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

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SUMMARY

The discussion provides a counterexample to the assertion that if the composition of two functions, f and g, is invertible, then both functions must also be invertible. The example uses the functions f(x) = x² defined over the reals and g(x) = √x defined for x > 0. The composition f(g(x)) results in the identity function over the domain x > 0, demonstrating that while the composition is invertible, f itself is not invertible over its entire domain. This illustrates that the invertibility of a composition does not guarantee the invertibility of the individual functions.

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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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