Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
@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
@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
@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
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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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