Let f, g: N → N such that f(n + 1) = f(n) + f(1) n ∈ N and g be any...

$f\left(n + 1\right) - f\left(n\right) = f\left(1\right)$

A.P. with common difference = $f ( 1 )$

General term = ${T}_{n} = f ( 1 ) + ( n - 1 ) f ( 1 ) = nf ( 1 )$

$⇒ f\left(n\right) = nf\left(1\right)$⇒ f$is one-one</p><p>Now, Let$f\left(g\left({x}_{2}\right)\right) = f\left(g\left({x}_{1}\right)\right)$</p><p>$⇒ g\left({x}_{2}\right) = g\left({x}_{1}\right)$(as$f$is one-one)</p><p>$⇒ {x}_{1} = {x}_{2}$(as$fog$is one-one)</p><p>$⇒ g$is one-one</p><p>For$f$to be onto,$f ( n )$should take all the values of natural numbers.</p><p>As$f ( x )$is increasing,$f ( 1 ) = 1$</p><p>$⇒ f ( n ) = n$</p><p>Now,$f$\left(g\left(n\right)\right) = g\left(n\right)f\left(1\right)$

may be many-one if

$g\left(n\right)$ is many-one

So “if $g$ is onto, then $fog$ is one-one” is incorrect.