how to find two right-inverse functions of a function i am stuck in this problem. i need to find two right-inverse functions of this function: 

$h: \Bbb N_0\times \Bbb N \to \Bbb N, (m,n)\mapsto m+n$. 

i know that the function h' is a right inverse of a function h if and only if: 

$h  \circ h' = m+n$. 

how can i say this in mathematical way and as for the h function given above?  i did this: 

$h': \Bbb N-&gt;\Bbb N_0\times\Bbb N, m+n\mapsto (m,n)$.
  $h \circ h' = h(h'(m+n))=m+n$

but now from here, i dont know how to finish the proof and how to show another right-inverse function of h

thanks for help in advance
As you point out, $h^{-1}$ maps naturals to pairs $(m, n)$. If you think about what $h^{-1}$ is doing, you'll see that you need to find a pair $(m, n)$, $m \in \mathbb{N}_0$, $n \in \mathbb{N}$ such that $m+n$ is the function $h^{-1}$'s input. That is, if $h^{-1}(k) = (m, n)$, then $m + n = k$. An easy approach is to fix some $m$ first, then let $n = k - m$. This will work provided $k &gt; m$. We can then handle the remaning cases separately. This gives $h^{-1}(n) = (0, n)$ as a simple example, since $(0, n) \in \mathbb{N}_0 \times \mathbb{N}$. So that gives one right inverse. Another can be given by, for example,

$$h^{-1}(n) = \begin{cases}(1, n-1), &amp; \mbox{if } n &gt; 1\\(0, 1), &amp; \mbox{if } n = 1\end{cases}$$

Can you devise a suitable function for $m = 2$? For $m = 3$?
Just to give a little more clarity. For each $m\in \mathbb{N}$, define $g_{m}:\mathbb{N}\rightarrow \mathbb{N}_{0}\times \mathbb{N}$ as
\begin{eqnarray*}
g_{m}(n)=\left \{
\begin{array}{cc}
(m,n-m)&amp;\text{ if } n\geq m\\
(0,n)&amp;\text{ if } n&lt;m
\end{array}\right.
\end{eqnarray*} 

Then each $g_{m}$ is a right inverse of $h$.