Set Theory, Functions. Injective/Surjective

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SUMMARY

The discussion focuses on the properties of functions in set theory, specifically exploring the conditions under which the composition of functions can be surjective or injective. The user seeks to construct examples of functions f and g, where f is surjective, g is injective, and their composition f*g is neither surjective nor injective. Additionally, the discussion includes a proof regarding the inverse of the composition of bijections, asserting that the inverse of the product of bijections is indeed the product of their inverses.

PREREQUISITES
  • Understanding of surjective and injective functions
  • Familiarity with function composition
  • Knowledge of bijections and their properties
  • Basic proof techniques in set theory
NEXT STEPS
  • Research examples of surjective and injective functions in ℝ and ℤ
  • Study the properties of function composition in set theory
  • Learn about the definitions and implications of bijections
  • Explore proof strategies for demonstrating properties of functions
USEFUL FOR

Students and educators in mathematics, particularly those studying set theory, functions, and their properties. This discussion is beneficial for anyone looking to deepen their understanding of injective, surjective, and bijective functions.

ktheo
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Homework Statement



Give f:A→A and g:A→A where f is surjective, g is injective, but f*g is neither surjective nor injecive

The Attempt at a Solution



I don't know why I can't really think of two... I assume it's easiest to do one in ℝ, but when it comes to producing surjective-non-injective functions in general I tend to do them in Z since I find it easiest. I was thinking to do something involving e^x but I'm not sure. How should I approach this? Should I just think of functions I know are neither onto or one-to-one and work with products to find something?

Homework Statement



Assume that f:A→B and g:C→D are bijections. Prove that f^-1 x g^-1 is the two sided inverse of f x g (and in particular, that f x g is a bijection as well).

The Attempt at a Solution



I was wondering if someone could direct me to a similar proof or point me in the direction of some definitions that can help me here. I don't even know how to structure this into a proof.
 
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