Question

Consider functions f and g such that composite gof is defined and is one are g both necessarily one-one.

Answer 1

Consider function $f$ and of such that composite $gof$ is definite of is one are $g$ both necessarily one-one.

Let $f:A\to B$ and $g:B\to C$ be two function such that $gof:A$ $\because C$ is defined.

We can given that $gof:A\to C$ in one-one.

we are to prove that $f$ is one -one if possible.

Suppose that $f$ is not one-one.

$x_1,x_2\in A$ $\because x_1\ne x_2$ but $f\left(x_1\right)=f\left(x_2\right)$

$\Rightarrow g\left(f\right(x_1\left)\right)=g\left(f\right(x_2\left)\right)$

$gof\left(x_1\right)=gof\left(x_2\right)$

$\because x_1,x_2\ne A\because x_1\ne x_2$ but $\left(gof\right)\left(x_1\right)=\left(gof\right)\left(x_2\right)$

$gof$ is not one which is against the given hypothysis that $g$ of is one -one superposition is wrong.

$f:\{1,2,3,4\}\ne \{1,2,3,4,5,6\}$ defined as $f\left(x\right)=x\forall x$

$g:\{1,2,3,4,5,6\}\to \{1,2,3,4,5,6\}$ as $g\left(x\right)=x$

and $g\left(5\right)=g\left(6\right),gof\left(x\right)=x$ which shows $gof$ is one-one $g$ is not one-one.