- #1
Mikemaths
- 23
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X1 = {(x; y; z) ∈ R^3 | x > 0}
just need to check my thinking
is pi1(X1) = {1} i.e. trivial
just need to check my thinking
is pi1(X1) = {1} i.e. trivial
I have done some more research and now I understand that the fundamental group of X is the group consisting of the homotopic equivalence classes of loops of base x in X. But this must be expressed in algebraic terms, now I understand that {2} is not a group but I beleiev there are two different equivalence classes of loops in X2 as described above as the loops either side of the x axis cannot be equivalent?
Am i talking rubbish or is this valid
Well exactly! Any loop in R^3\{(x; y; z) | x = 0; y = 0; 0 <= z <= 1} can be homotoped to the constant loop by avoiding the unit line on the z axis just as any loop in R^3\{(x; y; z) | x = 0; y = 0; z = 0} can be homotoped to the constant loop by avoiding the origin. So by this argument, both spaces have trivial fundamental group. (And it makes sense to speak of the fundamental group of these spaces without reference to a particular base point because they are path connected spaces, so the fundamental group is independant of the base point (up to isomorphy of course).)I am not sure since a loop from x to x in
R^3\{(x; y; z) | x = 0; y = 0; 0 <= z <= 1}
is a similar situation to
R^3\{(x; y; z) | x = 0; y = 0; z = 0}
As they can always avoid the unit line on z axis that is missing as it were.
This space is not isomorphic (we say homeomorphic) to the torus, why would you think that?Also is R^3\{(x; y; z) | x = 0; 0 <= y <= 1}
isomorphic to the Torus and therefore fundamental group of this is Z + Z (disjoint)?