# Identifying This Quotient Topology

Let $$\sim$$ relation be on $$[0,1]$$ as following.

$$x \sim y \Leftrightarrow x=y$$ or $$x,y \in \{0,1\}$$

Show that $$[0,1] / \sim$$ is homeomorphic to $$S=\{x\in \mathbb R^2 : \|x\|=1 \} \subseteq \mathbb R^2$$

I need to concrete this quotient set’s elements. What are equivalence classes and what do this set’s elements look like? Also I am not sure on whether I have shown that it is an equivalence relation truly. Showing only one of three conditions of equivalence relation will provide me verifying myself.

It is easy I know but I am confused on this. Thanks for any help

For the homeomorphism part, identify $$\mathbb{R}^2$$ with the complex plane.
A natural choice for a homeomorphism is $$f:[x]\mapsto e^{2\pi ix}$$. This function is surjective because the complex logarithm inverts it on it's image. The injectivity follows from $$f([x])=f([y]) \Rightarrow x-y\in\mathbb{Z} \Leftrightarrow x,y \in \left\{0,1\right\}\Leftrightarrow [x]=[y]$$.
$$f$$ is continuous, because the complex exponential function is holomorphic. The continuity of $$f^{-1}$$ follows from the complex inverse function theorem.

With kindest regards,
soucerer

To verify that $$\sim$$ is an equivalence relation, you have to check reflexivity, symmetry and transitivity. Note that $$x=x$$ for all $$x\in [0,1]$$ and hence $$\sim$$ is reflexive. Also not that "$$x=y$$ or $$x,y\in\{0,1\}$$" is equivalent to "$$y=x$$ or $$y,x\in\{0,1\}$$" and hence $$\sim$$ is symmetric. For transitivity you'll have to do a case distinction and I'll leave that to you.

Given that $$\sim$$ is indeed an equivalence relation, for $$x\in[0,1]$$ let us denote by $$\overline x \in [0,1]/{\sim}$$ the equivalence class of $$x$$ with respect to $$\sim$$. For $$x\notin\{0,1\}$$ we then have singletons $$\overline x = \{x\}$$ and furthermore $$\overline 0 = \overline 1 = \{0,1\}$$. Hence we may write $$[0,1]/{\sim} = \left.\bigg\{ \,\{x\}\,\middle|\, x\in (0,1)\,\right\} \cup \left.\bigg\{ \{0,1\} \right\}.$$

I will answer in multiple bite size parts (assuming the relation is actually an equivalence relation, proven in another answer on this post):

• Explicitly find the elements of $$A_1 = [0,1]/ \sim$$

• Find a homeomorphism from $$A_1$$ to $$A_2 = [0,1 )$$

• Find a homeomorphism from $$A_2$$ to $$A _3=[0,2\pi )$$

• Find a homeomorphism from $$A_3$$ to the unit circle $$S$$

The equivalence relation states that each $$x$$ in $$(0,1)$$ is equivalent to only itself, ie $$[x]=x$$. And that $$0$$ is equivalent to $$1$$ (so you can choose one of them to be the representative. I choose to denote this by $$ =0=$$). So we have that $$A_1 = (0,1) \cup $$ Now we can define the function
$$f : A_1 \to A_2, \qquad f ([x])= x$$ It is very easily injective and surjective. Also $$f$$ and its inverse are continuous because the image of open sets are open in both directions. So $$f$$ is homeomorphic. Now define the function
$$g : A_2 \to A_3, \qquad g (x)=2 \pi x$$ It is very easily injective and surjective. Also $$g$$ and its inverse are continuous because the image of open sets are open in both directions. So $$g$$ is homeomorphic. Finally let us define the function $$h : A_3 \to S, \qquad h (x)= (\cos x , \sin x)$$ It is very easily injective (since the trigomometric functions do not cycle in the range $$[0,2\pi)$$) and surjective (by definition of trigonometric function). Also $$h$$ and its inverse are continuous because the image of open sets are open in both directions since the trigonometric functions are continuous. So $$h$$ is homeomorphic

Finally we have that $$h \circ g \circ f$$ is a homeomorpjism from your quotient set to $$S$$