Question

Show that if $f:A\to B$ and $g:B\to C$ are $1-1$, then $gof:A\to C$ is also $1-1$

Answer 1

$\because f$ is one-one.

$\therefore$If $f\left(x1\right)=f\left(x2\right),thenx1=x2$

$\because G$ is also one-one.

So,$g\left(x1\right)=g\left(x2\right)\therefore x1=x2Now,gof\left(x1\right)=gof\left(x2\right)g\left(f\left(x1\right)\right)=g\left(f\left(x2\right)\right)$

$\because g$ is one-one.

$\therefore f\left(x1\right)=f\left(x2\right)$

As $f$ is one-one.

$x1=x2$

Hence $gof$ is also one-one.