Show whether composition is identity mapping 
Let $F:\mathbb{N} \rightarrow \mathbb{N}$ be the mapping defined by $F(n)=n+1$ for all $n \in \mathbb{N}=\{0,1,2,3, . . .\}$
and let $I:\mathbb{N}\rightarrow \mathbb{N}$ be the
identity mapping defined by $I(n) =n$ for all $n \in \mathbb{N}$.
(a) Show that there does not exist any mapping $G:\mathbb{N} \rightarrow \mathbb{N}$
which satisfies $F\circ G=I$.
(b) Construct one example (or several examples) of mapping $H:\mathbb{N} \rightarrow \mathbb{N}$ which satisfies $H\circ F=I$

For (a), is it correct that there will be no mapping $G$ that satisfies the requirement because if I let $n=0$, we cannot map $0$ to $-1$?
And for (b), can I say that $H(n)= n-1$ is one of the examples?
 A: $a)$ Your intuition about $G$ is right, but you have to prove that there does not exist any function $G$ such that $F \circ G = I$. And you haven't proved this, because you haven't show that if a function $G$ such that $F \circ G = I$ existed, then $G(0) = -1$ by necessity. You are just claiming that the function $G(n) = n-1$ would be a solution for the problem, and you cannot accept such a $G$ because it is not a function from $\mathbb{N}$ to $\mathbb{N}$ (since $G(0) = -1$). But, a priori, it does not exclude that there might be other solutions for your problem.
We want to prove that there does not exist any mapping $G \colon \mathbb{N} \to \mathbb{N}$ which satisfies $F \circ G = I$. I give you a proof by contradiction. Suppose that there exists a function $G \colon \mathbb{N} \to \mathbb{N}$ such that $F \circ G = I$. A basic theorem in the theory of functions (see here for an elementary proof) says that, since $F \circ G =I$, the function $F \colon \mathbb{N} \to \mathbb{N}$ must be a surjection, i.e. for every $n \in \mathbb{N}$ there is an $m \in \mathbb{N}$ such that $F(m) = n$. But $0 \in \mathbb{N}$ and there is no $m \in \mathbb{N}$ such that $F(m) = 0$ (indeed, the image of $F$ is $\mathbb{N} \smallsetminus \{0\} = \{1,2,3,\dots\}$). So, $F$ is not a surjection and we have a contradiction. Therefore, there is no function $G \colon \mathbb{N} \to \mathbb{N}$ which satisfies $F \circ G = I$.

$b)$ Your intuition about the definition of $H$ is right, but you have to be more rigorous. Indeed, if you define $H(n) = n -1$ for every $n \in \mathbb{N}$, then $H$ is not a function from $\mathbb{N}$ to $\mathbb{N}$ because $H(0) = -1 \notin \mathbb{N}$. A slight modification in the definition of $H$ gives you a function $H \colon \mathbb{N} \to \mathbb{N}$ such that $H \circ F = I$.
Let $H$ be a function whose domain is $\mathbb{N}$ and defined by
$$H(n) = 
\begin{cases}
0 &\text{if } n = 0 \\
n-1 &\text{otherwise}.
\end{cases}
$$
Clearly, also the image of $H$ is $\mathbb{N}$ because for every $n > 0$ we have $n -1 \in \mathbb{N}$.
Hence, $H \colon \mathbb{N} \to \mathbb{N}$ i.e. $H$ is a function from $\mathbb{N}$ to $\mathbb{N}$.
Moreover, for every $n \in \mathbb{N}$, we have
$$H(F(n)) = H(n+1) = (n+1)-1 = n$$
(the second equality holds because $n + 1 > 0$).
Therefore, $H \circ F = I$.
