0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.