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Ok, given $f: A\rightarrow B$ is bijective. How can I prove now that $f(f^{-1}(x))=x$? It must be injective and surjective, but how is it possible to pick an element from $A$ and show after applying $f(f^{-1}(\cdot))$ that is must be the same element?

Regards, Kevin

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    To make the domains of definitions a bit more explicit in the last comment, the inverse of $f$ is the function $f^{-1}$ such that $f \circ f^{-1} = \mathrm{id}_B$ and $f^{-1} \circ f = \mathrm{id}_A$.2011-10-24

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Here is the proof:

Let $x$ is an arbitrary point from $B$. As the function $f$ is bijective, it follows that there is a unique point $y$ in $A$ such that $f(y)=x$. Therefore $f(f^{-1}(x))=f(y)=x$.

Sincerely, Tigran