Proof of surjectivity 
Let $f_1:\mathbf{R}^2 \rightarrow\mathbf{R}$ and $f_2:\mathbf{R}^2 \rightarrow \mathbf{R}$ be functions and define $f:\mathbf{R}^2 \rightarrow \mathbf{R}^2$ by $f(x,y)=(f_1(x,y),f_2(x,y))$.
If $f$ is surjective, prove that $f_1$ and $f_2$ are also surjective.

My proof goes as follows.
Let $(a,b)\in\mathbf{R}^2$. Since $f$ is surjective, for every $(a,b)\in\mathbf{R}^2$ there exists $(x,y)\in\mathbf{R}^2$ such that $f(x,y)=(a,b)$.
Therefore $f_1(x,y)=a\in\mathbf{R}$ en $f_2(x,y)=b\in\mathbf{R}$. But $a$ and $b$ are random so $f_1$ and $f_2$ are surjective.
But I believe there is something missing.. How can I justify the last sentence better?
 A: One way to rewrite what you said more "formally" is to directly show that both $f_1$ and $f_2$ are surjective:

So what is for a function to be surjective?
Given $X, Y$ sets and a function $f:X \to Y$, we say that $f$ is surjective if for every $y \in Y$, there exist $x\in X$ such that $f(x)=y$

The problem in this case is to show that $f_1$ and $f_2$ are surjective. So we go back to our definition and try to see how $f_1$ and $f_2$ fit:
In both cases, the set $X = \Bbb{R}^2$ and $Y= \Bbb{R}$ and we have $f_i : \Bbb{R}^2 \to \Bbb{R}$
So take $x \in \Bbb{R}$. To see that $f_1$ is surjective, we want to see that there exist $(a,b) \in \Bbb{R}^2$ such that $f_1(a,b)=x$ (this is exactly the definition of surjectivity).
Now, since $f$ is surjective, given $(x,0) \in \Bbb{R}^2$ there exist $(a',b') \in \Bbb{R}^2$ such that $f(a',b')=(f_1(a',b'),f_2(a',b'))=(x,0)$.
Which means that $f_1(a',b')=x$ and $f_2(a',b')=0$. So we found our element $(a,b)\in \Bbb{R}^2$ such that $f_1(a,b)=x$. This element is precisely $(a',b')$. Thus, $f_1$ is surjective.
A: You can justify it better by instead choosing the $a$ and $b$ you want before-hand. 
"To prove $f_1$ is surjective, we claim that for all real $a$, there exist $x,y$ such that $f_1(x,y)=a$."
Then proceed. You can even combine the proofs of surjectivity for each if you wish.
