Jordan-Brouwer separation theorem

[quasar987]

For those who have the book, in corollary 19.6 page 234 of Bredon, the Jordan-Brouwer separation theorem has just been proven and so we know that if S is homeomorphic to S^(n-1), then S^n\S has two connected components, which are open in S^n and their boundary is S.

So we let V be one of the components of S^n\S and let x be in V. Then Bredon writes that excision implies the isomorphism

H_{i+1}(V,V\backslash \{x\})\cong H_{i+1}(\mathbb{D}^n,\mathbb{D}^n\backslash\{0\})

(where D^n is the closed n-disk).

It's like he said "V u S is homeomorphic to D^n (with x being sent to 0), so excising the boundary \partial\mathbb{D}^n\cong S gives the above isomorphism".

But for n>2, it is not true in general that V u S is homeomorphic to D^n as the (counter-) example of the Alexander horned disk shows...

So what is it that he's doing in that step?P.S. Don't bother with Google Book, page 234 is missing from the preview. :(

 

[quasar987]

Ah. V is an open of S^n and thus a manifold. It suffices to embed D^n in V around x and excise V\D^n.

 

[wofsy]

For those who have the book, in corollary 19.6 page 234 of Bredon, the Jordan-Brouwer separation theorem has just been proven and so we know that if S is homeomorphic to S^(n-1), then S^n\S has two connected components, which are open in S^n and their boundary is S.

So we let V be one of the components of S^n\S and let x be in V. Then Bredon writes that excision implies the isomorphism

H_{i+1}(V,V\backslash \{x\})\cong H_{i+1}(\mathbb{D}^n,\mathbb{D}^n\backslash\{0\})

(where D^n is the closed n-disk).

It's like he said "V u S is homeomorphic to D^n (with x being sent to 0), so excising the boundary \partial\mathbb{D}^n\cong S gives the above isomorphism".

But for n>2, it is not true in general that V u S is homeomorphic to D^n as the (counter-) example of the Alexander horned disk shows...

So what is it that he's doing in that step?


P.S. Don't bother with Google Book, page 234 is missing from the preview. :(

Could you explain the Alexander horned sphere counter example? I'd like to understand it.

This excision seems straight forward without any use of this supposed homeomorphism. The complement of any open ball contained in V that contains the point x is closed in V-{x} and is contained in the interior of V-{x}. I think this works.

 

[quasar987]

Could you explain the Alexander horned sphere counter example? I'd like to understand it.

The way I understand it is this. The Alexander horned sphere is a subspace X of R³ homeomorphic to S². By the Jordan-Brouwer separation theorem (corollary 19.6 in Bredon) then, the Alexander horned sphere separates R³ into two connected open sets having X as their boundary, one bounded, one unbounded. Now, it can be shown that the unbounded component has nontrivial fundamental group (see Hatcher p.170-172: http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf). So, passing to S³ through stereographic projection \sigma: \mathbb{R}^3\rightarrow \mathbb{S}^3 \backslash \{N\}, we get that the component of C \mathbb{S}^3\backslash \sigma(X) corresponding to the unbounded component of R³\X is not homeomorphic to \mathbb{B}^3 (the 3-ball) since it has nontrivial fundamental group, while the 3-ball does. And so the closure of C could not be homeomorphic to D³ because otherwise C would be homeomorphic to \mathbb{B}^3.

Does that sound right to you?

This excision seems straight forward without any use of this supposed homeomorphism. The complement of any open ball contained in V that contains the point x is closed in V-{x} and is contained in the interior of V-{x}. I think this works.

Yup, this is essentially the same thing I said in post #2.

 

[wofsy]

The way I understand it is this. The Alexander horned sphere is a subspace X of R³ homeomorphic to S². By the Jordan-Brouwer separation theorem (corollary 19.6 in Bredon) then, the Alexander horned sphere separates R³ into two connected open sets having X as their boundary, one bounded, one unbounded. Now, it can be shown that the unbounded component has nontrivial fundamental group (see Hatcher p.170-172: http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf). So, passing to S³ through stereographic projection \sigma: \mathbb{R}^3\rightarrow \mathbb{S}^3 \backslash \{N\}, we get that the component of C \mathbb{S}^3\backslash \sigma(X) corresponding to the unbounded component of R³\X is not homeomorphic to \mathbb{B}^3 (the 3-ball) since it has nontrivial fundamental group, while the 3-ball does. And so the closure of C could not be homeomorphic to D³ because otherwise C would be homeomorphic to \mathbb{B}^3.

Does that sound right to you?


Yup, this is essentially the same thing I said in post #2.

I think that it is right that the one point compactification of the open domain is not a ball but I don't know how to prove it without thinking more about the construction of the Alexander horned sphere. But a priori just because a set is not simply connected does not mean that its 1 point compactification can not be, for instance a punctured disc

 

[quasar987]

I think that this is where the hypothesis n>2 comes in. Namely, \sigma(\mbox{unbounded connected component of }\mathbb{R}^3\backslash X) =C\backslash \{N\} has nontrivial fundamental group. Now consider the stereographic projection with respect to the opposite pole, \widetilde{\sigma}:\mathbb{S}^3\backslash \{S\}\rightarrow \mathbb{R}^3, so that N is now mapped to the origin in R³. Then \widetilde{\sigma}(C\backslash \{N\}) is an open set (with nontrivial fundamental group) in R³ around 0 but not containing 0 and adding the origin to that set does not change its fundamental group (proof?).

What do you think?

 

[wofsy]

I think that this is where the hypothesis n>2 comes in. Namely, \sigma(\mbox{unbounded connected component of }\mathbb{R}^3\backslash X) =C\backslash \{N\} has nontrivial fundamental group. Now consider the stereographic projection with respect to the opposite pole, \widetilde{\sigma}:\mathbb{S}^3\backslash \{S\}\rightarrow \mathbb{R}^3, so that N is now mapped to the origin in R^3. Then \sigma(C\backslash \{N\}) is an open set (with nontrivial fundamental group) in R^³ around 0 but not containing 0 and adding the origin to that set does not change its fundamental group (proof?).

What do you think?

I don't think you have a proof although I see your intuition about N > 2.
What about taking a disc tangent to the z-axis and with a diameter lying on the x-axis and spinning it around the z axis to get a torus like solid centered at the origin. Then remove the origin. This has fundamental group equal to Z but compactifies to a contractible set (without the origin it deforms onto its central circle)

 

[quasar987]

Yes but your counter-example has little to do with \widetilde{\sigma}(C\backslash \{N\}), which has the form of an open neighborhood of 0 of which 0 has been removed.

Or did I miss your point?

 

[wofsy]

Yes but your counter-example has little to do with \widetilde{\sigma}(C\backslash \{N\}), which has the form of an open neighborhood of 0 of which 0 has been removed.

Or did I miss your point?

All I was saying is that your argument doesn't seem right. You say that an open set in R^3 that has non-trivial fundamental group and which has the origin as a limit point must still have non-trivial fundamental group after the origin is included. I just gave a counter example to that claim.

I still think your intuition is correct because the construction of the horned sphere shows that outside of a sufficiently large sphere the unbounded component is equal to R^3 minus a ball and is a ball after you add in the point at infinity. Since this large bounding sphere is simply connected and the ball is contractible, adding the point at infinity leaves the fundamental group unchanged.

 

[quasar987]

All I was saying is that your argument doesn't seem right. You say that an open set in R^3 that has non-trivial fundamental group and which has the origin as a limit point must still have non-trivial fundamental group after the origin is included. I just gave a counter example to that claim.

But unless I misunderstood you, the set in your counter-example is not open. More to the point, it is not of the type "an open neighborhood of 0 of which 0 has been removed".

 

[wofsy]

But unless I misunderstood you, the set in your counter-example is not open. More to the point, it is not of the type "an open neighborhood of 0 of which 0 has been removed".

well you can make it open by leaving off the boundary and retaining the interior. It is not an open neighborhood of zero but I don't see why that matters since your argument did not directly use that. but maybe I didn't understand your argument completely.

 

[wofsy]

I think that this is where the hypothesis n>2 comes in. Namely, \sigma(\mbox{unbounded connected component of }\mathbb{R}^3\backslash X) =C\backslash \{N\} has nontrivial fundamental group. Now consider the stereographic projection with respect to the opposite pole, \widetilde{\sigma}:\mathbb{S}^3\backslash \{S\}\rightarrow \mathbb{R}^3, so that N is now mapped to the origin in R³. Then \widetilde{\sigma}(C\backslash \{N\}) is an open set (with nontrivial fundamental group) in R³ around 0 but not containing 0 and adding the origin to that set does not change its fundamental group (proof?).

What do you think?

I just reread this proof and I agree that the fundamental group is unchanged but just because you brought the point at infinity to the origin doesn't change anything. You still need to show that adding this point doesn't matter and this follows because you can deformation retract the part of the unbounded component that is outside of some large sphere onto that sphere. All of the non-trivial homotopy is isolated away from infinity. Then as you rightly pointed out, for N>2, that sphere has trivial fundamental group so adding the point at infinity leaves the fundamental group unchanged.

 

[quasar987]

well you can make it open by leaving off the boundary and retaining the interior.

But then 0 does not belong to that set.

It is not an open neighborhood of zero but I don't see why that matters since your argument did not directly use that.

On the contrary, it entirely depends on that:

Then \widetilde{\sigma}(C\backslash \{N\}) is an open set (with nontrivial fundamental group) in R³ around 0 but not containing 0 and adding the origin to that set does not change its fundamental group (proof?).

And by the way, by "(proof?)" at the end, I meant "This assertion needs a proof.", not "Do you think I have a proof?".I will look at your retraction argument tomorrow, but on first reading, it has the ring of truth!

 

[wofsy]

But then 0 does not belong to that set.



On the contrary, it entirely depends on that:


And by the way, by "(proof?)" at the end, I meant "This assertion needs a proof.", not "Do you think I have a proof?".


I will look at your retraction argument tomorrow, but on first reading, it has the ring of truth!

I think our intuition was right and leads to this retraction argument. I am interested in what you think. Let me know.

By the way, though this isn't important, if you add the origin to that open solid torus thing then it becomes contractible.

 

[quasar987]

I just reread this proof and I agree that the fundamental group is unchanged but just because you brought the point at infinity to the origin doesn't change anything. You still need to show that adding this point doesn't matter and this follows because you can deformation retract the part of the unbounded component that is outside of some large sphere onto that sphere. All of the non-trivial homotopy is isolated away from infinity. Then as you rightly pointed out, for N>2, that sphere has trivial fundamental group so adding the point at infinity leaves the fundamental group unchanged.

I do not see how the fact that S² has trivial fundamental group implies that adding the point at infinity leaves the fundamental group of the unbounded component of R³\X unchanged.

 

[wofsy]

I do not see how the fact that S² has trivial fundamental group implies that adding the point at infinity leaves the fundamental group of the unbounded component of R³\X unchanged.

Quasar maybe I'm wrong about this but I think it is Van Kampen's Theorem. Let's think this through together.

We have the unbounded component with the north pole attached. Take a small open ball around the north pole. We know that its intersection with the unbounded component deformation retracts onto a sphere . Since the sphere is simply connected so is the intersection. Further the ball around the north pole is contractible and so is also simply connected.

As I understand Van Kampen's theorem, the fundamental group of the union is the free product of the fundamental groups of the ball around the north pole and the unbounded component modulo a normal subgroup generated by identifying the homotopy classes in each of the two components of the closed paths in the intersection. (I didn't say that very well.) But the intersection is simply connected so this group is trivial. Since the ball has zero fundamental group, the fundamental group of the union is the free product of the fundamental group of the unbounded component with the trivial group.

I think this works. what do you say?

 

[wofsy]

I do not see how the fact that S² has trivial fundamental group implies that adding the point at infinity leaves the fundamental group of the unbounded component of R³\X unchanged.

i was thinking about this argument a little more. You can see in the case off two dimensions why the argument doesn't work - as your intuition pointed out in the first place.

Take the Mobius band and add a point at infinity. A small disc around the point at infinity intersects the Mobius band in a set that deformation reacts onto a circle, a circle that winds twice around the band's equatorial circle. This curve is has homotopy class in the Mobius band that is twice the generator of the fundamental group since it winds around the equatorial circle twice. It is not null homotopic. But in the small disc it is null homotopic because the disc is contractible. Van Kampen's theorem says then that the fundamental group of the union (which is the projective plane) is Z/2Z because this intersection curve is null homotopic in the disc.

 

[quasar987]

Quasar maybe I'm wrong about this but I think it is Van Kampen's Theorem. Let's think this through together.

We have the unbounded component with the north pole attached. Take a small open ball around the north pole. We know that its intersection with the unbounded component deformation retracts onto a sphere . Since the sphere is simply connected so is the intersection. Further the ball around the north pole is contractible and so is also simply connected.

As I understand Van Kampen's theorem, the fundamental group of the union is the free product of the fundamental groups of the ball around the north pole and the unbounded component modulo a normal subgroup generated by identifying the homotopy classes in each of the two components of the closed paths in the intersection. (I didn't say that very well.) But the intersection is simply connected so this group is trivial. Since the ball has zero fundamental group, the fundamental group of the union is the free product of the fundamental group of the unbounded component with the trivial group.

I think this works. what do you say?

I certainly think it works!

If we call A the unbounded component of R³\X and A' the compactified A, then Van Kampen gives as you say \pi_1(A')\cong\pi_1(A)*0\cong\pi_1(A).

Well, that was a pretty fun exercice!