Purplemath and Paul's Web Notes are pretty good in my experience.
Hrm.
(a - b)^c
= a^c * (1 - b)^cA couple problems to work here (notes):
You can get it from a successive squaring algorithm but I'm not sure of other shortcuts via significances of the exponents; 2^6 = 64, and 2^7 = 128.
Are these the same as cosecant/secant/cotangent?
Ah then I've been working with arcsin and stuff already without knowing it
I think removing a point from the 2-sphere should just give you a disk with trivial fundamental group, no? Then removing k points is the same as having a disk with k-1 points removed, so I think you are just one off in your proposed solution.
You are correct that they are both trivial. There is no problem there; it just means that if you want to distinguish a 2-sphere from a punctured 2-sphere you must look to some other topological invariant instead. Every simply-connected space will have trivial fundamental group, and in fact you can look at the fundamental group as a measure of just how far a space is from being simply-connected.
You should be able to do that one constructively. From what you've said before, you already know that the free group on n generators corresponds to a wedge of n circles. So given a presentation of a group as n generators with some additional relations, begin with that wedge and add the relations to it. Remember that a relation a=b just means that you want the circles representing a and b to be homotopic, so you must add a surface across which they can be deformed. So just glue one disk to your bouquet for each of the relations. You will want to do this without consideration of ambient space, because for arbitrary groups you are bound to get many unwanted intersections between these surfaces if you work in, say, R^3. But this construction should work for any finitely generated group.
