The space of homeomorphisms of the closed unit interval

Question

Denote by $X$ the space of homeomorphisms of the unit interval $[0,1]$, equipped with the topology of uniform convergence. It is an exercise in a textbook I'm reading to show that $$X \cong \{ 0, 1 \} \times [0,1]^{\omega}.$$ The progress I've made on this so far is to note that homeomorphisms $f: [0,1] \to [0,1]$ are either increasing or decreasing, and so it should be the case that $f(0) = 0$ or $f(0) = 1$. I think this splits the space into two connected components, as suggested by the factor $\{ 0, 1 \}$. Can anyone provide further advice on how to establish the isomorphism? My only instinct for the remaining factor $[0,1]^\omega$ is that continuous functions on $[0,1]$ are determined by their values on a countable dense set, but I can't see how this would lead to a bicontinuous map.

Answer 1

It's not true. The space $\{0,1\} \times [0,1]^\omega$ is compact, but the space of homeomorphisms of $[0,1]$ is not compact in the topology of uniform convergence.

For $n \geqslant 2$, let

$$f_n(x) = \begin{cases}\qquad (2 - 2^{-n})x &, 0 \leqslant x \leqslant \frac{1}{4} \\ \quad\frac{1}{2} + 2^{-n}\bigl(x-\frac{1}{2}\bigr) &, \frac{1}{4} \leqslant x \leqslant \frac{3}{4} \\ 1 - (2-2^{-n})(1-x) &, \frac{3}{4} \leqslant x \leqslant 1.\end{cases}$$

Then $f_n$ is a homeomorphism of $[0,1]$, but $(f_n)_{n\geqslant 2}$ converges uniformly to

$$f(x) = \begin{cases} 2x &, 0 \leqslant x \leqslant \frac{1}{4} \\ \frac{1}{2} &, \frac{1}{4} \leqslant x \leqslant \frac{3}{4}\\ 2x-1 &, \frac{3}{4} \leqslant x \leqslant 1\end{cases}$$

which is not a homeomorphism.

Answer 2

I believe it is true that $X\cong \{0,1\}\times(0,1)^\omega$, and the following is an explicit homeomorphism. As you observe, we can split $X$ into two pieces according to orientation, so it suffices to produce a homeomorphism $Y\to (0,1)^\omega$, where $Y$ is the space of increasing homeomorphisms. Given an increasing homeomorphism $f:[0,1]\to[0,1]$, define a sequence $(r_n)\in(0,1)^\omega$ inductively as follows. First, $r_0=f(1/2)$. Inductively, suppose we have defined $r_0,\dots,r_{n-1}$ and in the process we have used the value of $f$ at $n$ different points. These $n$ points (and their images under $f$) split the domain and range of $f$ each into $n+1$ subintervals; say the domain is split into $I_0,\dots,I_n$ and the range is split into $J_0,\dots,J_n$ (with these subintervals listed in increasing order). At this point we have two choices. The first choice is to define $r_{n+i}$ ($i=0,\dots,n$) to be the image of the midpoint of $I_i$ under $f$, rescaled by linearly identifying $J_i$ with $(0,1)$. The second choice is to define $r_{n+i}$ to be the image of the midpoint of $J_i$ under $f^{-1}$, rescaled by linearly identifying $I_i$ with $(0,1)$.

Use this procedure to inductively define a sequence $(r_n)$, alternating between the two choices (it's a good exercise to figure out why this construction would not give a bijection if you just always made the first choice). It is not too hard to see that this is a bijection and in fact a homeomorphism from $Y$ to $(0,1)^\omega$. To construct the inverse, note that reversing the procedure starting from any $(r_n)$ will yield two countable dense subsets of $(0,1)$ and an order-preserving bijection between them, and this bijection will extend uniquely to an order-preserving bijection $[0,1]\to[0,1]$.