Question

Give examples of two functions $f:N\to N$ and $g:N\to N$ such that gof is onto but $f$ is not onto

Answer 1

$Letf:N\to Nbef\left(x\right)=x+1andg:N\to Nbeg\left(x\right)=\{x-1,x>11,x=1\}$

We will first show that $f$ is not onto.

Checking $f$ is not onto.

$Letf:N\to Nbef\left(x\right)=x+1Lety=f\left(x\right),whereyϵN\therefore y=x+1\Rightarrow x=y-1fory=1,x=1-1=0$

But $0$ is not a natural no.

$\therefore f$ is not onto.

Finding $gof:$

$f\left(x\right)=x+1andg\left(x\right)=\{x-1,x>11,x=1\}$

$forx=1f\left(x\right)=x+1;g\left(x\right)=1Sinceg\left(x\right)=1,g\left(f\left(x\right)\right)=1Sogof=1Forx>1f\left(x\right)=x+1,g\left(x\right)=x-1Sinceg\left(x\right)=x-1,g\left(f\left(x\right)\right)=f\left(x\right)-1gof=(x+1)-1\Rightarrow gof=xSo,gof=\{x,x>11,x=1\}Letgof=y,whereyϵNSo,y=\{x,x>11,x=1\}$

here $y$ is a natural no.As $y=x$.

So,$x$ is also a natural no.

hence, $gof$ is onto.