Showing that a function between $\Bbb R[x]/(x^2 - 1)$ and $ \Bbb R\times \Bbb R $ is surjective? I am fairly new to more abstract mathematical structures and frankly I'm quite new to detailed proofs in general. For context, this is not a homework problem, I am just trying to better understand how to convert conceptual understandings of things into parts of proofs.
I would like to show that the function  $$f: \Bbb R[x]/(x^2 - 1) \rightarrow \Bbb R \times \Bbb R $$
$$f(a + bx + (x^2 - 1)) := (a + b, a - b) $$
is surjective. I understand that this requires showing that for every $q \in \Bbb R \times \Bbb R$ there must exist some $p\in \Bbb R[x]/(x^2 - 1)  $ such that $f(p)=q$. It seems clear to me that this is the case, because two elements of $\Bbb R$ are being mapped to two elements of $\Bbb R$, but for one thing I don't know if this is always the case, and more importantly that's not even a remotely rigorous explanation. 
Assuming I could show that $f$ is a ring homomorphism, how exactly could I show surjectivity? And is there some direction to take for showing this in general or is the route to showing surjectivity dependent on the problem? For instance, I find showing injectivity very straightforward because it only requires one to consider what would cause $a=b$ in the case that $f(a) = f(b)$, which is usually even more straightforward if $f$ is a homomorphism. 
 A: This is the task:
Let $(x,y)\in \mathbb R^2$. Suppose that $(a+b,a-b)=(x,y)$ and try to solve for $a,b$.  
If you succeed, you will have your hands on something that maps to $(x,y)$
A: First, your statement of what you want to prove is wrong.  You want to prove that for each $q\in\mathbb{R}\times\mathbb{R}$, there exists $p\in \Bbb R[x]/(x^2 - 1)$ such that $f(p)=q$.  The order of the "for each..." and "there exists..." matters, since the way you wrote it, you pick $p$ before $q$, so you would have the same $p$ for every $q$ which is obviously impossible.
Now, as for how to prove it, just do it!  Take an element $q\in\mathbb{R}\times\mathbb{R}$; that is, an ordered pair $q=(c,d)$ where $c,d\in\mathbb{R}$.  Now your goal is to find some $p\in \Bbb R[x]/(x^2 - 1)$ such that $f(p)=q$.  Looking at the definition of $f$ as $$f(a + bx + (x^2 - 1)) = (a + b, a - b),$$ can you see how you could find $p$ that works?
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 You just need to solve for $(a,b)$ such that $(a+b,a-b)=(c,d)$, since then you can define $p=(a+bx+(x^2-1))$ and $f(p)$ will be $q$.  This is just a matter of simple algebra: you want to solve the system of equations $$a+b=c$$ and $$a-b=d$$ for $a$ and $b$.

In general, this is how you prove a map $f:X\to Y$ between two sets is surjective: you take an arbitrary $y\in Y$, and explain how to find an element $x\in X$ such that $f(x)=y$.  Really, much more generally, this is how you prove any statement of the form "for all... there exists...": you take an arbitrary case of the "for all...", and explain how to find the thing that needs to exist.
Sometimes, knowing that a map is a homomorphism does make it easier to prove it is surjective.  For instance, if $X$ and $Y$ are rings and $f:X\to Y$ is a ring-homomorphism, then the image of $f$ is always a subring of $Y$, and so to show $f$ is surjective, it suffices to show that the image of $f$ contains a set which generates $Y$ as a ring.  In other words, you only need to solve $f(x)=y$ for $x$ for a few special values of $y$ which you know are enough to generate all of $Y$.  For this particular problem, though, that doesn't really save any substantial amount of work.
