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# Homeomorphism

### From Mathematics Is A Science

## Contents

## 1 What maps preserve topological features?

Homeomorphisms are continuous functions that preserve topological properties.

As we know, path-connectedness is preserved under continuous functions, which means that no new path-components can appear. If we choose to think of continuous functions as mappings of the terrain, we are trying to learn how to answer the most basic question:

*Can we get from point $A$ to point $B$?*

The continuity of the mapping prevents it from tearing the map and guarantees that we will never give the answer “No” while, in fact, the terrain it depicts answers “Yes”:

However, even though we can't *tear* things, we might still end up *gluing* its pieces! We might glue -- by mistake -- two islands together (or even continents). A “glued” map might give the answer “Yes” to our question while, in fact, the terrain it depicts answers “No”:

So, while a continuous map can't bring the number of components up, it can bring it down.

What could be wrong with gluing if, indeed, it is a continuous operation? The problem is that *the inverse of gluing is tearing*!

That's why we will define a class of maps so that both the function $$f: X {\rightarrow} Y $$ and its inverse $$f^{-1}: Y {\rightarrow} X$$ are continuous.

Even though tearing isn't allowed, we can cut -- if you glue it back together exactly as before. For example this is how you can un-knot this knot:

One can achieve this without cutting by seeing the knot in the $4$-dimensional space.

**Exercise.** What is this, topologically?

## 2 Definitions

Let's make this idea more precise. Recall a couple of definitions.

A function $f: X {\rightarrow} Y$ is called *one-to-one*, or injective, if
$$f(x) = f(y) \Rightarrow x = y.$$
Or, the preimage of a point, if non-empty, is a point.

A function $f: X {\rightarrow} Y$ is called *onto*, or surjective, if

Or, the image of the domain space is the whole target space, $f(X) = Y$.

A function that is one-to-one and onto is also called bijective.

**Theorem.** A function $f:X {\rightarrow} Y$ is bijective then it has the *inverse*:

**Definition.** Suppose $X$ and $Y$ are topological spaces and $f:X {\rightarrow} Y$ is a function. Then, $f$ is called a *homeomorphism* if

- $f$ is bijective,
- $f$ is continuous,
- $f^{-1}$ is continuous.

Then $X$ and $Y$ are said to be *homeomorphic*, as well as *topologically equivalent*.

The word “homeomorphism” comes from the Greek words “homoios” for similar and “morphē” for shape or form. One may understand the meaning of this word as follows: the shapes of two geometric objects are *similar* by being topologically *identical*.

Let's examine informally how the definition may be applied to demonstrate that spaces are *not* homeomorphic.

Consider $X$ a segment and $Y$ two segments. We already know that the number of path-components has to be preserved. Therefore, these two are not homeomorphic.

Consider now $X$ letter “T” and $Y$ a segment. Both have two path-components, are they homeomorphic? One trick is to pick a point $a$ in $X$, remove it, and see if $X\setminus\{a\}$ is homeomorphic to $Y\setminus \{b\}$ for *any* $b\in Y$. Let's choose $a$ to be in the cross of “T”:

Removing this point creates a mismatch of components regardless what $b$ is.

So, the method is to examine how many components can be produced by removing various points from the space.

**Exercise.** Using just this approach classify the symbols of the standard computer keyboard:

- ` 1 2 3 4 5 6 7 8 9 0 - =
- q w e r t y u i o p [ ] \
- a s d f g h j k l ; '
- z x c v b n m , . /

and

- $\sim \hspace{2 mm} ! \hspace{2 mm} @ \hspace{2 mm} \# \hspace{2 mm}\$ \hspace{2 mm}\% \hspace{2 mm}\hspace{2 mm}\widehat{ } \hspace{2 mm}\& \hspace{2 mm}* \hspace{2 mm}( \hspace{2 mm}) \hspace{2 mm}\_ \hspace{2 mm}+$
- Q W E R T Y U I O P $\{ \}$ |
- A S D F G H J K L : “
- Z X C V B N M < > ?

**Exercise.** Prove that an *isometry*, i.e., a bijection between two metric spaces that preserves the distance, is a homeomorphism.

**Example.** It might be hard to quickly come up with an example of *a continuous bijection that isn't a homeomorphism*. The reason is that our intuition takes us to Euclidean topology. Instead, consider the identity function
$$f=Id:(X,\tau) \to (X,\kappa),$$
where $\kappa$ is the anti-discrete topology and $\tau$ isn't. Then, on one hand

- $f^{-1}(X)=X$ is open in $(X,\tau)$,

so that $f$ is continuous. On the other hand, there is a proper subset $A$ of $X$ open in $\tau$ but

- $f(A)=A$ isn't open in $(X,\kappa)$,

therefore, $f^{-1}:(X,\kappa)\to (X,\tau)$ isn't continuous. A specific example is:

- $X=\{a,b\},$
- $\tau=\{\emptyset,\{a\},\{b\},X\},$
- $\kappa=\{\emptyset,\{a\},X\}.$

This example suggests a conclusion: the identity function will be a desired example when $\kappa$ is a proper subset of $\tau$. Consider, for example, $X={\bf R}$, while $\tau$ is the Euclidean topology and $\kappa$ any topology with fewer open sets (it's “sparser”), such as the topology of rays. $\square$

However, sometimes we don't need to verify the continuity of the inverse. It is guaranteed when we impose certain -- mild and natural -- restrictions on the topological spaces involved (see compact and Hausdorff).

**Exercise.** We are given a continuous function $f:{\bf R}\to {\bf R}$. Define a function $g:{\bf R}\to {\bf R}^2$ by
$$g(x)=(x,f(x)).$$
Prove that $g$ is continuous and that its image, the graph of $f$, is homeomorphic to ${\bf R}$.

## 3 Examples

**Theorem.** Closed intervals of non-zero, finite length are homeomorphic.

**Proof.** Let $X = [a,b], Y = [c,d], b>a, d>c$. We will find a function $f: [a,b] {\rightarrow} [c,d]$ with $f(a) = c$ and $f(b) = d$. The simplest function of this kind is linear:

To find the formula, use the point-slope formula from calculus. The line passes through $(a,c)$ and $(b,d)$, so its slope is $m = \frac{d-c}{b-a}\ne 0$. Hence, the line is given by $$f(x) = c + m(x-a).$$ We can also give the inverse explicitly: $$f^{-1}(y) = a + \frac{1}{m} \cdot (y-b).$$ Finally we recall that linear functions are continuous. $\blacksquare$

Similarly...

**Theorem.** Open intervals of finite length are homeomorphic.

**Exercise.** Prove the theorem.

**Theorem.** An open interval is not homeomorphic to a closed interval (nor half-open).

**Exercise.** Prove the theorem. Hint: remove the end point.

A closed interval $[a,b]$ can't be homeomorphic to a point because this function can't be bijective.

**Theorem.** All open intervals, even infinite ones, are homeomorphic.

**Proof.** Tangent gives you a homeomorphism between $(- \pi /2, \pi /2)$ and $(- \infty , \infty )$.

$\blacksquare$

Another way to justify this conclusion is given by the following construction:

Here the “north pole” $N$ is taken out from a circle to form $X$. Then $X$ is homeomorphic to a finite open interval, and to an infinite interval, $Y$. The function $f:X \rightarrow Y$ is defined as follows:

**Exercise.** Prove that this $f$ is a homeomorphism.

The above construction is a 2D version of the *stereographic projection*:

which is, literally, a map.

This construction is used to prove that the sphere with a pinched point is homeomorphic to the plane.

In the $n$-dimensional case, we place ${\bf S}^n$ in ${\bf R}^{n+1}$ as the unit “sphere”: $${\bf S}^n=\{x\in {\bf R}^{n+1}:||x||=1\}$$ and then remove the north pole $N=(0,0,...,0,1)$. The stereographic projection $$P:{\bf S}^n \setminus {N} \to {\bf R}^{n}$$ is defined by $$P(x)=\left( \frac{x_1}{1-x_{n+1}}, \frac{x_2}{1-x_{n+1}},...,\frac{x_n}{1-x_{n+1}} \right), \forall x=(x_1,...x_{n+1}).$$ It's inverse is $$P^{-1}(y)=\frac{1}{1+||y||}(2y_1,2y_2,...,2y_n,||y||^2-1), \forall y=(y_1,...y_{n}).$$

**Exercise.** Prove that this is a homeomorphism.

**Exercise.** Show that the sphere and the hollow cube are homeomorphic. Hint: the idea is that if you insert a balloon inside a box you can then inflate it and fill the box from the inside:

We can instead concentrate on the inverse of $f$. Let's illustrate the idea in dimension $2$, i.e., square $Y$ and circle $X$. One can give an explicit representation of the function: $$f^{-1}(u) = \frac{u}{||u||},$$ where $||u||$ is the norm of $u$. The norm is known to be continuous and, therefore, so is its restriction to $Y$. Next, the ratio of two continuous functions is continuous, as we know from calculus, as long as the denominator isn't $0$ (it isn't on $Y$). Hence $f^{-1}$ is continuous. This function is essentially a radial projection.

**Exercise.** Let $a,b$ be two points on the sphere $X={\bf S}^2$. Find a homeomorphism of the sphere to itself that takes $a$ to $b$. What is $X$ is the plane or the torus?

## 4 Topological equivalence

**Theorem.** Homeomorphism creates an equivalence relation on the set of all topological spaces.

**Exercise.** Prove the theorem.

Then it makes sense to call two spaces *topologically equivalent* if they are homeomorphic. We will use the following notation for that:
$$X \approx Y.$$

There is more to it... Suppose $$f:(X,\tau_X)\to(Y,\tau_Y)$$ is a homeomorphism. From the definition,

- if $U$ is open in $Y$ then $f^{-1}(U)$ is open in $X$, and
- if $V$ is open in $X$ then $f(V)$ is open in $Y$.

In other words,
$$V \in \tau_X \Leftrightarrow f(V)\in \tau _Y.$$
Therefore, we can define a function between the two topologies:
$$f_{\tau}:\tau_X\to\tau_Y$$
by
$$f_{\tau}(V)=f(V).$$
It's a *bijection*!

The result is that whatever is happening, topologically, in $X$ has an exact counterpart in $Y$: open and closed sets, their unions and intersections, exterior, interior, closure of sets, and convergent and divergent sequences, continuous and discontinuous functions, etc. To put it informally,

*homeomorphic spaces are topologically indistinguishable*.

The situation is similar to that in algebra when isomorphic groups (or vector spaces) have the exact matching of every algebraic operation and, in that sense, they are algebraically indistinguishable.

**Exercise.** Prove that the classes of homeomorphic spaces form a category.

This idea justifies the common language about topological spaces that's often appropriate:

*is*$Y$.

This is the reason why we speak of some classes of homeomorphic spaces as if they are represented by specific topological spaces:

*the*circle ${\bf S}^1$,*the*disk ${\bf B}^2$*the*sphere ${\bf S}^2$,*the*torus ${\bf T}^2$, etc.

Properties “preserved” under homeomorphisms are called *topological invariants*. In anticipation of our future studies we state this

**Theorem.** If two topological spaces are homeomorphic then their homology groups are isomorphic.

In other words, homology is a topological invariant. However, the converse isn't true. Indeed, even though a point and a segment aren't homeomorphic, their homology groups coincide.

**Exercise.** Prove that for a given topological space, the set of all of its self-homeomorphisms form a group with respect to composition.