Homotopy Problem

1. Feb 11, 2007

Gagle The Terrible

I was asked to proove that R^m\R^k (m>k) has the same type of homotopy that S^(m-k-1) .
I know I can use two criterias : the definition and the rectract criterium. The latter is more appealing because of it's "simplicity" but the function

r:[0,1]*X --> X given by (1-t)x + tx/norm(x) yields no results and is in fact a pretty bad choice ...

I have tried in vain to use the definition but I can't find the two functions that will satisfy the conditions.

Gagle

2. Feb 11, 2007

Gagle The Terrible

If I use the definition (i.e. there existe f : R^m\R^k --> S^(m-k-1)
and g: S^(m-k-1) --> R^m\R^k
such as f o g is homotopic to the identity ( of S^(m-k-1) )
and g o f is homotopic to the identity (of R^m\R^k) .

To proove such homotopies , is the function h:[0,1]*X --> X given by (1-t)ID + tx/norm(x) sufficient to proove my point ?
here ID is the Identity. If I am not mistaking, it is the same for both spaces...

The reason I keep using such a function is that we are in R^n .
Is that a good idea ?

Thanks for any help

Gagle

Last edited: Feb 11, 2007
3. Feb 11, 2007

StatusX

That'll give you a retraction of R^n-R^k onto a subset of S^(n-1), namely S^(n-1)-S^(k-1). Can you show this is homotopy equivalent to S^(n-k-1)?

4. Feb 11, 2007

quasar987

Hi Gagle,

We're probably in the same class. You're the guy with the ponytail?

We've seen in class, using the "retract lemma" that $\mathbb{R}^{m-k}\setminus \{0\}\simeq \mathbb{S}^{m-k-1}$, right?

And we also know that $\simeq$ is an equivalence relation on the set of all topological spaces. This means that $\simeq$ is transitive, so if we can show that $\mathbb{R}^m\setminus \mathbb{R}^k \simeq \mathbb{R}^{m-k}\setminus \{0\}$, then it is also true that $\mathbb{R}^m\setminus \mathbb{R}^k \simeq \mathbb{S}^{m-k-1}$ and we will have won.

Now, obviously, we cannot use the retract lemma to prove this because neither $\mathbb{R}^{m-k}\setminus \{0\}$ or $\mathbb{R}^m\setminus \mathbb{R}^k$ is a subset of the other.

So we must fall back on the original definition of "homotopy type". That is to say, we must find continuous functions f and g such that their compositions are respectively homotopic to the identity on $\mathbb{R}^{m-k}\setminus \{0\}$ and on $\mathbb{R}^m\setminus \mathbb{R}^k$.

Try it this way and if you don't get it I can lend you my solution tomorrow. (I'm the guy sitting behind the guy with the ponytail :tongue:)

5. Feb 11, 2007

quasar987

P.S. I created a thread on #1 of the "problem sheet" at https://www.physicsforums.com/showthread.php?t=155623 in case you haven't noticed.

The solution we found is based on the fact that there exists a bijection btw R and R². I don't know about you, but before today I had always thought that there existed NO bijection btw these two sets. But then again, this result is considered one of the basic result of set theory so I guess M. Cornea though we knew about it. He told me the other day that where he went to school [Romania], they learn calculus early in (their equivalent of our) secondary school and by the end of secondary school they've covered algebra up to fields and modules [the equivalent of Algebra 2 at UdeM].

But if you found a another solution, I'd be glad to hear it.

6. Feb 11, 2007

Gagle The Terrible

Thanks a Ton Quasar ! And yes, I'm the guy sitting in front of you :P
In the mean time, I had considered that approach. The details ought to come.

EDIT : I got the first question right but thanks for your help. It assured me of my reasoning.

Last edited: Feb 11, 2007
7. Feb 13, 2007

quasar987

The solution:

The first thing is to identify what $\mathbb{R}^m\setminus\mathbb{R}^k$ means. Obviously this does not make sense formally because $\mathbb{R}^k$ is not a subset of $\mathbb{R}^m$. But it is easy to figure out what M. Cornea meant if you consider the two cases that have a natural geometrical interpretation, namely $\mathbb{R}^2\setminus\mathbb{R}$ and $\mathbb{R}^3\setminus\mathbb{R}^2$. The first is the plane minus the real line... that is to say, $\mathbb{R}^2\setminus\{(x,0)\in\mathbb{R}^2:x\in \mathbb{R}\}$. The second is 3D space minus the Oxy plane, that is, $\mathbb{R}^3\setminus\{(x,y,0)\in\mathbb{R}^3:x,y\in\mathbb{R}\}$. This easily generalizes to $\mathbb{R}^m\setminus\mathbb{R}^k:=\mathbb{R}^m\setminus\{(x_1,...,x_k,0,...0)\in\mathbb{R}^m:x_1,...x_k\in\mathbb{R}\}$

Alright, so now picking up where post #4 ended, we must find continuous functions $f:\mathbb{R}^m\setminus\mathbb{R}^k\rightarrow\mathbb{R}^{m-k}\setminus \{0\}$ and $g:\mathbb{R}^{m-k}\setminus \{0\}\rightarrow \mathbb{R}^m\setminus\mathbb{R}^k$ such that

$$h_1:g\circ f\simeq \mbox{id}_{\mathbb{R}^m\setminus\mathbb{R}^k}$$
$$h_2:f\circ g\simeq\mbox{id}_{\mathbb{R}^{m-k}\setminus \{0\}}$$

(for some homotopies h_1 and h_2 to be determined also). Consider $f(x_1,...,x_m)=(x_{k+1},...x_m)$ and $g(x_1,...,x_{m-k})=(0,...,0,x_1,...,x_{m-k})$. These functions are obviously continuous. Remains to match them with a pair of homotopies.

Consider $h_1:\mathbb{R}^m\setminus\mathbb{R}^k\times [0,1]\rightarrow \mathbb{R}^m\setminus\mathbb{R}^k$ and $h_2:\mathbb{R}^{m-k}\setminus\{0\}\times [0,1]\rightarrow \mathbb{R}^{m-k}\setminus\{0\}$ defined by

$$h_1(x_1,...,x_m,t)=(g\circ f)(x_1,...,x_m)+t(x_1,...,x_k,0,...,0)$$

$$h_2(x_1,...,x_{m-k},t)=(f\circ g)(x_1,...,x_{m-k})$$

(Indeed, $f\circ g$ is already the identity on $\mathbb{R}^{m-k}\setminus\{0\}$, so the trivial homotopy does the trick).