Prove that there exists a bijective function $i:S \to S$ such that $i \circ g = g$ and $g \circ i = g$ for all bijections $g: S \to S$.

functions

Let $S$ be a set.

($a$) Prove that there exists a function $i:S \to S$ such that $i \circ g = g$ and $g \circ i = g$ for all bijections $g: S \to S$.

($b$) Prove that $i$ is a bijection.

Proof of ($a$):

Suppose $y \in S$. Then by surjectivity of $g$, there exists $x \in S$ such that $g(x)=y$. Suppose $i:S \to S$ is defined as $i(x_{0})=x_{0}$ for any $x_{0} \in S$. Then,

$y=g(x)=g(i(x)) = (g \circ i)(x)$,

since $x = i(x) \in S$.

By definition of $i$,

$y = g(x) = i(g(x)) = (i \circ g)(x)$.

Thus, $(g \circ i)(x) = (i \circ g)(x)$ for any $x \in S$. QED

Proof of ($b$):

For $x_{1},x_{2} \in S$, $x_{1} \ne x_{2}$, we have that $i(x_{1})=x_{1}$ and $i(x_{2}) = x_{2}$ both imply $i(x_{1}) \ne i(x_{2})$. So, $g$ is injective. Then for any $x \in S$, there exists $x$ in $S$ such that $i(x) = x$. This implies that $g$ is surjective. Since we have shown that $g$ is both injective and surjective, it follows that $g$ is a bijective function. QED

I wanted to know if both proofs are correct.

Best Answer

(b) is largely correct with the following stylistic flaws:

  • "both" should be replaced by "together"
  • "there exists $x$ in $S$ such that $i(x)=x$" should be replaced by "there exists $y$ in $S$ such that $i(y)=x$, namely $x$".

(a) is incomprehensible to me:

  • The definition of $i$ should be the start of the proof.
  • $g$ should be introduced by a sentence like "Let $g$ be a bijection of $S$ onto $S$".
  • You introduce $y$ as if you want to prove a statement of the form $\forall y\in S:\ldots$, but end up proving something of the form $\forall x\in S:\ldots$.
  • You only try to prove $g\circ i=i\circ g$, but not that both terms are equal to $g$.

A correct proof of (a) would be:

Define $i:S\rightarrow S$ by $i(x_0)=x_0$ for all $x_0\in S$. Let $g$ be a bijection of $S$ onto $S$. Then, for all $x\in S$, $$ g(x)=g\big(i(x)\big)=(g\circ i)(x) $$ and $$ g(x)=i\big(g(x)\big)=(i\circ g)(x). $$

Note that I've not used that $g$ is bijective. That's because it isn't necessary. We have $g\circ i=g=i\circ g$ for all functions $g:S\rightarrow S$. The problem statement is misleading in that regard.

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