There are 2 functions H,G $\colon \mathbb N \to \mathbb N$ which defined like that:
$$H(x) = \begin{cases} 2x + 1 \end{cases} $$
$$G(x)=\begin{cases} x - 1 & if & x > 1\\ 3 & if & x = 1\\ \end{cases}$$
we will define F $\colon \mathbb N^N \to \mathbb N^N$ like that: $F(f) = G ∘ f ∘ H $
you need to prove that $F$ is Surjective.
Okay, you need to prove that if there is a function
$h:\mathbb N \to \mathbb N$
Then there is an $f:\mathbb N \to \mathbb N$ so that
$F(f) = G\circ f \circ H = h$.
So $F(f)(x) = G(f(H(x))) = h(x)$.
$G(f(2x+1)) = $
$\begin{cases} f(2x+1) - 1 & if & f(2x+1) > 1\\ 3 & if & f(2x+1) = 1\\ \end{cases}\\ =h(x)$
Let's map this out:
Case 1:
$n \to 2n+ 1 \to f(2n+1)\to f(2n+1) - 1 = h(n)$
We can have this happen by defining. If $m = 2n + 1$ is odd. Then $f(m) = h(\frac {m-1}2) + 1=h(n) + 1$
Note: as $h(n) + 1$ by definition is greater than $1$ so we don't need to worry about $f(2n+1) = 1$ and branching to $G(f(2n+1)) = 3 \ne h(n)$.
So..... there is no Case 2.
We haven't defined $f$ for even numbers or for the $m =1$. We can do what ever we want in those case as $H(n)$ is never even or equal to $1$.
So define if $m = 2n$ then $f(m) = 3,567$. Why not?
If $m = 1$ then $f(1) = 29$. Why not?
And again if $m = 2n + 1$ is odd then $f(m) = h(\frac {m-1}2) + 1=h(n) + 1$
So:
$G(f(H(n)) = G(f(2n + 1))=g(h(n) + 1)= (h(n) + 1) -1 = h(n)$.
So There does exist an $f$ so that $F(f) = h$. And so $F$ is surjective. (But definitely not injective)