Homeomorphism of $\vert K\vert_d$ and $S^1$

Question

Consider $K=\{v_0,v_1,v_2,v_0<v_1,v_1<v_2,v_0<v_2\}$, where this can be drawn as a triangle with vertex $v_0,v_1,v_2$ and you write $v_0<v_1,v_1<v_2,v_0<v_2$ in each of their sides.

Given that $H_1(K)=Z_0(K)/B_1(K)=\mathbb Z\times\mathbb Z\times\mathbb Z/\mathbb Z\times\mathbb Z\times\{1\}=\mathbb Z$

and $\vert K\vert_d\approx S^1$ (where $\vert K\vert_d$ is the geometric realization of $K$ and it is defined as $\displaystyle\vert K\vert_d=\{\alpha:V_k\to[0,1]\vert\sum_{v\in V_k}\alpha (v)=1\ \&\ \alpha^{-1}(0,1]\in K \}$) then $H_1(S^1)=\mathbb Z.$

My doubt is in the definition of homeomorphism $\vert K\vert_d\approx S^1$, which should I consider?

I'd appreciate your help, thanks.

Answer 1

Ok, so let me use more formal notation. So $S=\{v_0, v_1, v_2\}$ is the underlying set of vertices and

$$K=\big\{\{v_0\}, \{v_1\}, \{v_2\}, \{v_0, v_1\}, \{v_1, v_2\}, \{v_0, v_2\}\big\}$$

is the simplicial complex. Now the geometric realization is given by

$$|K|=\big\{f:S\to[0,1]\ \big|\ \sum_{s\in S}f(s)=1\text{ and }f^{-1}(0,1]\in K\big\}$$

The topology on $|K|$ is inherited from $[0,1]^S$ which is compact. In our case $|K|$ is a subset of a three dimensional cube. And you can draw it to see that $|K|$ is a triangle.

Now we are ready to construct the homeomorphism. I will treat $S^1$ as the unit sphere in the complex numbers $\mathbb{C}$. Define

$$\Theta:S^1\to|K|$$

as follows: with $z\in S^1$ associate $n\in\{0,1,2\}$ such that

$$\frac{2}{3}\pi n \leq Arg(z) < \frac{2}{3}\pi (n+1)$$

In other words when $z$ is in one out of three equal arcs of $S^1$. Let $I(z)$ be this $n$. Then normalize $Arg(z)$ as follows:

$$N(z)=\bigg(Arg(z)- \frac{2}{3}\pi I(z)\bigg)\bigg/\bigg(\frac{2}{3}\pi (I(z)+1) - \frac{2}{3}\pi I(z)\bigg)$$

You can simplify this, but it doesn't really matter. The point is that $N(z)\in [0, 1)$ and $N(z)$ is a continuous bijection on each one of three arcs.

And now we are good to go. Define

$$\Theta(z)(v_n)=\begin{cases} 0 & \text{if }n=I(z) \\ N(z) & \text{if }n=(I(z)+2)\text{ mod }3 \\ 1-N(z) & \text{otherwise} \\ \end{cases}$$

I leave it as an exercise that this is well defined, continuous and bijective (and hence homeomorphism since $S^1$ is compact).

Now this looks complicated. That's because we have to deal with lots scaling and formal transformations. But the the idea behind it is very simple. Take the sphere $S^1$. Mark $3$ points on it. And then squeeze what is left so that it becomes a triangle.