Linear Algebra Proof on Composition of One-to-One Functions

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SUMMARY

The discussion centers on proving that the composition of one-to-one functions is also a one-to-one function. It establishes that if functions f and g are one-to-one, then their composition (f(g(x))) maintains the one-to-one property. The proof utilizes the definition of one-to-one functions, where f(x1) = f(x2) implies x1 = x2, and suggests a proof by contradiction to reinforce the argument. The conclusion is that the composition of one-to-one functions is indeed one-to-one, supported by logical reasoning and definitions.

PREREQUISITES
  • Understanding of one-to-one functions and their properties
  • Familiarity with function composition
  • Knowledge of proof techniques, particularly proof by contradiction
  • Basic concepts in set theory related to functions
NEXT STEPS
  • Study the properties of one-to-one functions in more detail
  • Learn about function composition and its implications in mathematics
  • Explore proof techniques, focusing on proof by contradiction
  • Review set theory concepts relevant to functions and mappings
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Mathematics students, educators, and anyone interested in understanding function properties and proof techniques in linear algebra.

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Homework Statement




Prove that the composition of one-to-one functions is also a one-to-one function.



Homework Equations




A function is one-to-one if f(x1)=f(x2) implies x1=x2. Composition is (f*g)(x)=f(g(x)). Proof-based question.



The Attempt at a Solution




A one-to-one function does not repeat the image. If we have two one-to-one function f(x) and g(x), then f and g do not repeat their images. Then, when then the composition, for example f(g(x)), for all x, g(x) does not repeat the image and after that applying f(x) also does not repeat the image, therefore the composition of the function is one-to-one as well.

Is this a good proof for the question?
 
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I would suggest a proof by contradiction. Let g be a one-to-one function from set A to set B, f a one-to-one function from set B to set C. Suppose the statement were not true- that f(g(x)) is not one-to-one. Then there exist a, b, a not equal to b, in A such that f(g(a))= f(g(b)). Since f is one-to-one, there must exist a unique x in b such that f(x)= f(g(a))= f(g(b)). Can you complete this?
 
If in doubt, proof by contradiction!
 

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