# 2.07 Homeomorphisms

## Video

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

## 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

1. 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?