Understanding Mappings: Injective, Surjective, and Homomorphisms

  • Context: Graduate 
  • Thread starter Thread starter STEMucator
  • Start date Start date
  • Tags Tags
    Mapping
Click For Summary

Discussion Overview

The discussion centers on the properties of mappings, specifically injective, surjective, and homomorphic mappings, with a focus on their definitions and implications in the context of algebraic structures such as groups. Participants explore the conditions under which these mappings hold and the terminology associated with them.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant outlines the definitions of injective, surjective, and homomorphic mappings, suggesting that a bijective homomorphism is an isomorphism with an inverse.
  • Another participant agrees with the definitions but questions the necessity of specifying that A and B are groups for the homomorphism definition to hold, indicating that the context matters.
  • A later reply clarifies that A and B were intended to be groups, explaining the choice of A = B as a means to indicate a mapping from a group to itself.
  • A question is raised about whether being bijective is sufficient for the existence of an inverse, to which another participant responds affirmatively, stating that bijection implies the existence of an inverse.

Areas of Agreement / Disagreement

Participants generally agree on the definitions of injective, surjective, and homomorphic mappings, but there is some debate regarding the necessity of specifying the algebraic structure of A and B. The discussion on the sufficiency of bijection for the existence of an inverse appears to reach consensus.

Contextual Notes

The discussion assumes familiarity with algebraic structures such as groups, rings, and modules, but does not explicitly define these terms. The implications of the mappings may depend on the specific properties of these structures, which are not fully explored.

STEMucator
Homework Helper
Messages
2,076
Reaction score
140
So I want to clarify if what I'm thinking is correct.

Suppose we have a mapping f : A → B and we have a in A and b in B.

If f is an injective map, then f(a) = f(b) implies that a = b or conversely a≠b implies f(a)≠f(b).

If f is a surjective map, then for b in B, there exists an a in A such that f(a) = b.

If A = B then f is a homomorphism from A to B if it is operation preserving. That is f(ab) = f(a)f(b) for all a and b in A.

If f is both injective, surjective, and operation preserving, then it is a bijective homomorphism, also known as an isomorphism, and thus has an inverse f-1 : B → A.

If f is an injective homomorphism, it is called a monomorphism.

If f is a surjective homomorphism, it is called an epimorphism.

If A = B and f is a homomorphism, then it is called and endomorphism.

Also a bijective endomorphism is an automorphism.

I'm hoping that those are correct ^
 
Physics news on Phys.org
Zondrina said:
So I want to clarify if what I'm thinking is correct.

Suppose we have a mapping f : A → B and we have a in A and b in B.

If f is an injective map, then f(a) = f(b) implies that a = b or conversely a≠b implies f(a)≠f(b).

If f is a surjective map, then for b in B, there exists an a in A such that f(a) = b.

Correct.

If A = B then f is a homomorphism from A to B if it is operation preserving. That is f(ab) = f(a)f(b) for all a and b in A.

It depends. What are A and B?? Are they groups? rings? modules?? You should say that. If A and B are groups, then your definition is correct. But we usually call that a group homomorphism (although we use homomorphism when the structure of group is understood).
Also, I see no reason why we should take A=B.

If f is both injective, surjective, and operation preserving, then it is a bijective homomorphism, also known as an isomorphism, and thus has an inverse f-1 : B → A.

If f is an injective homomorphism, it is called a monomorphism.

If f is a surjective homomorphism, it is called an epimorphism.

If A = B and f is a homomorphism, then it is called and endomorphism.

Also a bijective endomorphism is an automorphism.

I'm hoping that those are correct ^

That is all correct.
 
Yes I intended for A and B to be groups. The only reason I took A = B is to imply that f was mapping from a group to itself.
 
Last edited:
Out of curiosity, is a bijection sufficient for an inverse or must the map also be a homomorphism for an inverse to happen.
 
Being bijective is equivalent to the existence of an inverse. So yes, it is sufficient.
 

Similar threads

Undergrad Question about "A Group Epimorphism is Surjective"
  • · Replies 13 ·
Replies
13
Views
2K
Undergrad How do we distinguish two different notations for cokernel and coimage?
  • · Replies 41 ·
2
Replies
41
Views
4K
Undergrad Trouble with a passage in Zariski & Samuel's Commutative algebra text
  • · Replies 5 ·
Replies
5
Views
2K
Undergrad Correct constructed example of Abelian Monoid?
  • · Replies 0 ·
Replies
0
Views
1K
Undergrad Meaning of "passage to the quotient"?
  • · Replies 3 ·
Replies
3
Views
1K
Undergrad Notation questions about kernel, cokernel, image and coimage
  • · Replies 7 ·
Replies
7
Views
2K
Undergrad Showing that inverse of an isomorphism is an isomorphism
  • · Replies 3 ·
Replies
3
Views
4K
Undergrad Meaning of "identify" & variations in different math context
  • · Replies 5 ·
Replies
5
Views
1K
Undergrad Did Artin make a mistake regarding isomorphisms?
  • · Replies 8 ·
Replies
8
Views
2K
Undergrad ##(A/\mathfrak{a})_{\mathfrak{p}/\mathfrak{a}}## and its isomorphism?
  • · Replies 31 ·
2
Replies
31
Views
3K