|Mar21-12, 01:32 PM||#1|
Linear Algebra Proof on Composition of One-to-One Functions
1. The problem statement, all variables and given/known data
Prove that the composition of one-to-one functions is also a one-to-one function.
2. Relevant 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.
3. 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?
|Mar21-12, 02:02 PM||#2|
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?
|Mar21-12, 02:04 PM||#3|
If in doubt, proof by contradiction!
|composition, linear algebra, one-to-one, proof, prove|
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