Injectivity and Surjectivity in Function Compositions

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SUMMARY

This discussion focuses on the properties of injectivity and surjectivity in function compositions, specifically for functions f: A->B and g: B->C. It establishes that if both f and g are injective, then their composition g o f is also injective. Conversely, if g o f is injective, it implies that both f and g must be injective. Additionally, if both functions are surjective, then their composition g o f is surjective, and the surjectivity of g o f indicates that both f and g are surjective as well. The discussion culminates in summarizing these findings in a theorem format.

PREREQUISITES
  • Understanding of function composition
  • Knowledge of injective (one-to-one) functions
  • Knowledge of surjective (onto) functions
  • Familiarity with set notation and proofs
NEXT STEPS
  • Study the definitions and properties of injective functions in detail
  • Learn about surjective functions and their implications in function theory
  • Explore the concept of function composition and its properties
  • Investigate formal proof techniques in mathematics
USEFUL FOR

Mathematics students, educators, and anyone studying advanced function theory or preparing for exams in abstract algebra or real analysis.

tomboi03
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Let f: A->B and g: B->C
a. C0\subsetC, show that (g o f)-1(C0))=f-1(g-1(C0)).
b. If f and g are injective, show that g o f is injective
c. If g o f is injective, what can you say abuot injectivity of f and g?
d. If f and g are surjective, show that g o f is surjective.
e. If g o f is surjective, what can you say about surjectivity of f and g.
f. Summarize your answers to (b)-(e) in the form of a theorem

I don't know how to do this... can someone else me..?

Thank You
 
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Hi tomboi03! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help.

Start with part a. … take a typical member of the LHS, and prove that it is in the RHS … then do it the other way round. :smile:
 

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