Understanding Homomorphisms: The Relationship Between A and B

  • Thread starter tgt
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In summary, a homomorphism f:A->B can be used to define a new map, f_{Ker(f)}, which maps the kernel of f to the range of f. However, this map is only the identity map f(x)= 1B, which may not be very interesting.
  • #1
tgt
522
2
Suppose there exists a homomorphism f:A->B then does it make sense to have

[tex]f_{Ker(f)}:Ker(f) \to B[/tex] ?

why doesn't my tex show up?

Offcourse, [tex]Im(f_{Ker(f)})=1_{B}[/tex]

Moderator Note: Fixed LaTeX.
 
Last edited:
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  • #2
tgt said:
Suppose there exists a homomorphism f:A->B then does it make sense to have

why doesn't my tex show up?

Offcourse, Im(f_{Ker(f)})=1_{B}
You need to close the tex enviroment. e.g.
Code:
[tex]\frac{dy}{dx}[ /tex]
(without the space in the square brackets of course)
 
Last edited:
  • #3
What are you trying to say?
 
  • #4
Are you trying to say that the elements in A that are in the kernel map into B. Well yes that's true since they all map to the zero element within B.
 
  • #5
morphism said:
What are you trying to say?

Given a map, we can define a new map, mapping the kernel of the map to the range of the map.
 
  • #6
Yes, but what's the point?
 
  • #7
tgt said:
Suppose there exists a homomorphism f:A->B then does it make sense to have

[tex]f_{Ker(f)}:Ker(f) \to B[/tex] ?

why doesn't my tex show up?

Offcourse, [tex]Im(f_{Ker(f)})=1_{B}[/tex]

Moderator Note: Fixed LaTeX.
Yes, that makes sense. It isn't very interesting, however, since [itex]f_{Ker(f)}[/itex] is just the identity map f(x)= 1B, as you say.
 
Last edited by a moderator:

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