I have to prove this:
Let $A$ and $B$ be sets and suppose $f: A \rightarrow B$ is one-to-one and onto. Then $f^{-1} \circ f = id_{A}$.
So far I have this:
We know that dom$(f^{-1} \circ f) =$ dom$(f)=A=$ dom$(id_{A})$. For any $a \in A$, $f^{-1} \circ f(a)=f^{-1}(f(a))=a=id_{A}$.
Is this enough to prove this (because the domains are equal and $f^{-1}(a) = f(a)$)?
Best Answer
You showed that domains of both functions are equal and that for all $a \in A,$ $$f^{-1} \circ f (a) = a = id(a)$$ Thus, you are done. The only step that may be missing is to justify why $f^{-1}(f(a))=a$. Besides that, everything looks good.