MAS61015 Algebraic Topology 3 Topological spaces 5 Paths

4. Homeomorphism

Video (Definition 4.1 to Example 4.8)

Definition 4.1.

Let $X$ and $Y$ be topological spaces, and let $f\colon X\to Y$ be a function. We say that $f$ is a homeomorphism if

(a)

$f$ is a bijection, so there is an inverse map $f^{-1}\colon Y\to X$ , satisfying $f^{-1}(f(x))=x$ for all $x\in X$ and $f(f^{-1}(y))=y$ for all $y\in Y$ .
(b)

Both $f$ and $f^{-1}$ are continuous.

We say that $X$ and $Y$ are homeomorphic if there exists a homeomorphism from $X$ to $Y$ . We will write $X\simeq Y$ to indicate that $X$ and $Y$ are homeomorphic.

Remark 4.2.

Later we will introduce a different notion called homotopy equivalence, and write $X\cong Y$ (with an extra bar) to indicate that $X$ and $Y$ are homotopy equivalent. It is important to distinguish between these two ideas. However, you should be aware that there is no consistency in the literature about the notation used. It is safest to say “homeomorphic” or “homotopy equivalent” in words, to avoid confusion.

Remark 4.3.

There are very few indirect techniques for proving that two spaces are homeomorphic. Instead, we just have to find a specific homeomorphism. Many examples will be given below.

Proposition 4.4.

Suppose that $a,b,c,d\in{\mathbb{R}}$ with $a<b$ and $c<d$ . Then $[a,b]\simeq[c,d]$ and $(a,b)\simeq(c,d)$ and $[a,b)\simeq[c,d)$ and $(a,b]\simeq(c,d]$ .

Proof.

We can define maps $[a,b]\xrightarrow{f}[c,d]\xrightarrow{g}[a,b]$ by

f(t)=c+\frac{d-c}{b-a}(t-a)\hskip 40.0ptg(t)=a+\frac{b-a}{d-c}(t-c).

We find that $f(g(t))=t$ and $g(f(t))=t$ , so $g$ is inverse to $f$ . It is clear that $f$ and $g$ are continuous, so $f$ is a homeomorphism. The same formulae also give homeomorphisms $(a,b)\to(c,d)$ and so on. ∎

Proposition 4.5.

The formulae

f(x)=\frac{x}{\sqrt{1-x^{2}}}\hskip 40.0ptg(y)=\frac{y}{\sqrt{1+y^{2}}}

give continuous maps $(-1,1)\xrightarrow{f}{\mathbb{R}}\xrightarrow{g}(-1,1)$ which are inverse to each other. Thus, they are both homeomorphisms, and $(-1,1)\simeq{\mathbb{R}}$ . Moreover, the same maps restrict to give homeomorphisms between $[0,1)$ and $[0,\infty)$ .

Proof.

First, note that when $x\in(-1,1)$ we have $0\leq x^{2}<1$ so $0<1-x^{2}\leq 1$ , so dividing by $\sqrt{1-x^{2}}$ does not cause any problems. Thus, $f$ is a continuous function from $(-1,1)$ to ${\mathbb{R}}$ . Similarly, for all $y$ we have $1+y^{2}\geq 1>0$ so $\sqrt{1+y^{2}}\geq 1>0$ , so dividing by $\sqrt{1+y^{2}}$ does not cause a problem, and $g$ is a continuous map from ${\mathbb{R}}$ to ${\mathbb{R}}$ . Note also that

g(y)^{2}=y^{2}/(1+y^{2})=1-1/(1+y^{2})<1,

so $g(y)\in(-1,1)$ , so $g$ can in fact be regarded as a continuous map $g\colon{\mathbb{R}}\xrightarrow{}(-1,1)$ .

Note that if $y=f(x)=x/\sqrt{1-x^{2}}$ then

1+y^{2}=1+\frac{x^{2}}{1-x^{2}}=\frac{1}{1-x^{2}}

so $\sqrt{1+y^{2}}=(1-x^{2})^{-1/2}$ , so $g(y)=y/\sqrt{1+y^{2}}=x$ . This shows that $g(f(x))=x$ .

Conversely, suppose we start with $y$ and put $x=g(y)=y/\sqrt{1+y^{2}}$ . We then have

1-x^{2}=1-\frac{y^{2}}{1+y^{2}}=\frac{1}{1+y^{2}},

so $\sqrt{1-x^{2}}=(1+y^{2})^{-1/2}$ , so $f(x)=x/\sqrt{1-x^{2}}=y$ . This shows that $f(g(y))=y$ , so $f\colon(-1,1)\xrightarrow{}{\mathbb{R}}$ and $g\colon{\mathbb{R}}\xrightarrow{}(-1,1)$ are inverses of each other, as required. ∎

Remark 4.6.

Suppose we are trying to define a map $p\colon U\to V$ , and we give a formula for $p(u)$ . For this to be valid, we must check two things.

(a)

The formula must be meaningful for all elements $u\in U$ . It must never involve division by zero, or square roots of negative numbers in a context where a real result is required, for example.
(b)

The resulting value $p(u)$ must lie in $V$ . For example, $V$ might be a subspace of ${\mathbb{R}}^{3}$ defined by various equations or inequalities. Typically it will only be obvious from the formula that $p(u)$ lies in ${\mathbb{R}}^{3}$ , so we need to check that the equations or inequalities are satisfied as an extra step.

You should observe how these checks were done in the proof of Proposition 4.5. It is sadly common for them to be omitted in homework or exam answers submitted by students. Do not let that be you.

Remark 4.7.

There is some interesting geometry behind the maps $f$ and $g$ in Proposition 4.5: we have $f(x)=y$ and $g(y)=x$ if and only if $x$ and $y$ are related as in the diagram below.

In more detail, let $C$ be the circle of radius one centred at $(0,1)$ . Given a point $x\in(-1,1)$ , let $(x,z)$ be the point on the lower half of $C$ lying directly above $(x,0)$ . Then draw a line from $(0,1)$ through $(x,z)$ , and let $y$ be the point where it meets the axis. I claim that $y=x/\sqrt{1-x^{2}}=f(x)$ . Indeed, as the point $(x,z)$ lies on $C$ , the distance from $(x,z)$ to $(0,1)$ must be one, so $x^{2}+(1-z)^{2}=1$ , so $1-z=\sqrt{1-x^{2}}$ . On the other hand, the diagram contains a triangle of base $y$ and height $1$ , and a nested triangle of base $x$ and height $1-z$ . These triangles have the same angles, so we must have $y/1=x/(1-z)$ , so $y=x/\sqrt{1-x^{2}}$ as claimed.

Example 4.8.

The previous example can be generalised as follows. Put $OB^{n}=\{x\in{\mathbb{R}}^{n}\;|\;\|x\|<1\}$ (so in particular $OB^{1}=(-1,1)\subset{\mathbb{R}}$ ). We can define $f\colon OB^{n}\xrightarrow{}{\mathbb{R}}^{n}$ and $g\colon{\mathbb{R}}^{n}\xrightarrow{}OB^{n}$ by

	$\displaystyle f(x)$	$\displaystyle=x/\sqrt{1-\\|x\\|^{2}}$
	$\displaystyle g(y)$	$\displaystyle=y/\sqrt{1+\\|y\\|^{2}}.$

Using essentially the same argument as given above, we see that $f$ and $g$ are mutually inverse homeomorphisms, so $OB^{n}$ is homeomorphic to ${\mathbb{R}}^{n}$ .

Example 4.9.

The map $f(x)=(\|x\|,x/\|x\|)$ gives a homeomorphism ${\mathbb{R}}^{n}\setminus\{0\}\to(0,\infty)\times S^{n-1}$ , with inverse $f^{-1}(t,y)=ty$ .

Interactive demo

Example 4.10.

Video

Note that Definition 4.1 specifies that both $f$ and $f^{-1}$ must be continuous. It is necessary to make the definition this way, because it can easily happen that we have a bijection $f\colon X\to Y$ where $f$ is continuous but $f^{-1}$ is not, and we do not want to count maps like that as homeomorphisms. We give such an example here.

Put $X=(-\infty,0]\cup(1,\infty)={\mathbb{R}}\setminus(0,1]$ and $Y={\mathbb{R}}$ . Define maps $X\xrightarrow{f}Y\xrightarrow{g}X$ by

f(x)=\begin{cases}x&\text{ if }x\leq 0\\ x-1&\text{ if }x>0.\end{cases}\hskip 40.0ptg(x)=\begin{cases}x&\text{ if }x% \leq 0\\ x+1&\text{ if }x>0.\end{cases}

It is easy to see that $f$ and $g$ are inverse to each other. We claim that $f$ is continuous but that $g$ is not. To see this, we introduce the subsets

	$\displaystyle U$	$\displaystyle=(-\infty,0]=X\cap(-\infty,\tfrac{1}{2})$
	$\displaystyle V$	$\displaystyle=(1,\infty)=X\cap(\tfrac{1}{2},\infty).$

The sets $(-\infty,\tfrac{1}{2})$ and $(\tfrac{1}{2},\infty)$ are open in ${\mathbb{R}}$ , so $U$ and $V$ are open in the subspace topology on $X$ . The map $f$ is given by $f(x)=x$ on $U$ and by $f(x)=x-1$ on $V$ , so the restrictions to $U$ and $V$ are both continuous, so $f$ is continuous on $X$ by open patching (Proposition 3.34). On the other hand, we have $g^{-1}(U)=(-\infty,0]$ , which is not an open subset of the space $Y={\mathbb{R}}$ ; this proves that $g$ is not continuous. We therefore have a continuous bijection whose inverse is not continuous, so it does not count as a homeomorphism.

Example 4.11.

Video

Define $f\colon S^{1}\times\{1,-1\}\xrightarrow{}O_{2}$ by

f(x,y,z)=\left[\begin{matrix}x&-yz\\ y&xz\end{matrix}\right]

f(\cos(\theta),\sin(\theta),+1)=\left[\begin{matrix}\cos(\theta)&-\sin(\theta)% \\ \sin(\theta)&\cos(\theta)\end{matrix}\right]=\text{ rotation matrix }

and

f(\cos(\theta),\sin(\theta),-1)=\left[\begin{matrix}\cos(\theta)&\sin(\theta)% \\ \sin(\theta)&-\cos(\theta)\end{matrix}\right]=\text{ reflection matrix }

It is easy to see that this is a homeomorphism, with inverse

f^{-1}\left[\begin{matrix}a&b\\ c&d\end{matrix}\right]=(a,c,ad-bc).

We can also define $g\colon{\mathbb{R}}^{3}\times S^{1}\times\{1,-1\}\to GL_{2}({\mathbb{R}})$ by

g(u,v,w,x,y,z)=\left[\begin{matrix}e^{u}&0\\ 0&e^{v}\end{matrix}\right]\left[\begin{matrix}1&t\\ 0&1\end{matrix}\right]\left[\begin{matrix}x&-yz\\ y&xz\end{matrix}\right].

This can also be shown to be a homeomorphism. The inverse is

g^{-1}\left[\begin{matrix}a&b\\ c&d\end{matrix}\right]=(u,v,w,x,y,z),

where

$\displaystyle\Delta$	$\displaystyle=ad-bc$	$\displaystyle w$	$\displaystyle=(ac+bd)/\|\Delta\|$
$\displaystyle u$	$\displaystyle=\ln\|\Delta\|-\tfrac{1}{2}\ln(c^{2}+d^{2})$	$\displaystyle x$	$\displaystyle=\operatorname{sgn}(\Delta)d/\sqrt{c^{2}+d^{2}}$
$\displaystyle v$	$\displaystyle=\tfrac{1}{2}\ln(c^{2}+d^{2})$	$\displaystyle y$	$\displaystyle=c/\sqrt{c^{2}+d^{2}}$
$\displaystyle z$	$\displaystyle=\operatorname{sgn}(\Delta).$

Remark 4.12.

We could try to define a map $g\colon S^{1}\to S^{1}$ by the rule

g((\cos(\theta),\sin(\theta))=(\cos(\theta/2),\sin(\theta/2)).

There could be two different things wrong with this definition, depending on how we interpret it.

(a)

Any point in $S^{1}$ can be expressed as $(\cos(\theta),\sin(\theta))$ for infinitely many different values of $\theta$ , differing by multiples of $2\pi$ . For example, the point $(-1,0)$ can be represented as $(\cos(\pi),\sin(\pi))$ or as $(\cos(-\pi),\sin(-\pi))$ . The first representation gives $g((1,0))=(\cos(\pi/2),\sin(\pi/2))=(0,1)$ , and the second representation gives $g((-1,0))=(\cos(-\pi/2),\sin(-\pi/2))=(0,-1)$ . Thus, equation (A) does not give us a well-defined function $S^{1}\to S^{1}$ .
(b)

We could instead note that any point in $S^{1}$ can be expressed as $(\cos(\theta),\sin(\theta))$ for a unique choice of $\theta$ satisfying the auxiliary condition $-\pi<\theta\leq\pi$ , and we could define $g$ by saying that formula (A) holds for this choice of $\theta$ . This gives a well-defined function, which satisfies $g((-1,0))=(0,1)$ . However, for points $(x,y)\in S^{1}$ lying just below $(-1,0)$ , we find that $\theta\approx-\pi$ and so $g(x,y)$ is close to $(0,-1)$ . This shows that $g$ is discontinuous, despite the apparent continuity of all ingredients in equation (A).

For this kind of reason, it is generally a bad idea to define functions in terms of $\theta$ . This is why our initial definition of $f\colon S^{1}\times\{1,-1\}\to O_{2}$ in Example 4.11 was given directly in terms of the coordinates $x$ , $y$ and $z$ .

Example 4.13.

Let $N$ be the “north pole” of the sphere $S^{2}$ , in other words the point $(0,0,1)$ . There are mutually inverse continuous maps $S^{2}\setminus\{N\}\xrightarrow{f}{\mathbb{R}}^{2}\xrightarrow{g}S^{2}% \setminus\{N\}$ given by

f(x,y,z)=\frac{(x,\;y)}{1-z}\hskip 40.0ptg(u,v)=\frac{(2u,\;2v,\;u^{2}+v^{2}-1% )}{u^{2}+v^{2}+1},

so $S^{2}\setminus\{N\}$ is homeomorphic to ${\mathbb{R}}^{2}$ . This is called stereographic projection. Geometrically, $f(x,y,z)$ is the unique point where the line joining $N$ to $(x,y,z)$ meets the plane $z=0$ , and $g(u,v)$ is the unique point where the line joining $N$ to $(u,v,0)$ meets $S^{2}$ :

Interactive demo

We will leave it to the reader to check directly from the formulae that $f(g(u,v))=(u,v)$ and $g(f(x,y,z))=(x,y,z)$ . Essentially the same formulae can be used to prove that $S^{n}\setminus\{P\}\simeq{\mathbb{R}}^{n}$ for any $n\geq 1$ and any $P\in S^{n}$ .