Composite Function Homework: Proving One-to-One & Onto

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SUMMARY

The discussion focuses on proving properties of composite functions, specifically that if the composite function \( g \circ f \) is one-to-one, then \( f \) must also be one-to-one, and if \( g \circ f \) is onto, then \( g \) must be onto. The key equations discussed include \( g \circ f(x) = g(f(x)) \) for every \( x \) in set \( A \). The participant expresses uncertainty about approaching the problem but recognizes the need to verify the one-to-one and onto properties of the functions involved.

PREREQUISITES
  • Understanding of functions and their properties, specifically one-to-one and onto functions.
  • Familiarity with composite functions and their definitions.
  • Basic knowledge of set theory and notation.
  • Experience with mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the definitions and properties of one-to-one and onto functions in detail.
  • Learn about composite functions and their implications in function theory.
  • Explore mathematical proof techniques, particularly for function properties.
  • Practice problems involving composite functions to solidify understanding.
USEFUL FOR

Students studying mathematics, particularly those focusing on functions, set theory, and proof techniques. This discussion is beneficial for anyone looking to deepen their understanding of composite functions and their properties.

nikie1o2
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Homework Statement


let A,B,C be sets, and let f : A--> B and g : B--> C be functions. The composite function denoted by g o f is a function from A to C defined as follows:



Homework Equations



g o f(x)=g(f(x)) for every x in A.

Prove that if g o f is one-to one, then f is one-to one
Prove that is g o f is onto, then g is onto

The Attempt at a Solution


I really don't know how to approach this problem. I didnt things similar with binary structures determining if they are an isomorphism by not familiar with composite functions
 
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I think I know how to approach it...check me if I'm right though

check if its one-to-one and onto

if it is one-to-one that implies if f(a)=f(b) then a=b
 

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