# Theorem about inverse of an inverse f

• issacnewton
In summary, the conversation discusses how to prove that if a function f is bijective, then f=(f^{-1})^{-1}. The speaker goes through their attempted proof and asks for suggestions. Another person suggests using a theorem stating that if f is a bijection, then its inverse is also a bijection. The speaker then uses this theorem to show that f=(f^{-1})^{-1}. The other person confirms that their logic is correct.
issacnewton
Hi

I have to prove that if $f:A\rightarrow B$ is bijective (onto and one to one)
then

$$f=(f^{-1})^{-1}$$

Following is my attempt. Since $f:A\rightarrow B$, we have $f^{-1}:B\rightarrow A$.

Now there is another theorem which says that function f is a bijection (onto and one to one)
iff inverse of f is also bijection. Let $g:A\rightarrow B$ be the inverse of
$f^{-1}$. We know that g exists since $f^{-1}$ is a one to one.
No let $x \in A$ be arbitrary , then

$$\exists \,\, y \in B \backepsilon$$

$$g(x)=y$$ but since

$$y \in B \Rightarrow \exists \,\, x_1 \in A \backepsilon$$

$$f(x_1)=y \Rightarrow f(x_1)=g(x)$$

So if I can show that $x=x_1$ and since x is arbitrary , I will complete the proof.
But I am stuck here. Can anybody suggest something ?

thanks
Newton

Let $g=(f^{-1})^{-1}$ and examine $g\circ f^{-1}$ and $f^{-1}\circ g$ what does this say about g?

Hi mat

Thanks for the hint. Here's my improvement. There is another theorem given in the section
in which I am doing this problem.

Theorem: Suppose that $f:A\rightarrow B$ is a bijection, then

(a)$$(f^{-1}\circ f)(x) = x \;\;\forall x \in A$$

and

(b) $$(f \circ f^{-1})(y)=y \;\;\forall y \in B$$

So I think I can use this theorem here, since $f^{-1}$ and $g$ are inverses
of each other. and both of them are bijections since f itself is a bijection. We have that

$$g:A\rightarrow B \;\; \mbox{and}\;\; f^{-1}:B\rightarrow A$$

So using the stated theorem, we have

$$f^{-1}(g(x))=x \;\;\forall x \in A$$

$$\Rightarrow g(x) = f(x) \;\;\forall x \in A$$

In my last post , I got the equation $f(x_1)=g(x)$. So from here can I claim that
$x=x_1$

If I can claim that, then I can say that

$$f(x)=(f^{-1})^{-1}(x) \;\;\forall x \in A$$

$$\Rightarrow f=(f^{-1})^{-1}$$

is it ok ?

mat , can you confirm my logic in the last post ?

Looks good.

thanks for helping, mat

## 1. What is the Theorem about inverse of an inverse f?

The theorem states that if a function f has an inverse function, and that inverse function also has an inverse, then the original function f is equal to the inverse of its inverse. In other words, the inverse of the inverse function is the original function.

## 2. How is the Theorem about inverse of an inverse f proven?

The theorem can be proven using algebraic manipulation and the definition of inverse functions. By setting up equations for the original function, its inverse, and the inverse of the inverse, it can be shown that they are all equivalent.

## 3. What is the importance of the Theorem about inverse of an inverse f?

This theorem is important because it allows us to say that a function and its inverse are essentially the same function. It also helps in solving equations and understanding the relationship between a function and its inverse.

## 4. Can the Theorem about inverse of an inverse f be applied to all functions?

No, the theorem only applies to functions that have an inverse. Not all functions have an inverse, for example, a horizontal line has no inverse because it does not pass the vertical line test.

## 5. What are some examples of functions that follow the Theorem about inverse of an inverse f?

Some examples include logarithmic functions, exponential functions, and trigonometric functions such as sine and cosine. These functions have inverses that are also functions, and the theorem can be applied to them.

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