Bijective functions questions. I'm having some problems with these questions on functions being bijective or not and I would appreciate some help.

Are the following functions bijective?
a) $f\colon \Bbb N \to \Bbb N$ defined as: $f(n)=n^2$.
b) $f\colon [-1,1]\to[0,1]$ defined as: $f(n) = n^2$.
c) $f\colon \Bbb R^2 \to \Bbb R^2$ defined as: $f(x,y)=(x+y,x-y)$.
d) $f\colon \Bbb Z^2 \to \Bbb Z^2$ defined as:
$$f(x,y)=\begin{cases}(x,y+1), &\text{if $x$ and $y$ are both even or odd}\\(x+1,y),&\text{otherwise}\end{cases}$$

I have two questions from these sets of questions.
For (b) I said the $0$ from the $[0, 1]$ set does not have a pre-image because no number from the $[-1, 1]$ set squared equals $0$. Therefore it's not surjective so it's not bijective. I am not sure if I am going about this question the right way, or if it's even right.
And also for (c), from what I understand, the input is a set of pairs of rational numbers and so is the output, however, I don't know how to prove it through disproving injection or surjection. Any help would be appreciated.
UPDATE: I attempted to do D and this is what I got. I first tried to prove surjectivity. So, if I have an element [20, 10] for example, you can't get to there because the function only outputs pairs of numbers that are 1 apart. So this means its not surjective so it cannot be bijective. Is this correct?
 A: I’ll do (c) as a model and leave (d) for you to try.
To show that $f$ is bijective, you must show that it is both injective and surjective.


*

*$f$ is injective. Suppose that $\langle x_0,y_0\rangle,\langle x_1,y_1\rangle\in\Bbb R^2$ and $f(x_0,y_0)=f(x_1,y_1)$; then $\langle x_0+y_0,x_0-y_0\rangle=\langle x_1+y_1,x_1-y_1\rangle$, so $x_0+y_0=x_1+y_1$ and $x_0-y_0=x_1-y_1$. Adding these two equations, we find that $2x_0=2x_1$ and hence that $x_0=x_1$; subtracting this from $x_0+y_0=x_1+y_1$, we find that $y_0=y_1$. Thus, $\langle x_0,y_0\rangle=\langle x_1,y_1\rangle$, and $f$ is therefore injective. (This is the standard approach to proving injectivity: assume that the function takes two things in the domain to the same thing in the range, and show that the things in the domain were actually the same thing.)

*$f$ is surjective. Let $\langle u,v\rangle\in\Bbb R^2$ be arbitrary; we must find $\langle x,y\rangle\in\Bbb R^2$ such that $f(x,y)=\langle u,v\rangle$, i.e., such that $\langle x+y,x-y\rangle=\langle u,v\rangle$, or, finally, $x+y=u$ and $x-y=v$. Adding these two equations, we find that $2x=u+v$, so we must set $x=\frac12(u+v)$. Subtracting the second from the first gives us the equation $2y=u-v$, from which we see that we must set $y=\frac12(u-v)$. You can now check that with these values for $x$ and $y$ we do indeed find that $f(x,y)=\langle u,v\rangle$, and it follows that $f$ is surjective.
