Question

Let $f:A\to B$ and $g:B\to C$ be one-one onto functions. prove that $\left(gof\right):A\to C$ which is one-one onto

Answer 1

$f$ is one-one

If $f\left(x_1\right)=f\left(x_2\right)$

then $x_1=x_2$

$g$ is one-one

If $g\left(x_1\right)=g\left(x_2\right)$

then $x_1=x_2$

Suppose $g.f\left(x_1\right)=g.f\left(x_2\right)$

As $g$ is one-one.

So,$f\left(x_1\right)=f\left(x_2\right)$

As $f$ is one-one.

$x_1=x_2$

Hence,$g.f$ is one-one.

Since $g:B\to C$ is onto

Suppose $z\in C,$ then there exists a pre-image in $B$

Let the pre-image be $y$

Hence, $y\in B$ such that $g\left(y\right)=z$

Similarly,Since $f:A\to B$ is onto

If $y\in B,$ then there exists a pre-image in $A$

Let the pre-image be $x$

Hence, $x\in A$ such that $f\left(x\right)=y$

Now,$g.f=g\left(f\left(x\right)\right)=g\left(y\right)=z$

So, for every $x$ in $A,$ there is an image $z$ in $C,$ thus,

$gof$ is onto.