Let $g : B \rightarrow B$ be a function satisfying the condition $(g \circ g)(x) = x$ for all $x \in B$. Show that $g$ is a bijection.
For $g$ is onto
we want to prove that there exists an $a \in B$ such that $g(a) = b$ for arbitrary element of $B$.
Let $b \in B$, then since $(g \circ g)(b)=b$, then $ g(g(b))=b$.
There exists $a \in B$ , $ g(b)=a \in B$ so that $g(g(b))=g(a)=b$ as required.
Hence $g$ is onto.
How we want to prove that there exists $g(x_1)=g(x_2)$ for $x_1,x_2 \in B$, where $x_1=x_2$.
Let $x_1,x_2 \in B$ and suppose $g(x_1)=g(x_2)$ , so $g(g(x_1))=g(g(x_2)) $.
Then since $g \circ g(x)=x$ and $x$ is in $B$, $x_1=x_2$, hence $g$ is one to one.
Is this right? Or is there better way to prove this?