Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I thought it was closed, under the usual topology $\mathbb{R}$, since its compliment $(-\infty, 0) \cup (1,\infty)$ is open.

However, then then intersection number would not agree mod 2, since it can arbitrarily intersect a compact manifold even or odd times.

P.S. The corollary.

$X$ and $Z$ are closed submanifolds inside $Y$ with complementary dimension, and at least one of them is compact. If $g_0, g_1: X \to Y$ are arbitrary homotopic maps, then we have $I_2(g_0, Z) = I_2(g_1, Z).$

The contradiction (my question):

Let [0,1] be the closed manifold $Z$, and then it can intersect an arbitrary compact manifold any times, contradicting with the corollary.


Aneesh Karthik C's comment answered my question, so just to clarify:

I was thinking $g_0$ is one wiggle of [0,1] such that it intersects a compact manifold once, and $g_1$ is some other sort that [0,1] intersect twice. Then it contradicts with the corollary. But apparently it doesn't, because [0,1] does not satisfy the corollary as a closed manifold. By definition, a closed manifold is a type of topological space, namely a compact manifold without boundary.

Since [0,1] is not a closed manifold, it can intersect a compact manifold as much as it want, without contradicting with the theorem.

I didn't realize that [0,1] is not a closed manifold. So I thought it contradicts and that's why I ask the question.

share|improve this question
7  
The first line is absolutely correct. I can't make any sense of the second. –  Zach L. Jul 13 '13 at 6:01
    
One of the assertions you're making is wrong, and it's not the one about $[0,1]$ being closed... –  Potato Jul 13 '13 at 6:12
2  
A closed manifold is a compact boundaryless manifold. So the last line "Let $[0,1]$ be the closed manifold $Z$" is wrong –  Host-website-on-iPage Jul 13 '13 at 6:19
    
OH, that's exactly what confuses me, thanks so much @AneeshKarthikC! So [0,1] is not a closed manifold because it has boundary points, 0 and 1, right? Thanks again~ –  1LiterTears Jul 13 '13 at 6:21
3  
Yes. Quite right. In fact some of us call manifolds that are not closed, as open manifolds. Basically when you talk of closed manifolds you should not make reference to a mother space, for that is when confusion sets in! –  Host-website-on-iPage Jul 13 '13 at 6:25
show 5 more comments

1 Answer 1

up vote 2 down vote accepted

A closed manifold is a compact boundaryless manifold. So the last line "Let [0,1] be the closed manifold Z" is wrong, for $\partial[0,1]\ne\phi$.

share|improve this answer
    
@Jellyfish: Thanks :-) –  Host-website-on-iPage Jul 14 '13 at 6:13
    
Thank you very much Aneesh!! –  1LiterTears Jul 14 '13 at 7:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.