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I am trying to solve the following topology problem:

In $\mathbb{R^2}$ consider the unitary square $\mathbb{Q}:=[0,1]^{2}$. Defined is an equivalence relation by: $(x,y)\sim (x', y')\Leftrightarrow \left\{\begin{matrix}(x,y)=(x',y') \\\left \{ x,x' \right \}=\left \{ 0,1 \right \} , y=y' \\ \left \{ y,y' \right \}=\left \{ 0,1 \right \}, x=x' \end{matrix}\right.$

Prove that $\mathbb{Q}/\sim $ is homeomorphic to $\mathbb{S^{1}} \times \mathbb{S^{1}}$ where $\mathbb{S^{1}}$ is the unitary circle given by $\left \{ x\in \mathbb{R^2} \mid \left | x \right |=1\right \}$

Thank you in advance!

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Hint: Do the case of $[0,1]$ with $x\sim x'\iff \{x,x'\} = \{0,1\}$ first. – Jason DeVito Nov 12 '12 at 18:33
@JasonDeVito Pedantic comment: $x\sim x'\iff \{x,x'\}=\{0,1\}\vee x=x'$ :-) – Mariano Suárez-Alvarez Nov 12 '12 at 20:05
Your relation $\sim$ is not an equivalence relation because it fails to be transitive: By the second case, $(0,1)\sim(1,1)$, by the third case $(1,1)\sim(1,0)$, but none of the cases gives us $(0,1)\sim(1,0)$. – Hagen von Eitzen Nov 12 '12 at 20:51
up vote 2 down vote accepted

We want to prove that $\mathbb{Q} /\sim$ is homeomorphic to $\mathbb{S}^1 \times \mathbb{S}^{1}$. That is, we want to show that there exists a bijective function $f: \left(\mathbb{Q} / \sim \right) \rightarrow \mathbb{S}^1 \times \mathbb{S}^1$ that is continuous and has $f^{-1}: \mathbb{S}^1 \times \mathbb{S}^1 \rightarrow \mathbb{Q} / \sim$ is continuous.

We can first start off by showing that for $X:=[0,1]$ and $x \sim x' \Leftrightarrow \{x,x'\}=\{0,1\}$, that $X^{*}=X / \sim$ is homeomorphic to $\mathbb{S}^1$. By the equivalence relation we have a partition of $X$ into the following sets: $$X = \{\{x\} \; | \; x \in (0,1)\} \coprod \{0,1\}$$

That is, we have an equivalence class for each element in the open interval $(0,1)$ and an equivalence class containing the elements $0$ and $1$. Define $X^{*}$ to be a quotient space of $X$ (under the equivalence class partition just mentioned) and let $p: X \rightarrow X^{*}$ be the surjective map that carries each point of $X$ to the element of $X^{*}$ containing it. We want to consider the quotient topology induced by $p$, where we have that $p$ is a quotient map and $X^{*}$ is a quotient space. We now want to construct a function $f: X^{*} \rightarrow \mathbb{S}^1$ which shows that $X^{*}$ and $\mathbb{S}^1$ are homeomorphic.

We now want to take points in the interval $X=[0,1]$ and map them to $\mathbb{S}^1$ such that the identified point $0 \sim 1$ on the interval is respected and the points in the open interval $(0,1)$ are mapped nicely to a point on $\mathbb{S}^1$.

With this motivation, we can define $f: X^{*} \rightarrow \mathbb{S}^1$ as, $$f(x)= \begin{cases} (\cos(2\pi x),\sin(2\pi x)) \;\;\;\;\; \text{if } x \notin \{0,1\}\\ (1,0) \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{if } x \in \{0,1\} \end{cases} $$

Clearly $f: X^{*} \rightarrow \mathbb{S}^1$ is a bijection as $(1,0)$ is mapped to the equivalence class $\{0,1\}$ and all other points on the interval defined as $(\cos(2\pi x),\sin(2\pi x))$ are mapped to the equivalence class $[x]$, and vice versa.

We first show that $f: X \rightarrow \mathbb{S}^1$ is continuous. Well, let $U$ be an arbitrary open set in the subspace topology on $\mathbb{S}^1$. We want to have that $f^{-1}(U)$ is open in the quotient topology on $X^{*}$. If $(1,0) \notin U$ then $f^{-1}(U) \subseteq (0,1)$ which means that there is an equivalence class corresponding to each element $x \in f^{-1}(U)$ with the union of the equivalence classes as a subset of $(0,1)$, which is open in $X$ and thus $f^{-1}(U)$ is open in the quotient topology on $X^{*}$. If $(1,0) \in U$, then $f^{-1}(U)$ is the union of the equivalence class $\{0,1\}$ and some number of equivalence classes corresponding to each single element on $\mathbb{S}^1$ which was not equal to $(1,0)$ contained in $U$. For this reason, $f^{-1}(U)$ is open in the quotient topology on $X^{*}$. As a summary, we have that $$f^{-1}(U) = \begin{cases} \bigcup [x], x \in (0,1) \; \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \text{if } (1,0) \notin U \\ \left(\bigcup [x]\right) \cup \{0,1\}, x \in (0,1) \;\;\;\;\;\;\;\;\; \text{if } (1,0) \in U \end{cases} $$

We now show that $f^{-1} : \mathbb{S}^1 \rightarrow X^{*}$ is continuous. So, let $V$ be an arbitrary open set in the quotient topology on $X^{*}$. We want to show that $(f^{-1})^{-1}(V) = f(V)$ is open in the subspace topology on $\mathbb{S}^1$. It is straightforward to see that you simply work in the opposite direction as I just explained for showing $f$ is continuous.

Therefore, we have constructed a homeomorphism $f$ between $[0,1]/\sim$ and $\mathbb{S}^1$. Given this outline, you should be able to generalize to look at the case of showing $[0,1]^2 /\sim$ is homeomorphic to $\mathbb{S}^1 \times \mathbb{S}^1$. Let me know if you run into troubles for that case and I can fill in the details, but it would be best for yourself.

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Dear Samuel Thank you very much for the help! I'll try to make the details of the proof by myself. :) – Lullaby Nov 12 '12 at 21:12

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