Largest domain $A$ such that $f$ is bijective I have just started a course that teaches elementary set theory, being a novice the notation confuses me. Usualy when I figure out what they are asking for, finding it is easy. Now I am having problems once again figuring out what the problem asks for.
Could someone please explain to me what the problem asks me to find in layman terms? 
This is taken from the first homework exercises here

I understand that the function $f(x) = x^3$ is bijective if we look at $x\geq0$ or $x\leq0$, but how would I write it as an answer? If someone could solve the first one for me, I will try to solve the remaining ones myself. 
 A: I would say that largest possible would be an $A$ (possible many of them) such that $f(A)$ is as large as possible.

For the first one, as you mentioned $f(x) = x^3$ is a bijection as a function $\mathbb{R} \rightarrow \mathbb{R}$. Since $A \subset \mathbb{R}$, letting $A = \mathbb{R}$ is as "large" as possible. $f(A) = \mathbb{R}$ and $f^{-1}(x) = x^{\frac{1}{3}}$.
To understand what "as large as possible means" : note that since $f(x) = x^3$ is bijective on all of $\mathbb{R}$. It is still bijective on its image if you restrict to any $A \subset \mathbb{R}$. For example, if you let $A = [0,1]$, then $f : A \rightarrow A$ is a still a bijection. However $A$ is not "as large" as possible since $f(A) = [0,1]$. 
For the second one, you can let $A = [-\frac{\pi}{2}, \frac{\pi}{2}]$. Using some basic understanding of sine, it easy to see that $\sin(x) : A \rightarrow [0,1]$ is a bijection. $f^{-1}(x) = \sin^{-1}(x)$, restricted to $A$ of course.  I leave it to you to think of other $A$ such that the $f(A) = [-1,1]$ and $f$ is bijective.
A: Suppose that you’re given a function $f:X\to Y$ and asked to find a set $A\subseteq X$ on which $f$ is a bijection. $A$ is as large as possible if there is no $B\subseteq X$ such that 


*

*$f$ is a bijection on $B$, and  

*$A\subsetneqq B$.


In other words, if you expand $A$ to include even one more point of $X$, $f$ is not a bijection on the expanded set. This means that if $x$ is any point of $X$ that does not belong to $A$, there is already some point $a\in A$ such that $f(a)=f(x)$: if there were no such $a$, you could expand $A$ to $B=A\cup\{x\}$, and $f$ would still be a bijection on $B$.
Note that if $f$ is a bijection on all of $X$, then $X$ is automatically the largest possible subset of itself on which $f$ is a bijection.
Another point to note is that there may be more than one set $A$ that is as large as possible. To use an example that is not part of the problem, consider the function $f:\Bbb R\to\Bbb R:x\mapsto\cos x$. You know that the possible values of $\cos x$ are precisely the real numbers in the interval $[-1,1]$, so if this $f$ is a bijection on some $A\subseteq\Bbb R$, and $f[A]=[-1,1]$, then $A$ is as large as possible. Three such sets $A$ are $[0,\pi]$, $[\pi,2\pi]$, and $(-\pi/2,0]\cup[\pi/2,\pi]$, as you can see fairly easily if you draw the graph of $y=\cos x$.
By the way, there is a typo in part (c) of the problem: the function should read
$$\begin{bmatrix}x\\y\end{bmatrix}\mapsto\begin{bmatrix}1&0\\1&2\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x\\x+2y\end{bmatrix}\;.$$
