JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Sets Relations and Functions

The recurrence $f(n+1)=f(n)+f(1)$ gives $f(n)=nf(1)$, so $f(n)=cn$ with $c=f(1)\in \mathbb{N}$. Such an $f$ is one-one, and if it is also onto then $1$ has to be a value, which forces $c=1$ and $f(n)=n$. Since $f$ is injective, $f\circ g$ is one-one whenever $g$ is, and conversely $f(g(n_1))=f(g(n_2))$ gives $cg(n_1)=cg(n_2)$ and hence $g(n_1)=g(n_2)$, so $f\circ g$ being one-one forces $g$ to be one-one. A surjection of $\mathbb{N}$ onto itself need not be injective: $g(1)=g(2)=1$ with $g(n)=n-1$ for $n\ge 3$ is onto, and then $f\circ g$ takes the same value at $1$ and $2$. The statement that is not true is that if $g$ is onto, then $f\circ g$ is one-one.