Can any function be parametrised? I'm going over surface integrals for my Calc 2 exam in May and the questions rely heavily on parametrization. Can it be proven that any function $\mathbb{R^n}\rightarrow \mathbb{R^m}$ (more specifically focusing on the case $m=1$) can be written as a parametrization and changed into a function $\mathbb{R}\rightarrow\mathbb{R}$? 
Also any tips or hints on trying to find the parametrization would really help, on the more difficult questions I sometimes struggle to find a suitable one.
Thanks 
 A: In the simplest case, we have curves with implicit equations like $y=f(x)$. Obviously these can be parameterised -- you just take $x=t$ and $y=f(t)$. Not very interesting.
Moving to two dimensions, we have curves given by implicit equations like $f(x,y)=0$. In other words we have sets of the form $S = \{(x,y) \in \mathbb{R}^2 : f(x,y)=0\}$, and it is reasonable to ask whether such a set can be parameterised. In other words, can we find an interval $I$ and a function $g: I \to \mathbb{R}^2$ such that $g(I) = S$.
If $g$ is continuous, then $g(I)$ will be a connected set. So, if the original curve has multiple disjoint pieces, then obviously you're going to have trouble. For example, the "curve" with equation $x^2 = 1$ is a pair of straight lines, which can not be parameterised using a single parametric equation. The hyperbola $xy=1$ is a more interesting example.
In these sorts of situations, you may have to settle for parameterising just one connected piece of the curve. I expect that this will be possible in calculus problems, but I can't prove that its always possible.
Another point is that you can often find several simple parameterizations, and choosing a good one can make your work (like computing derivatives) significantly easier. For example, if you want to parameterize the unit semi-circle where $y \ge 0$, you have your choice of
$$
x = t \quad ; \quad y = \sqrt{1-t^2} \quad (-1 \le t \le 1)
$$
$$
x = \cos(t) \quad ; \quad y = \sin(t) \quad (0 \le t \le \pi)
$$
$$
x = \frac{2t}{1+t^2} \quad ; \quad y = \frac{1-t^2}{1+t^2} \quad (-1 \le t \le 1)
$$
Any of these might be the most suitable, depending on your purpose. A slightly more interesting example is the hyperbola $x^2 - y^2 = 1$. Some possible parameterizations of the right-hand half are:
$$
x = \sec(\theta) \quad ; \quad y = \tan(\theta) \quad (-\pi/2 < \theta < \pi/2)
$$
$$
x = \cosh(\phi) \quad ; \quad y = \sinh(\phi) \quad (-\infty < \phi < \infty)
$$
$$
x = t + \frac1t \quad ; \quad y = t - \frac1t \quad (-\infty < \theta < \infty)
$$
It's hard to offer any universal advice for finding parameterizations, but a technique that often works is "moving lines". You imagine a straight line that is either moving or rotating, so that its position or angle is controlled by a parameter $t$. Suppose you can find an intersection point $P(t)$ of this line with your given curve. Then the mapping $t \mapsto P(t)$ gives you a parameterization (of some portion of the curve, anyway).
The situation with surfaces in three dimensions is largely analogous. The "moving lines" idea uses a two-parameter family of lines. One obvious family is formed by taking the vertical line through the point $(x,y,0)$, which gives you surfaces of the form $z = f(x,y)$. Another choice is lines parameterized by polar angles $(\theta,\phi)$. I generally don't venture into higher dimensions.
A: If your question is whether any subset of $\mathbb R^n$ can be described as the graph of a function, then the answer is no .The graph of , e.g., $S^n; n\geq 1$ cannot be expressed as a continuous map $f: \mathbb R \rightarrow \mathbb R $, since this would be a continuous bijection ( when restricted to a map from the unit interval $I$ into the image $S^1$),  between a compact space and a Hausdorff space, and so a homeomorphism/embedding, and this is not possible by, e.g, Borsuk-Ulam's theorem. I think for more general results you could look up both the inverse function theorem and the implicit function theorem.
