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Homeomorphism

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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”:

California as an island 3.png

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”:

World map with errors.png

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:

Trefoil unknotted.png

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

Exercise. What is this, topologically?

Double twist.png

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

for every $y \in Y$ there is $x \in X$ such that $y = f(x)$.

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:

there is a unique function $f^{-1}: Y {\rightarrow} X$ such that $f^{-1}f = {\rm id}_X$ and $ff^{-1} = {\rm id}_Y$.

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”:

Segment vs T.png

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:

Point-point line.png

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 $(- \infty , \infty )$.

Tangent graph.png

$\blacksquare$

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

Stereographic projection dim1.png

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:

given $x \in X$, draw a line through $x$ and $N$, find its intersection $y$ with $Y$, then $y = f(x)$.

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

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

Stereographic projection.png

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:

Square and circle homeomorphic.png

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!

Homeomorphism.png

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:

if $X$ is homeomorphic to $Y$ then $X$ 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.