MHB Proving Inverse of 1-1 Function $f$

[evinda]

Hi! (Wave)

Could you give me a hint how I could show that if $f$ is a function, that is $1-1$, then, it stands that:

$$(\forall x \in dom(f)) f^{-1}(f(x))=x$$

? (Thinking)

 

[caffeinemachine]

Hi! (Wave)

Could you give me a hint how I could show that if $f$ is a function, that is $1-1$, then, it stands that:

$$(\forall x \in dom(f)) f^{-1}(f(x))=x$$

? (Thinking)

Firstly, there is a problem with your statement as such.

Note that $f^{-1}(y)$ is a subset of the domain of $f$ whenever $y$ is in the target of $f$. So what you intend to write is this:

$$(\forall x\in \text{dom}f)\ f^{-1}(f(x))=\{x\}$$

Now. Assume that there is some pesky $x$ in the domain of $f$ such that $f^{-1}(f(x))=S\neq \{x\}$.

Note that $x$ is clearly in $S$.

By assumption, there exists $x'\in S$ such that $x'\neq x$. But then $f(x')=f(x)$ where $x'\neq x$. So $f$ is not $1$-$1$.

This is merely language. If you find this cryptic, try thinking what a $1$-$1$ looks like in terms of pictures. Make the domain a a "bubble" and put a few dots in it. Do the same for the target. From here it should be intuitively clear what's happening.