Let $f:A\to B$ be a bijection for any set A and B. Show there exists $g:B\to A$ which is also a bijection. Show $g=f^-1$.

This is what I have done: Proof: Suppose $g(y_1)=g(y_2)$ for some $y_1$ and $y_2$ in B. Since f is surjective, $ \exists$ $x_1$ and $x_2$ such that $y_1=f(x_1)$ and $y_2=f(x_2)$. By assumption, $g(f(x_1))=g(f(x_2))$. Hence, $x_1=x_2$. Then $f(x_1)=f(x_2)$, therefore $y_1=y_2$. Hence, g is injective.

To show g is surjective: For any x in A, $ \exists$ y in B such that $y_1=f(x_1)$. Then for that y, $g(y_1)=g(f(x_1))=x_1$.

Therefore, g is bijective.

Now, I do not even know where to start to show taht $g=f^{-1}$. Any help is appreciated. Thank you.


You need to show the existence of a function $g:B\to A$; therefore you must start by constructing such a function.

Let $y\in B$. By surjectivity of $f$, there exists some $x\in A$ such that $f(x)=y$. Furthermore such an element $x$ is unique given $y$, for if $x_1,x_2\in A$ and $f(x_1)=y=f(x_2)$, then $x_1=x_2$ by injectivity of $f$. We therefore define $g(y)$ to be the element $x\in A$ such that $f(x)=y$.

It is now straightforward to show that $g$ is a bijection from $B$ to $A$. To show that $g=f^{-1}$, show that $g\circ f$ is the identity function on $A$ and $f\circ g$ is the identity function on $B$.


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