Question

Let $g:N\to N$ be defined as
$g(3n+1)=3n+2$,
$g(3n+2)=3n+3$,
$g(3n+3)=3n+1$, for all $n\ge 0$.
Then which of the following statements is true?

• A. $gogog=g$
• B. There exists a one-one function $f:N\to N$ such that $fog=f$
• C. There exists a function $f:N\to N$ such that $gof=f$
• D. There exists an onto function $f:N\to N$ such that $fog=f$

Answer 1

The correct option is D There exists an onto function $f:N\to N$ such that $fog=f$

$\because g(3n+1)=3n+2,g(3n+2)=3n+3\text{and}g(3n+3)=3n+1$
$\therefore gogog(3n+1)=g\left(g\right(g(3n+1)\left)\right)=g\left(g\right(3n+2\left)\right)=g(3n+3)=3n+1$
So we can see that $gogog\left(x\right)=x$ (identity)
For $fog=f$ to hold $f$ must be an onto function.