# Bijection proof (intro analysis)

1. Nov 24, 2013

### zelmac

I'm wondering whether my solution to this problem is correct, since the answer in the answer sheet says that this is provable, and I think I found a counterexample... Any help is appreciated :)

1. The problem statement, all variables and given/known data

Prove or disprove: $f:D->K$, where $D,K != empty$, is a bijection iff there exists a unique function $g:K->D$ such that:
$f\circ{g}\circ{f}=f$

2. Relevant equations

3. The attempt at a solution

Lets look at the following function: $f:\{1\}-->\{1,2\}$, defined by $f(1)=1$. Clearly the function$g:\{1,2\}-->\{1\}$, defined by $g(1)=1$ and $g(2)=1$ satisfies $f\circ{g}\circ{f}=f$, since $f(g(f(1)))=f(g(1))=f(1)$, and 1 is the only element in the domain of f. Lets assume that $g':\{1,2\}-->\{1\}$ is a function that satisfies $f\circ{g'}\circ{f}=f$. The only way that g' is even a function is if we define it as $g'(1)=1$and $g'(2)=1$, so $g'=g$, so $g$ is a unique function $K-->D$ which satisfies the condition.
But f is not a surjection, so we have found a function $f:D-->K$ which is not a bijection, but for which there exists a unique function $g:K-->D$ such that $f\circ{g}\circ{f}=f$, so the proposition does not stand.

Last edited: Nov 24, 2013
2. Nov 24, 2013

### Dick

Sounds correct to me. There is only one function g:{1,2}->{1}, but f is not a surjection.