One to one and onto in composite function

  • Thread starter BlackDeath
  • Start date
  • #1

Homework Statement



I just want to make sure that I am correct. if we have a composite function f(g(x)).

Homework Equations


f(g(x)) is onto if and only if both f(x) and g(x) are onto
f(g(x)) is one to one if and only if or both f(x) and g(x) are one to one


The Attempt at a Solution



when I try to make f(x) is onto, but not one to one. And g(x) is one to one but not onto, f(g(x)) is not onto
 

Answers and Replies

  • #2
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This is false: The fact that f(g(x)) is one to one only guarantees that f(x) is one to one.

For example, let [itex] f: \mathbb R \to \mathbb R^2 [/itex] by [itex] f(x) = (x,0) [/itex] and [itex] g: \mathbb R^2 \to \mathbb R [/itex] be [itex] g(x,y) = x+y [/itex]. f is one-to-one but g is not one-to-one. However, the function g(f(x)) = x is just the identity function and is injective.

Precisely the same example shows that this does not work for onto functions. g(x,y) is onto but f(x) is not. However, g(f(x)) = x is onto, so again it breaks.
 

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