Logic Flawed? Is Rn Homotopic to Rm

[dumbQuestion]

Rn homotopic to Rm?!?

I am so confused about something simple

Ok, I know that Rn is contractible, just by a straightline homotopy sending all points to the origin. So this means Rn has the homotopy type of a point. So Rm, for a different integer m, has the homotopy type of a point. Since homotopy equivalence is an equivalence relation, this means that Rm is homotopic to Rn? But this is not possible right?

Can someone tell me where my logic is flawed?

 

[dumbQuestion]

(another example being, (0,1) homeomorphic to R, and we know (0,1) is homotopic to S1, and also we know S1 is homotopic to R2, so that would mean R is homotopic to R2)

 

[dumbQuestion]

well... now that I think about it... it makes sense. If I have say R2, I could homotope it to R by just a straight line homotopy (every point (x1,y1) traveling straight along the line x=x1 until it reaches (x1,0)). Nevermind. I answered my own question.

 

[Fedecart]

There is a HUGE difference in the concept of homotopy and the concept of homeomorphism.
You are right: ℝ^n has indeed the same homotopy type of ℝ^m, still they are not homeomorphic.

Maybe this is the reason why you were confused at firsts.

Furthermore, you are wrong when you say that (0,1) is homotopic to S^1. This is false. It is also false that S^1 is homopotic to R^2.
Where did you read that? However, it is true that ℝ is homotopic to ℝ^2.

PS The fact that ℝ^n is not homeomorphic to ℝ^m, despite being quite intuitive, is hard to prove and relies on this theorem http://en.wikipedia.org/wiki/Invariance_of_domain

Hope I clarified a little...

 

[dumbQuestion]

You're right, I am not sure why I claimed that (0,1) is homotopic to S1. I put an extra step in there and simply meant to say:

"(another example being, (0,1) homeomorphic to R, and we know (0,1) is homotopic to R2, so that would mean R is homotopic to R2)"

 

[micromass]

What do you mean with "homotopic" in the first place? Do you mean to say that the spaces are homotopy equivalent?

 

[dumbQuestion]

Yes, I'm sorry, my prof tends to throw around the term "homotopic" broadly for both maps as well as spaces and I think that's a bad habit. (Not criticizing him because he's a great prof, this material is so obvious to him he probably doesn't even think about it because he knows the material so well) But as someone new to this material it's caused me a lot of grief because there is a difference between these two concepts. I mean to say they are homotopy equivalent yes.

 

[Zelyucha]

Yes, I'm sorry, my prof tends to throw around the term "homotopic" broadly for both maps as well as spaces and I think that's a bad habit. (Not criticizing him because he's a great prof, this material is so obvious to him he probably doesn't even think about it because he knows the material so well) But as someone new to this material it's caused me a lot of grief because there is a difference between these two concepts. I mean to say they are homotopy equivalent yes.

The answer to your original question is Yes. The reason for this is that the topological space $$\mathbb{R}^n$$ is simply connected for every positive integer n and thus it is contractible. And so for any 2 positive integers m≠n $$\mathbb{R}^m \sim \mathbb{R}^n$$ (and by the '~' I mean they are homotopy equivalent even though as Fedecart pointed out they are not homeomophic).

 

[micromass]

TThe reason for this is that the topological space $$\mathbb{R}^n$$ is simply connected for every positive integer n and thus it is contractible.

I want to warn the OP that not every simply connected space is contractible. So saying "it is simply connected and thus contractible" is not a valid implication. The valid implication would be "it is contractible and thus simply connected".

 

[Bacle2]

You may also want to be careful saying _where_ it is that your space is contractible,
i.e., a loop may be deformed to a point within, say R^2 , but it is not contractible as
a stand-alone space, i.e., π₁(S¹) ≠ {0} (Say your loop does not
intersect itself, etc. so that it is an S¹).

And, as a followup to micromass, take any Sⁿ with n>1 as an example of simply-connected but not contractible (as a stand-alone).

And, BT
W, compactness is not a homotopy invariant, since, e.g., the intervals [a,b] and [a,b) on the real line are homotopy-invariant

EDIT In above line, homotopy-invariant should be homotopy-equivalent.