I have actually lost the theme of this thread. I even missed - sign on one of your posts.
Here is what I read as the OP:
$f: (\mathbb{Z}\times\mathbb{Z})\to\mathbb{Z}$ defined by the rule $(m,n)\mapsto m^2+n^4$.
If again I have not missed something, it is clearly $f(m,n)\ge0$ so can it be onto?
It can't be cause they don't map to anything, correct?
In post #14 I mentioned
And then I took a shot at it by stating this
Sorry for getting you mixed up. When I said OP I wasn't trying to refer to the question as a whole, but to understanding the domain/codomain, which you explained to me very well.
Suppose we want to prove $f: \Bbb Z \times \Bbb Z \to \Bbb Z$ given by $f(m,n) = 2m+n$ is onto.
Then we must show that for ANY integer $k$, there is at least one pair $(m,n)$ with $f(m,n) = k$.
We can pick $m,n$ any way we want, as long as they are both integers. Suppose we pick: $m = 0, n = k$?
I can tell you that $f$ is NOT injective. See if you can find two different pairs $(m_1,n_1),(m_2,n_2)$ such that:
$f(m_1,n_1) = f(m_2,n_2)$. I suggest finding two such pairs with both images of $f$ equal to 3.
Let's look at the last function:
$f(n) = n^2 + 1$.
Can we find an $n$ such that $f(n) = -1$? Suppose that (just for the time being) one existed. Then:
$n^2 + 1 = -1$
$n^2 = -2$.
If $n = 0$, then $n^2 = 0 \neq -2$. So $n$ is not 0.
If $n > 0$, then $n^2 > 0$ so $n^2 \neq -2$, since $-2 < 0$.
If $n < 0$ then $n = -k$ for some $k > 0$, and thus $n^2 = (-k)(-k) = (-1)(k)(-1)(k) = (-1)(-1)k^2 = k^2 > 0$, so again $n^2 \neq -2$.
So there is NO integer $n$ with $f(n) = -2$, so $f$ CANNOT be onto, since the value -2 is not in the range of $f$.
Can $f$ be injective? Again, no, because $f(-1) = (-1)^2 + 1 = 1 + 1 + 2 = 1^2 + 1 = f(1)$, so $f$ maps -1 and 1 to the same value. That is:
$f(-1) = f(1)$ but $-1 \neq 1$, and an injective function must satisfy:
$f(n) = f(m) \implies m = n$ for ALL $m,n$ and this simply isn't true for n= -1, and m = 1.