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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.

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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
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
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
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 $\partial[0,1]\ne\phi$.

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@Jellyfish: Thanks :-) – Host-website-on-iPage Jul 14 '13 at 6:13
Thank you very much Aneesh!! – 1LiterTears Jul 14 '13 at 7:25

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