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I'd like to know if the following proof is correct.

Prove that for every bijection $f : S \to S$, there exists a function $g : S \to S$ such that $f \circ g = i$ and $g \circ f = i$

proof: Let $f$ be a bijection from $S$ onto $S$, and let $i:S \to S$ be defined as $i(x)=x$ for any $x \in S$. By definition, each $y \in S$ is mapped to by exactly one element in $S$. Then let $g: S \to S$ be defined as $g(x_{0}) = z_{0}$ and $g(y_{0})=x_{0}$, where $f(x_{0})=y_{0}$ and $f(z_{0})=x_{0}$. It follows that

$i(x_{0})=x_{0}=g(y_{0})=g(f(x_{0}))=(g \circ f)(x_{0})$

and

$i(x_{0})=x_{0}=f(z_{0})=f(g(x_{0})) = (f \circ g)(x_{0})$

Skm
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    Yup, that looks good. One could expand a bit on why this completely defines a unique $g$, but other than that it is fine, I think. – Watercrystal Jul 03 '20 at 00:04
  • It looks like your set consists of at most 3 elements $x_0, y_0, z_0$. You probably need to avoid that impression. – markvs Jul 03 '20 at 01:06
  • @JCAA: Would I add "for arbitrary $x_{0},y_{0},z_{0} \in S$" at the end of the sentence "Then let $g: S \to S$ be defined as $g(x_{0}) = z_{0}$ and $g(y_{0})=x_{0}$, where $f(x_{0})=y_{0}$ and $f(z_{0})=x_{0}$?" – Skm Jul 03 '20 at 02:19
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    Arbitrary elements are rarely denoted $x_0$. – markvs Jul 03 '20 at 02:20

1 Answers1

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Your definition of $g$ has some problem.

Since $f$ is bijection, each element $x$ of $S$ has exactly one preimage under $f$. Now define $g:S\rightarrow S$ by, $g(x)=y$, where $y$ is the preimage of $x$ under $f$. Then we see that $g(x)=y\Leftrightarrow f(y)=x$. From this $f\circ g=i$ and $g\circ f=i$ follow easily.

Edit: Proof of $g\circ f=i$.
Let $y\in S$ then $g\circ f(y)=g(f(y))=g(x)=y$. Gives $f\circ g=i$.

  • So, I see how $f\circ g = i$ follows from the given definition. But, I'm not sure how we get $g \circ f = i$? – Skm Jul 03 '20 at 16:12