Jordan-Brouwer Separation Theorem and Invariance of Domain

1. Sep 2, 2011

Anonymous217

So I'm having trouble understanding how these two are related, i.e., how one proves the other.

I understand the ideas behind both of them: For J-B, you're basically taking R^n and throwing in a sphere, so the inside of the sphere is bounded and everything outside the sphere is unbounded. For Invariance of Domain, it's pretty obvious just by the definition (the image of an open subset of R^n is open).
However, I don't really see a relationship between the two. Can anyone give some insight?

Also, I was curious why we only need injectivity for both of them, where surjectivity is basically unnecessary in any possible proof. What makes 1-1 important?

Last edited: Sep 2, 2011
2. Sep 2, 2011

quasar987

I was unaware that these theorems were equivalent. Where did you hear such a claim?

The only way I know them to be related is that they are both obvious-looking statement that no one has been able to prove for a long time. That is, until homology came along.

3. Sep 2, 2011

mathwonk

(zeroth) homology measures components of a space hence can be used to prove the separation thm. components of an open set in Euclidean space are open, so knowing about components can be used to prove inv of domain.

If S is an n-1 sphere and B an n ball, and f an injective continuous map, separation implies the complement of the image of f(S) in R^n has 2 components and the complement of f(B) has only one.

since f(B-S) is connected, it is thus a component of the open complement of f(S), hence f(B-S) is open.

Since every open set is a union of such balls B, this shows that Jordan separation implies invariance of domain.

4. Sep 2, 2011

mathwonk

as a weak version of the other direction, suppose f is an injective continuous map from B to R^n, and we assume f(B) has connected complement. Then if we knew that f(B-S) were open, it would follow, since f(B-S) is also connected, that the complement of f(S) has 2 components. I.e. invariance of domain does imply that the complement of f(S) has one more component than the complement of f(B).

5. Sep 2, 2011

Anonymous217

^^ This is the type of answer I was looking for. It makes complete sense now; thanks.

6. Sep 2, 2011

mathwonk

notice the weak direction is weak because it is not clear, indeed probably not true, that an injection from S^n-1 to R^n extends to an injection on B^n.

7. Sep 2, 2011

mathwonk

you are welcome.

8. Sep 2, 2011

Bacle

Mathwonk:

I think we may be able to use Tietze's theorem that a continuous, real-valued map from a closed subspace of a normal space (I think the ball B^n, as a metric space, is normal, and its boundary S^(n-1) is closed in B^n--the interior seems to be open) X, extends into the whole space, tho I am not sure if there is a version for maps into R^n; maybe we can argue component-wise to get a continuous map.

Last edited: Sep 3, 2011
9. Sep 3, 2011

mathwonk

injectivity?

10. Sep 3, 2011

Bacle

I am not sure we can extend an injection into an injection; let me see.

11. Sep 3, 2011

mathwonk

i believe the alexander horned sphere is a famous counterexample.