Proving Bijections in $\mathbb{R}^n$ I have a question that may seem trivial or silly, however I'll try to make my point clear. I'm a student of Mathematical Physics and unfortunately my college doesn't offer a set theory course, so that we are not taught to prove bijections between arbitrary sets. The single variable calculus course we have teaches us to prove bijections by looking for one inverse function in a very clear way: we write $y = f(x)$, try to solve for $x$ and then create the function desired (it's of course more formal than that, first we prove injectivity showing that $f(x)=f(y)$ implies $x=y$ and then we look for the inverse to prove surjectivity).
The multivariable calculus course on the other hand doesn't talk about proving bijections from subsets of $\mathbb{R}^n$ to subsets of $\mathbb{R}^m$. On the other hand we have the analysis course, the topology course and the differential geometry course that all presumes we know how to prove those bijections and so on.
Now, I'm confused, because none of our courses teaches to prove that kind of thing. If we have some function $f: \mathbb{R}^n \to \mathbb{R}^m$, proving injectivity is the easiest part, but for surjectivity, seeking for some inverse seems kind of painful. By now I'm studying differential geometry of curves and surfaces (it uses Do Carmo's book), and the only thing I'm needing is this basic notion on how to prove bijections in these cases (for instance, when we come to show that something is a chart, proving continuity is rather easy, we have been taught on how to do it, but proving the bijection is complicated).
Is there any general guidelines as there is in the case of functions from $\mathbb{R}$ to $\mathbb{R}$? Is there any reference that can be given that treats this in a good way? Again, I know this is basic and that we all should know before coming to things like differential geometry, but since the course doesn't teach this, I came to ask here to fill this gap.
Thanks very much in advance, and sorry if this question is silly.
 A: For example, $\tan$ gives a (smooth) bijection $(-\frac\pi2,\,\frac\pi2)\to \Bbb R$.  By linear maps and shifting we can get bijections between any two open intervals.
Similarly, there is a bijection from the open $n$ dimensional (unit) ball to $\Bbb R^n$, using the previous bijection on each ray. These bijections (and their inverses) are also continuous, so they are homeomorphims.

An $n$ dimensional manifold is a topological space that has an $n$ dimensional ball around each of its point. 

To formulate it, we need the bijections to the balls (or either to $\Bbb R^n$).
A: Let $X$ and $Y$ be sets, and let $f: X \to Y$ be a function.   We say that $f$ is injective if for all $x, y \in X$,  $x \ne y \implies f(x) \ne f(y)$.   
We say that $f$ is surjective if for every $y \in Y$, there exists $x \in X$ such that $f(x) = y$.  A function is bijective if it is both injective and surjective.  
One does not need to find an inverse to prove that a function is surjective, and in general, there will be no inverse unless the function is bijective.  
Does that clear things up?
