- #1

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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

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- Thread starter Mikemaths
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- #1

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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

- #2

quasar987

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yes

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- #3

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Also does this mean that if X2 = {(x; y; z) ∈ R^3 | x not equal to 0}

then pi1(X2) = {2} since it splits R^3 into two sections

- #4

quasar987

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I say it X1 has trivial fundamental group because it obviously does. If this is not obvious to you, then it means your thinking relative to fundamental groups is deficient somewhere.

Also, pi1(X2) = {2} does not make sense since {2} is not a group!

- #5

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- #6

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Am i talking rubbish or is this valid

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- #8

quasar987

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Am i talking rubbish or is this valid

Like you said, the fundamental group of a space X is the group consisting of the homotopic equivalence classes of loops

But in general, the fundamental group is something that is computed

In the case of your example X2, for points p in the section where x<0, then [itex]\pi_1(X,p)=\{1\}[/itex] because clearly every loop there is homotopic to the constant loop. Notice that a loop based at y cannot wander in the region x>0 because it would have to pass through a point where x=0 to get there but those points are not part of X2.

And similarly, every loop based at a point q in the region where x>0 is clearly homotopic to the constant loop, so [itex]\pi_1(X,q)=\{1\}[/itex].

- #9

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So for example R^3\(0,0,0)?

This I believe is path connected however all loops are not homotopic to the constant loop because ones that go round the origin cannot be contracted is that true?

- #10

quasar987

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But in R^2\{0,0} (path connected), a path that goes around the origin is not homotopic to one that does not go around the origin. For a given base point x in R^2\{0,0}, there are, so to speak, [itex]|\mathbb{Z}|[/itex] homotopy classes of loops based at x:

...

-2) there are those that go around the origin 2 times clockwise

-1) there are those that go around the origin 1 time clockwise

0) there are those that do not wind around the origin. (those are homotopic to the constant loop)

1) there are those that go around the origin 1 time counter clockwise

2) there are those that go around the origin 2 times counter clockwise

...

- #11

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What about R^3\{(x,y,z)|x=0,y=0,0<=z<=1}

as this is path connected

- #12

quasar987

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What do you think?

- #13

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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.

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)?

- #14

quasar987

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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)?

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