Understanding Mappings: Injective, Surjective, and Homomorphisms

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SUMMARY

This discussion clarifies the definitions and properties of mappings in mathematics, specifically focusing on injective, surjective, and homomorphic mappings. An injective map satisfies the condition that f(a) = f(b) implies a = b, while a surjective map ensures that for every b in B, there exists an a in A such that f(a) = b. The discussion also defines a homomorphism as an operation-preserving mapping when A equals B, and introduces terms such as isomorphism, monomorphism, epimorphism, endomorphism, and automorphism. The conclusion confirms that a bijection is sufficient for the existence of an inverse, provided the map is also a homomorphism.

PREREQUISITES
  • Understanding of basic mapping concepts in mathematics
  • Familiarity with group theory, specifically group homomorphisms
  • Knowledge of injective and surjective functions
  • Concept of bijections and their properties
NEXT STEPS
  • Study the properties of group homomorphisms in detail
  • Learn about the implications of bijections in various mathematical structures
  • Explore the differences between monomorphisms and epimorphisms
  • Investigate the applications of automorphisms in algebraic structures
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Mathematicians, students of abstract algebra, and anyone interested in the foundational concepts of mappings and their properties in group theory.

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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 ^
 
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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.
 

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