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Ok, this thread is for any questions people might have on graphs in 3-dimensional space. I know I have many questions to ask.

Do you have any versing in the so-called "set-theoretic" function notations, like f:A->B? It's really not all that complicated, and in this case very illuminative.

So, in 2 dimensions, we have a function y=f(x). We can write this as f:A->B, where A is the domain of the function and B is the codomain; for these purposes, let both sets just be the real numbers. Then you get a graph like you normally would.

In trying to find the inverse of this function, you replace x and y, like you said. But this only gives a true "inverse" if the new function you get is actually a function. Consider a parabola, f(x)=y=x^2. If you just replace y and x, you get x=y^2, which is NOT a function of x. In this case, you can only get the inverse g(x)=Sqrt(x) if you restrict the domain and/or codomain in some way.

This generalizes to any dimension, and isn't too hard to understand for smooth functions. A function f:A->B (where A and B are, for simplicity, nice subsets of R^n and R^m) has an inverse if it is bijective. All this means is that the codomain B is the image of f - f covers all of B - and that each member of A has a *unique* image in B. If this is the case, then you get an inverse function which is also bijective; g:B->A. Really, this is a heuristic argument for the lemma "A function is invertible iff it is bijective."

So now you can easily go to a "3-dimensional graph", which is a function from R^2 (x and y) to R^1 (z). First, notice that if this function is going to be injective (each (x,y) having a unique image), the domain must be 1-dimensional. So in this case, "inverse" can only apply to lines in space, not to surfaces. Of course there are other restrictions, but this is the clearest one.

Now, you can define the inverse image of a set (of points) to be...the set of points that map to that set of points xp . This gives us another way to see if a function is injective: if every point in the image of f has a preimage consisting of *only one* point. But in general, the preimage defined this way isn't going to give you an "inverse function".

Note: Anyone else who reads this, feel free to check for accuracy...this was kind of a rush job sweatdrop