# 2.07 Homeomorphisms

## Video

Below the video you will find accompanying notes and some pre-class questions.

- Previous video:
**Hausdorffness**. - Index of all lectures.

## Notes

### Definition

*(0.15)* A continuous map \(F\colon X\to Y\) is a *homeomorphism* if
it is bijective and its inverse \(F^{-1}\) is also continuous. If
two topological spaces admit a homeomorphism between them, we say
they are *homeomorphic*: they are essentially the same topological
space.

*(1.13)* Consider the half-open interval \([0,2\pi)\) and the
continuous map \(F\colon [0,2\pi)\to S^1\) defined by
\(F(\theta)=e^{i\theta}\). This is continuous and
bijective. However, it is **not** a homeomorphism! We will see that
the circle has fundamental group \(\mathbf{Z}\) and the interval is
simply-connected, so they cannot be homeomorphic.

### Criterion for a map to be a homeomorphism

*(3.33)* Let \(X\) be a compact space and let \(Y\) be a Hausdorff
space. Then any continuous bijection \(F\colon X\to Y\) is a
homeomorphism.

*(5.00)* We need to show that \(F^{-1}\) is continuous, i.e. that
for all open sets \(U\subset X\) the preimage \((F^{-1})^{-1}(U)\)
is open in \(Y\). But \((F^{-1})^{-1}(U)=F(U)\), so we need to show
that *images* of open sets are open. It suffices to show that
complement of \(F(U)\) is closed.

*(6.23)* \(U\subset X\) is open, so \(X\setminus U\) is closed, and
since \(X\) is compact, this means \(X\setminus U\) is closed
(closed subsets of compact spaces are compact). The image of a
compact set is also compact, so \(F(X\setminus U)\) is compact. A
compact subset of a Hausdorff space is closed, so \(F(X\setminus
U)\) is closed, so \(F(U)=Y\setminus F(X\setminus U)\) is open, as
required.

### Example

*(8.46)* In the session on the subspace topology, we saw that the
2-torus \(T^2\) can be thought of as a subset of \(\mathbf{R}^3\)
and also as a subset of \(\mathbf{R}^4\). More precisely,

- write \(T\subset\mathbf{R}^3\) for the standard torus in \(\mathbf{R}^3\).
- define \(T'=\{(\cos\theta,\sin\theta,\cos\phi,\sin\phi)\ :\ \theta,\phi\in[0,2\pi)\}\subset\mathbf{R}^4\) to be the torus in 4-d.

I will show that they are both homeomorphic to \(S^1\times S^1\).

*(10.22)* We need a map \(F\colon S^1\times S^1\to T'\) which will
be \((e^{i\theta},e^{i\phi})\mapsto
(\cos\theta,\sin\theta,\cos\phi,\sin\phi)\). It is a continuous map
(we saw that [[./topsp03.org][\(\cos\) and \(\sin\) are continuous
functions on the circle) and it is bijective. The circle is a closed
and bounded set in \(\mathbf{R}^2\), so it is compact; the product
\(S^1\times S^1\) is compact by Tychonoff's theorem. The image
\(T'\) is a subspace of a Hausdorff space, hence
Hausdorff. Therefore \(F\) is a homeomorphism.

*(11.53)* We need to do the same for \(T\), and the same argument
will apply provided I can give you a continuous bijection \(G\colon
S^1\times S^1\to T\). I claim that the following map will do:
\[G(e^{i\theta},e^{i\phi})=
\left(\begin{array}{ccc}\cos\phi&-\sin\phi&0\\ \sin\phi&\cos\phi&0\\ 0&0&1\end{array}\right)
\left(\begin{array}{c}0\\2+\cos\theta\\\sin\theta\end{array}\right).\]
In other words, I am taking the unit circle in the \(yz\)-plane
centred at \((0,2,0)\) (angle coordinate \(\theta\)) and rotating it
by an angle \(\phi\) around the \(z\)-axis.

## Pre-class questions

- Let \(X\) be the set \(\{0,1\}\) equipped with the discrete topology and let \(Y\) be the set \(\{0,1\}\) equipped with the indiscrete topology. Write down a continuous bijection \(X\to Y\). Are these spaces homeomorphic? If not, why does the theorem from the video fail to apply in this case?

## Navigation

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**Hausdorffness**. - Index of all lectures.