Counter example to: if f*g is inv then f & g are inv

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A counterexample to the assertion that if the composition of two functions f and g is invertible, then both functions must be invertible is provided using f(x) = x^2 and g(x) = sqrt(x). The composition f(g(x)) results in x, which is invertible over the domain x > 0, but f is not invertible over its entire domain since it is not one-to-one. The discussion clarifies that the invertibility of f should not be restricted to the reduced domain defined by g's image. It emphasizes that the composition can be invertible even if one of the functions is not. The conclusion reinforces that the problem only requires demonstrating a non-invertible function that becomes invertible through composition.
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Homework Statement



Provide a counter example to the false assertion:
Suppose that f and g are functions and f ◦ g is invertible. Then f and g are invertible.

Homework Equations



Definitions:
An invertible function is 1-1 and onto
If the image of g is not contained in the domain of f then f ◦ g is not a legal expression.

The Attempt at a Solution



Let f(x)=x^2 over R and g(x)=sqrt(x) over x>0|R, then f(g(x))=x which is 1-1 and onto over x>0|R and is therefore invertible.
f(x) is not 1-1 over R thus it is not invertible providing the counter example to f and g are invertible.

My question: The function I defined as f is not invertible but the composition works over a subset of f's domain due to g. Does this mean that f's invertibility should only be considered over the reduced domain too?

I suspect not because I provided that exact wording of the problem and it says I only need to provide a function definition which is not invertible on its own but which is invertible in a composition.

I have a strong hunch I am over thinking this...
 
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You are doing it right. The way of getting a non-invertible function f is to consider a function that is invertible when restricted to the image of g but not invertible on its own domain. For this to be the case you need a g whose image is not the entire domain of f.
 
Thanks Orodruin!
 
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