# 2.07 Homeomorphisms

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.

*(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 \(\cos\) 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?

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