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

In summary, a counter example to the false assertion that f and g are invertible if f ◦ g is invertible can be provided by considering a function f that is not invertible on its own domain, but is invertible when restricted to the image of g, where the image of g is not the entire domain of f. This shows that f's invertibility should not be considered only over the reduced domain, as the composition with g allows for f to be invertible in this scenario.
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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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  • #2
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.
 
  • #3
Thanks Orodruin!
 

1. What does "f*g is inv" mean in this context?

"f*g is inv" means that the product of two functions, f and g, is invertible. This means that there exists another function, h, such that h(f*g) = (f*g)h = I, where I is the identity function.

2. Can you provide an example of two functions, f and g, where f*g is invertible but f and g are not invertible?

Yes, consider the functions f(x) = 2x and g(x) = 1/2x. These functions are not individually invertible, but their product, f*g(x) = 1, is invertible as it can be multiplied by any number to get back to the identity function.

3. Is it possible for both f and g to be invertible when their product, f*g, is not invertible?

Yes, it is possible for both f and g to be invertible when their product is not invertible. For example, consider the functions f(x) = x and g(x) = -x. Both f and g are invertible, but their product, f*g(x) = -x^2, is not invertible as it is not a one-to-one function.

4. Can you explain why the statement "if f*g is inv then f & g are inv" is not always true?

The statement is not always true because the invertibility of a product does not necessarily imply the invertibility of the individual functions. It is possible for the product to be invertible while one or both of the individual functions are not invertible.

5. How can this counter example be applied in real-world situations?

In real-world situations, this counter example can be applied in scenarios where two processes or systems are combined to form a new process or system. The invertibility of the combined process or system does not necessarily imply the invertibility of the individual processes or systems. This is important to consider when analyzing the functionality and potential limitations of the combined process or system.

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