# Prove that if $f$ is bijective, then $f^{-1}$ is bijective. [Verification]

Let $$f: X \to Y$$ be bijective, and let $$f^{-1}: Y \to X$$ be it's inverse. Conclude that $$f^{-1}$$ is also invertible.

Suppose that $$f^{-1}(f(x)) = f^{-1}(f(x')) \nRightarrow x=x'$$ (not injective), then $$x=x' \nRightarrow x=x'$$ which is a contradiction. Hence it is injective.

For any $$x$$ there exists an $$f(x)$$. Suppose that there exists an $$x$$ such that $$\nexists x' \in X: f^{-1}(f(x'))=x.$$ But that means that for some $$x$$, $$\nexists x'\in X: x'=x$$. But that $$x'$$ is simply $$x$$. This means that for every $$x$$, there is a corresponding $$x$$ value that satisfies surjectivity.

## 1 Answer

Injectivety:

Your solution is not correct, you have to suppose $$f^{-1}(x)=f^{-1}(x')$$ and not $$f^{-1}(f(x))=f^{-1}(f(x'))$$. Here is correct procedure:

Suppose we have $$x$$ and $$x'$$ such that $$f^{-1}(x)=f^{-1}(x')$$. Then we have: $$f(f^{-1}(x))=f(f^{-1}(x'))\implies x=x'$$ and thus a conclusion.

Surjectivity:

There is no need to suppose not existence of $$x'$$ and involving $$f$$ in first place with $$f^{-1}(f(x')) =x$$. Remember you have $$b$$ and you have to find $$a$$ such that $$f^{-1}(a)=b$$. Here is faster solution:

Take any $$b$$, and let $$a=f(b)$$. Then $$f^{-1}(a) = f^{-1}(f(b)) =b$$ and thus a conclusion.