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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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How do you define $f^{-1}$? The definition I know essentially states that $f(f^{-1}(x)) = x$ for $x \in B$ and $f^{-1}(f(x))= x$ for $x \in A$. (EDIT: Incidentally, your last sentence is incorrect. The expression $f(f^{-1}(x))$ makes sense for $x \in B$, not $x \in A$.) –  Srivatsan Oct 24 '11 at 11:26
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@Kevin: The function $f(x) = x^2$ is not invertible as a function from $\mathbb{R}$ to $\mathbb{R}$, so "$f^{-1}(4)$" is meaningless in that situation. You'd have to restrict the domain to get a different, invertible function. The identity $f^{-1}(f(x)) = x$ follows by the definition of $f^{-1}$, but $f^{-1}$ only exists when $f$ is bijective. –  Carl Mummert Oct 24 '11 at 11:34
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@Kevin. You're confusing two things. The function $f: \mathbb{R} \longrightarrow \mathbb{R}$ defined by $f(x) = x^2$ has NOT an inverse, since it's not injective: $(-2)^2 = 4 = 2^2$. Right? The (positive) square root $g(x) = \sqrt{x}$ is the inverse of the function $h: \mathbb{R}_+ \longrightarrow \mathbb{R}$, $h(x) = x^2$, that is $f(x)$ restrited to the positive real numbers. Remember: a "function" is not just a formula -it includes its domain. So, the functions $f$ and $h$ are different functions: $f$ has no inverse, $h$ has an inverse. –  a.r. Oct 24 '11 at 11:37
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@Kevin. Finally: by definition, the inverse of $f$ is the function $f^{-1}$ sucht that $f\circ f^{-1} = \mathrm{id} $ and $f^{-1}\circ f = \mathrm{id}$. –  a.r. Oct 24 '11 at 11:40
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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 $f^{-1} \circ f = \mathrm{id}_A$. –  Srivatsan Oct 24 '11 at 11:47

1 Answer 1

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

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