Is the space of distinct triples homeomorphic to a union of products?

Question

$\newcommand{\S}{\mathbb{S}^1}$Let $M=\{(x,y,z) \in (\S)^3 \, |\,\, x,y,z \,\,\text{are distinct}\}$.

Is $M$ homeomorphic to a finite union of products of one-dimensional manifolds?

I think $M$ is not connected, so it cannot be homeomorphic to a product.

$M$ is not connected, since if $x-y-z$ are ordered clockwise, you cannot move to $x-z-y$ being clockwise.

So I think that there are two connected components.

---

The case where considering pairs follows easily from the group structure of $\mathbb{S}^1$.

Answer 1

$\newcommand\S{{\mathbb S}^1}\newcommand\quad{ \ \ \ \ }$This answer is very similar to Greg's answer but I think it becomes slightly clearer if we insted use the multiplicative structure of the circle. We therefore view $\S$ as the unit circle in the complex plane, which is a group under multiplication.

Consider the homeomorphism $$ \varphi : (x,y,z)\in (\S)^3 \mapsto (z^{-1}x,z^{-1}y,z)\in (\S)^3 $$ and observe that $$ x=z \Leftrightarrow \varphi (x,y,z)\in \{1\}\times \S\times \S, $$ $$ y=z \Leftrightarrow \varphi (x,y,z)\in \S\times \{1\}\times \S, $$ $$ x=y \Leftrightarrow \varphi (x,y,z)\in \quad \ \Delta \quad\ \times \S, $$ where $\Delta $ is the diagonal of $\S\times \S$, that is, $\Delta =\{(u, v)\in \S\times \S: u=v \}$. It follows that $$ \varphi (M) = X\times \S, $$ where $X$ is the subset of $\S\times \S$ that we get after throwing away $$ \big (\{1\}\times \S\big ) \cup \big (\S\times \{1\}\big ) \cup \Delta . $$ Viewing $\S\times \S$ as a closed square, where we identify the bottom and top sides, as well as the left and right sides,

• removing $\{1\}\times \S$ corresponds to removing the left (and hence also the right) side of the square,

• removing $\S\times \{1\}$ corresponds to removing the bottom (and hence also the top) side, while

• removing the diagonal is, well, removing the diagonal.

We are then left with an open square minus the diagonal, hence two disjoint open triangles!

Answer 2

A similar linear algebra approach works here. Thinking of $S^1$ as $[0,1]$ modulo the identification of $0$ and $1$ and the group operation as addition, the function $(x,y,z) \mapsto (x-y,y-z,z)$ is an automorphism of $(S^1)^3$ that maps $M$ homeomorphically onto $\Big( \bigl( (0,1)\times(0,1) \bigr) \setminus \bigl\{(t,1-t)\colon t\in(0,1)\bigr\} \Big) \times S^1$. The first factor is the disjoint union of two open triangles, each of which is homeomorphic to an open square and thus the product of one-dimensional manifolds; therefore its product with the remaining $S^1$ is as well.