Question

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

Answer 1

$Letf\left(x\right)=xandg\left(x\right)=\left|x\right|where:f:N\to zandg:z\to zg\left(x\right)=\left|x\right|=\{x,x\ge 0-x,x<0\}$

checking $g\left(x\right)$ injective (one-one).

$g\left(1\right)=\left|1\right|=1g(-1)=\left|1\right|=1$

Since,different elements $1,-1$ have the same image $1,\therefore g$ is not injective (one-one).

checking$gof$ injective (one-one).

$f:N\to z\&g:z\to zf\left(x\right)=xandg\left(x\right)=\left|x\right|gof\left(x\right)=g\left(f\left(x\right)\right)=\left|f\left(x\right)\right|=\left|x\right|=\{x,x\ge 0-x,x<0\}$

here $gof\left(x\right):N\to z$

So,x is always natural no.

Here$\left|x\right|$ will always be a natural number.

So,$gof\left(x\right)$ has a unique image.

$\therefore gof\left(x\right)$ is injective.