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Suppose that $f$ is an injection. Show that $f^{-1}(f(x))=x$ for all $x$ in $D(x)$, and $f(f^{-1}(y))=y$ for all $y$ in $R(f)$. I understand the algebra behind it, and can show this with a random one-to-one function, but don't know where to start such a general proof.

For the first half:
Let $f(x_1)=y_1$ and $f(x_2)=y_2$
$\rightarrow$ $f^{-1}(y_1)=x_1$ and $f^{-1}(x_2)=y_2$.
Using composition, $f^{-1}(f(x_1))=f^{-1}(f(x_2))$
$\rightarrow$$f^{-1}(y_1)=f^{-1}(y_2)$
$\rightarrow$ $x_1=x_2$

I used similar logic for the second half. Is this correct, or am I just spinning in circles?

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5  
What is your definition of inverse function? –  Hagen von Eitzen Feb 3 '13 at 20:47
1  
Yes, it seems itself as the definition of $f^{-1}$. –  Berci Feb 3 '13 at 20:48
    
(Again, I apologize, I'm trying to learn how to code this, and don't have a clue!). Inverse: If f mapping A onto B is a bijection of A onto B, then g:={(b, a) element of BxA: (a,b) element of f} is a function of B into A. –  MrsMillz Feb 3 '13 at 20:57

1 Answer 1

Suppose that $f^{-1}(f(x)) \neq x$ for some $x \in D(f)$. Then $(f(x), x) \notin f^{-1}$. But since $$\{f^{-1} = \{(b,a): (a,b) \in f\}$$ this means $(x, f(x)) \notin f$ which is a contradiction. The proof of the second assertion is similar.

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