectivity: it's whether it's well-defined. You could have and . Then my function isn't well-defined. Hmm. I wonder if there's a way to fix my example. I was thinking maybe you could use equivalency classes, but then would you really be mapping from to ?
 

Similar threads

Back
Top