Determine injective or surjective of these functions I have the following functions:
$$
\begin{align}
f&: \mathbb{N} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N} \\
g&: \mathbb{N} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N}
\end{align}
$$
defined by $f(x, y) = (x + y, x)$ and $g(x, y) = (x - y, y)$.
a) Calculate $g\circ f$ for the originals $(2, 2), (3, 5)$ and $(4, 1)$.
$$
\begin{align}
f(2, 2) &= (2 + 2, 2) = (4, 2) \text{ and then }  g(4, 2) = (4 - 2, 2) = (2, 2)\\
f(3, 5) &= (3 + 5, 3) = (8, 3) \text{ and then } g(8, 3) = (8 - 3, 3) = (5, 3)\\
f(4, 1) &= (4 + 1, 4) = (5, 4) \text{ and then } g(5, 4) = (5 - 4, 4) = (1, 4)
\end{align}
$$
b) Give a function description for the function $h: \mathbb{N} \times \mathbb{N} \to \mathbb{N} \times \mathbb{N}$ with $h = g \circ f$ and show that it counts for all $x$, $y$ element of $\mathbb{N}$.
I have no idea how to tackle these two questions, don't even know where to begin.
c) Show that $f$ is injective, but not surjective.
To show that something is injective, I would need to find an element of the codomain that does not have an element in the domain. To find something that is surjective, I would need to find an element in the codomain that has more than one original.
Some sample input/output data:
input  ->  output
$$
\begin{align}
(0, 0) &\to (0, 0)\\
(0, 1) &\to (1, 0)\\
(0, 2) &\to (2, 0)\\
...&\text{etc...}\\
(1, 0) &\to (1, 1)\\
(1, 1) &\to (2, 1)\\
...&\text{etc...}\\
(2, 0) &\to (2, 2)\\
(2, 1) &\to (3, 2)\\
\end{align}
$$
...etc...
Now, if I have e.g. the element $(0, 1)$ in the codomain, then there is no corresponding element in the domain. In element $(0, 1), x = 1$, but then $y$ has to be $-1$ to make the $0$, and since the domain is $\mathbb{N} \times \mathbb{N}$, this cannot be.
Another example, this time with element $(1, 0)$ in the codomain. This means that $x = 0$, and that $y = 1$. So the corresponding element in the domain should then be $(0, 1)$, but when this element is put into $f$, it goes to $(1, 1)$. In other words, $(1, 0)$ also does not have an original.
Is this evidence enough to say that this function is not surjective, or do I still need to prove it further?
d) We confine the function $f$ now to
$$
f: \mathbb{N} \times \mathbb{N} \to \{(n, m)\lvert n \geq m\}
$$
still with
$$
f(x, y) = (x + y, x).
$$
Show that the inverse of $f$ now does exist, and calculate this inverse.
My problem here is that there are elements that satisfy $n \geq m$, but these are not inverse. For example, if I put $(2, 1)$ into $f$, the answer is $(3, 2)$. This is not the inverse of $(2, 1)$ What is the thinking mistake I'm making here?
 A: For (b):
Note that $$h = g\circ f(x,y) = g(x+y, x) = (y, x).$$
For (c):
Surjective? No. For this to be true you would have to be able to find $x,y$ such that $f(x,y) = (x + y, x)= (1,2)$. That would mean $x = 2$ and $y = -1 \notin \mathbb{N}$. So that is not possible. 
Injective? Say that $f(x,y) = f(x', y')$. We are assuming that two different inputs give the same output. For $f$ to be injective we need to prove that the inputs actually are the same. So we have $f(x,y) = f(x',y')$ and we need to prove that $x= x'$ adn $y = y'$. That $f(x,y) = f(x', y')$ means that $(x+y, x) = (x' + y', x')$. But if this is true then we certainly have that  $x = x'$. And if $x=x'$ and $x + y = x' + y'$, then it follows that  $y = y'$. Hence $f$ is injective. Note that $x'$ (the backtick or prime or what ever one calls it) is just another element. It isn't doesn't mean that it is necessarily related tot $x$. One could have chosen another variable name instead.
For (d):
Now say that $(n,m)\in \mathbb{N} \times \mathbb{N}$ with $n\geq m$. To prove that $f$ now is surjective, you want to find $x,y$ such that $$f(x,y) = (n,m)$$. But since $n\geq m$ you can write $n = m + a$ for $a\geq 0$. So you want $(x,y)$ such that 
$$x + y = n\quad \text{and}\quad x = m.$$ Well pick $x = m$. Then left is to find $y$ such that $x+y = n$, and with the choice of $x$, that means that we want $y$ so that $m + y = n$. That is $m + y = m + a$. This you have if you exactly if pick $y = a$. So $f$ is both injective and surjective here. It is clearly surjective and the injectivity follows from the fat that there was only one way to pick $x$ and $y$.
(I guess that there is the subtle assumption that $0\in \mathbb{N}$.)
