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 have one doubt that may be too general, I don't know, so sorry if this is not a good place to ask it. I've also seem many other people with the same problem that I have, so I think that if this question fits this site, it'll help other people to. I've been studying multivariable analysis, metric spaces and manifolds, and in all of them I find the same problem: although I already understood the main definitions and results I find myself a little lost when it comes to construct and to prove homeomorphisms.

For instance, in linear algebra when it comes to prove isomorphism of vector spaces I know a "procedure", I have a line of thought that even though can be dificult in some cases, will end up giving what I was seeking for. One point of this "procedure" is that we now that once we have the map it suffices to show that it's linear (a simple check of a property), show that it's kernel is just the null vector and to show surjectivity we look at the dimensions.

However, when it comes to construct and prove homeomorphism it seems like "the only way is to make a good guess", without some procedure and something like that. For example, it's not intuitive, at least to me that to show that the open ball is homeomorphic to $\mathbb{R}^n$ we would need to take the map $f(x) = x/(1+|x|)$. It's just that I look to this map and I think: "I would never have thought of it".

Anyway, if finding the map seems a problem, proving that the map indeed is homeomorphism seems even worse, because the most common way: find an $\epsilon$ also seems like depending on "good guesses". Even for simple function on the real line I look at the $\epsilon$'s that usually are used to prove continuity and I think: "I would never have thought of such a thing".

My question is: there is a systematic way of attacking those problems? Is there a procedure to find and prove hoemorphisms like there's in linear algebra to find and prove isomorphisms? Is there a way to make this less dependent on guesses? In the real line people often draw the small intervals, however, this kind of thing seems not too good, since we won't have this "graphical resource" to find a way to prove homeomorphisms between higher dimensional manifolds. Where can I really learn those things?

My interest is really the study of manifolds, and I'm working with Spivak's "A Comprehensive Introduction to Differential Geometry Vol. 1", however I'm feeling the need to better understand these questions about how to construct and how to prove homeomorphisms, since all the charts for the manifolds must be construct as homeomorphisms.

Thanks in advance for your help.

share|improve this question
2  
There are certainly heuristics akin to those you described for Linear Algebra that go into thinking about a homeomorphism between top. spaces, and hopefully someone else will write a few up. In general, though, it is often the case that one says two spaces are homeomorphic without explicitly giving the homeomorphism/proving it works. "Obviously" a cube is homeomorphic to a sphere or a solid tetrahedron or lots of other somewhat similar shapes you could get by messing with play-doh. Writing down the homeomorphism, though, is often no easy task. [Sorry if this comment is too general or silly!] –  Benjamin Dickman Apr 15 '13 at 22:51
2  
In your example, and many others, the geometry is a guide. We want to push the boundary of the ball to "$\infty$." For example, using complex numbers and the unit ball, $z$ goes to $\frac{1}{1-|z|}$ should do the job. –  André Nicolas Apr 15 '13 at 22:53
    
Contrary to linear algebra where a morphism is characterized by its value on a basis, there is no simple way to describe continuous maps. That's why there are so much books about algebraic topology in the litterature. –  Taladris Apr 15 '13 at 23:02

2 Answers 2

up vote 3 down vote accepted

Here's one way to think of the homeomorphism from $\mathbb{R}^n$ to the unit ball. Think of shrinking space radially so that a radius of infinity becomes a radius of one. That is, we want to keep the direction but collapse the radius. How do you collapse $[0,\infty)$ to $[0,1)$? Well, one way to get a small set from a big set is to invert numbers near infinity. At the same time, you don't want to invert numbers near zero. So we can first translate away from $0$, mapping $[0,\infty)$ to $[1,\infty)$. Then, inverting $[1,\infty)$ gives us $(0,1]$. Now the map $s\mapsto 1-s$ maps this to $[0,1)$. In conclusion, $r\mapsto (1-\frac{1}{1+r})$, or upon simplifying, $r\mapsto \frac{r}{1+r}$ is a map from $[0,\infty)$ to $[0,1)$ which sends $0$ to $0$ and collapses $\infty$ to $1$. Think of this as taking a radius in $[0,\infty)$ and spitting out a "new" radius in $[0,1)$. We got this function by playing around, but if we look at it, it makes sense. We shrink the radius by a factor that grows at the same rate as the radius, so that in the limit their ratio approaches $1$. In fact, for any $c>0$, $r\mapsto \frac{r}{c+r}$ would have worked as well. If you wanted to practice proving maps are homeomorphisms, you can try finding a continuous inverse.

But we will use this to find a homeomorphism from $\mathbb{R}^n$ to the unit ball. Well, if $x$ is a vector in $\mathbb{R}^n$, $\frac{x}{|x|}$ is a unit vector pointing in the direction of $x$ (assuming for the moment $x$ is nonzero so we can divide by its length). Now all we have to do is multiply by the "new" radius, that is, by $\frac{|x|}{1+|x|}$. In conclusion, our map is $x\mapsto x\cdot \frac{\frac{|x|}{1+|x|}}{|x|}$. Simplifying, you get $x\mapsto \frac{x}{1+|x|}$. Again, this answer makes sense for the same reason as before: you take $x$ and shrink it by a factor that grows at the same rate as its length. You can describe the inverse to this map by a similar process. I know I could have derived this much more quickly, but I just wanted to show you how you can go about attacking the problem if you only had a vague intuition.

Edit: Note that to prove the above map is continuous, you don't need to use an $\epsilon$-$\delta$ argument. If you have a map $f:\mathbb{R}^n\to \mathbb{R}^m$ given by $f(x)=(f_1(x),\dots,f_m(x))$ and you wish to show it is continuous, it is enough to show each of the $f_i$ is continuous. In showing real-valued functions are continuous, the idea is that any function which is a composition, sum, product, or ratio of continuous maps will end up being continuous. In the example above, the component functions are $f_i(x)=\frac{x_i}{1+\sqrt{x_1^2+\cdots+x_n^2}}$. Since the constant function $1$ is continuous, projecting onto the $i$th coordinate is continuous, squaring is continuous, and taking the squareroot is continuous, this function is continuous.


I'll just give one more example, which is stereographic projection.

We want to show that a sphere minus a point is homeomorphic to $\mathbb{R}^2$. Imagine a sphere sitting in space so that the equator is in the $x$-$y$ plane. The idea of stereographic projection is that, for any point $(x,y,z)$ on the sphere (other than the north pole itself), we can draw a line from the north pole and through $(x,y,z)$ until it hits the $x$-$y$ plane. Thus, we associate to each point on the sphere minus the north pole a point on the plane. You can see that there's an inverse: starting with a point on the plane, draw a line to the north pole and see where it intersects the sphere. Both these processes are intuitively continuous: if you jiggle a point only a little bit, its projection will also be jiggled only a little bit.

Thus we have (intuitively) convinced ourselves this mapping is a homeomorphism. Now finding the formula is just algebra: the line between the north pole and $(x,y,z)$ can be represented parametrically by $(0,0,1)+t(x,y,z-1)$. This line intersects the $x$-$y$ plane when the $z$ coordinate is zero, that is, when $1+t(z-1)=0$, so $t=\frac{-1}{z-1}=\frac{1}{1-z}$. Plugging in this value of $t$, we see that this point is $(\frac{x}{1-z},\frac{y}{1-z},0)$. For a map to $\mathbb{R}^2$, we omit the $z$ coordinate: $(x,y,z)\mapsto (\frac{x}{1-z},\frac{y}{1-z})$. You can now show this map is continuous by a similar analysis as above. And you can find the inverse map and show it is continuous similarly.


In short, it helps to first visualize a homeomorphism as a geometric transformation, and then to formalize it using algebraic expressions. And as a rule of thumb, most functions that can be described using algebraic expressions (or even analytic expressions involving $sin$, etc.) are continuous.

share|improve this answer

One other easy way to construct homeomorphisms is to use the fact that any map from a compact space to a Hausdorff space is closed. This implies that any continuous bijection from a compact space to a Hausdorff space is a homeomorphism. This helps a lot, since most bijections you can write down are continuous.

This doesn't help for your specific example, though.

share|improve this answer

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.