Fundamental group of a sphere with 6 points removed

[PsychonautQQ]

This space is homotopy equivalent to the complement of the three coordinate axes in $R^3$.

This is in the chapter about the Seifert-Van Kampen Theorem, so I'm expecting to invoke that theorem.

The thing is, how should we choose our open sets such that the intersection is path connected and that we can compute the fundamental groups of each open set?

Anyone want to help me get started on this one?

 

[WWGD]

I don't know if you're considering other approaches, but, since the sphere ( I assume you're talking 'bout the 2-sphere ) is the plane with 5 points removed. If you deformation -retract you get 5 loops attached pairwise at a point. I think that gives you a free product. EDIT: You basically start "peeling off" about a removed point until you cannot go further in a continuous way once you hit the part peeled about another removed point. Think n=1, then you don't have any obstacles and you can remove "everything" , leaving a circle. With two points removed, you hit a barrier, ending up with two circles wedged at one point, etc. This gives you a homotopy equivalence with a wedge of circles.

 

[PsychonautQQ]

I don't know if you're considering other approaches, but, since the sphere ( I assume you're talking 'bout the 2-sphere ) is the plane with 5 points removed. If you deformation -retract you get 5 loops attached pairwise at a point. I think that gives you a free product. EDIT: You basically start "peeling off" about a removed point until you cannot go further in a continuous way once you hit the part peeled about another removed point. Think n=1, then you don't have any obstacles and you can remove "everything" , leaving a circle. With two points removed, you hit a barrier, ending up with two circles wedged at one point, etc. This gives you a homotopy equivalence with a wedge of circles.

Yeah, I see. I don't think this is a free product on 6 generators though, because that would be if all the circles share a common point. In this case, they will be connected pairwise. I don't know what this group would look like.

 

[WWGD]

Yeah, I see. I don't think this is a free product on 6 generators though, because that would be if all the circles share a common point. In this case, they will be connected pairwise. I don't know what this group would look like.

I think this is your bouquet of circles. Each loop in one of the circle can be combined with a loop in one of the others as $a^nb^mc^pd^q e^r$ , a free group on 5 ( n, in general) elements $\{ a,b,c,d,e \}$. But let @lavinia verify and give you a more rigorous account on this.

 

[lavinia]

Think of the sphere as the surface of a cube. Its edges are a strong deformation retract of the sphere minus the six centers of the cube's faces.

 

[WWGD]

Think of the sphere as the surface of a cube. Its edges are a strong deformation retract of the sphere minus the six centers of the cube's faces.

The two, cube , sphere are actually even homeomorphic, not just homotopic. Though, of course, not diffeomorphic.

 

[lavinia]

The two, cube , sphere are actually even homeomorphic, not just homotopic. Though, of course, not diffeomorphic.

Right. Here we are thinking of the sphere as a topological manifold.

@PsychonautQQ Each face minus a point retracts continuously onto its edge square. If one takes away a point from each of the six faces then the whole thing retracts onto the framework of edge squares,the 1 skeleton of the cube. This is a homotopy equivalence.

BTW: A quick homology calculation on the 1 skeleton shows that the fundamental group cannot be the free group on six generators - as you argued above. The same calculation works for the fundamental group but it is a little messier.

- Some other numbers of removed points can be handled in the same way. For instance, the sphere minus four points retracts onto the 1 skeleton of a tetrahedron.

 

[WWGD]

Right. Here we are thinking of the sphere as a topological manifold.BTW: A quick homology calculation on the 1 skeleton shows that the fundamental group cannot be the free group on six generators - as you argued above. The same calculation works for the fundamental group but it is a little messier.

Notice I mentioned the free group on 5 generators, not six.

 

[lavinia]

Notice I mentioned the free group on 5 generators, not six.

Yes i know.

 

[WWGD]

Yes i know.

But you claimed that my statement that the fundamental group on 6 generators is incorrect

BTW: A quick homology calculation on the 1 skeleton shows that the fundamental group cannot be the free group on six generators - as you argued above. The same calculation works for the fundamental group but it is a little messier.

Is it also incorrect that the fundamental group is the free group on 5 generators?

 

[lavinia]

But you claimed that my statement that the fundamental group on 6 generators is incorrect Is it also incorrect that the fundamental group is the free group on 5 generators?

Not completely sure.

The 1 skeleton of the cube has six squares each of which is a cycle. It is easy to show that the difference of any two of them that are on opposite sides of the cube are homologous to the sum of other four. So the first homology has at most five generators. If the 1 skeleton were a bouquet of six loops then the first homology would be the free abelian group on six generators. But I haven't worked out the whole thing - leaving something for the OP. But It seems that you are right.

 

[disregardthat]

The space is, like WWGD explained, homotopy equivalent with a bouquet of 5 circles, and the fundamental group is the free product of 5 copies of $\mathbb{Z}$. Regarding the first homology group, this is the abelianization of $\pi_1$, hence the free abelian group $\mathbb{Z}^5$ (which can independently be verified using Alexander duality).

 

[lavinia]

The space is, like WWGD explained, homotopy equivalent with a bouquet of 5 circles, and the fundamental group is the free product of 5 copies of $\mathbb{Z}$. Regarding the first homology group, this is the abelianization of $\pi_1$, hence the free abelian group $\mathbb{Z}^5$ (which can independently be verified using Alexander duality).

Proof?

 

[disregardthat]

Proof?

Of the homotopy equivalence? WWGD explained it perfectly in post #2.

 

[lavinia]

Of the homotopy equivalence? WWGD explained it perfectly in post #2.

Right. I got that.

But I don't see the retraction off the top of my head. If you draw the five loops around the removed points the interiors retract onto the loops. What about the exterior? Maybe I am stupid but I didn't see how to do this.

 

[disregardthat]

Right. I got that.

But I don't see the retraction off the top of my head. If you draw the five loops around the removed points the interiors retract onto the loops. What about the exterior? Maybe I am stupid but I didn't see how to do this.

One can consider a disk containing all 5 points on the interior, and deformation retract down to its boundary. Then draw 5 disks containing each point but none of the others, and deformation retract to their boundaries. Now you're left with a big closed disk with 5 holes in it (it is a closed subset of $\mathbb{R}^2$). Now it should be easy to visualize how this can be deformed into a wedge of 5 circles. For example, arrange the smaller disks homeomorphically so that their centers form small neighborhoods of the vertices of a regular pentagon. Let P be the center of the pentagon, and deform the boundary of the smaller disks so that they intersect at P. The boundaries of the smaller (deformed) disks now form a wedge of 5 circles, so next its simply a matter of deformation retracting the large disk onto it.

EDIT: More easily perhaps, and in a way that makes the wedge of circles a deformation retract of the initial space, is to start out with $\mathbb{R}^2$ with 5 points removed, and homeomorphically align the points to the vertices of a regular pentagon, and then draw a 5 loops around each point intersecting only at the center P. Then one deformation retracts down to this.