How to deal with Homeomorphisms?

Solution 1:

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.

Solution 2:

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.