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Start with the interval $[0, 2]$ and glue the points $0, 1$ and $2$. Describe the equivalence relation $\mathcal{R}$ encoding this gluing and let $X = [0, 2]/\mathcal{R}$. Describe an embedding of $X$ in $\mathbb R^2$.

Equivalence relation is \begin{align*} x \mathcal{R}y \Leftrightarrow \begin{cases} x=y \\ x, y \in \{0, 1, 2\}. \end{cases} \end{align*}

It is easy to see that $X$ is homeomorphic to the double circle. But how to describe the embedding?

PaulichenT
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    Can you embed a "double circle" in $\mathbb R^2$? – Lee Mosher Oct 22 '20 at 14:06
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    If so, you should be able to write down a quotient map from $[0,2]$ to that embedding, whose point pre-images are the equivalence classes of $\mathcal R$. – Lee Mosher Oct 22 '20 at 14:07
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    The quotient map is this? $f: [0, 2] \rightarrow \mathbb R^2$, \begin{align} f(t):=\begin{cases} (-1 + \cos(2\pi t), \sin(2 \pi t)) \ \mbox{if} \ t\in [0, 1)\ (1 -\cos(2\pi t), \sin(2 \pi t)) \ \mbox{if} \ t \in (1, 2] \ (0, 0) \ \mbox{if} \ t=1 \end{cases} \end{align} – PaulichenT Oct 22 '20 at 14:54
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    No, this is the quotient map followed by the embedding, which is thus described by this formula. – Berci Oct 22 '20 at 15:33
  • Right. What should i do to find the quotient map described by Lee? – PaulichenT Oct 22 '20 at 15:59
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    Note that $\mathbb R^2$ is Hausdorff and $[0,2]$ is compact, so it suffices to find a continuous injection. –  Oct 22 '20 at 17:13
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    @Berci: This map $f$ is indeed a quotient map as asked for in my first comment. It is not, of course, the quotient map from $[0,2]$ to $X$, but see my answer. – Lee Mosher Oct 22 '20 at 17:16

2 Answers2

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In the comments, you've written out a formula for a quotient map $f$ from $[0,2]$ to the double circle, whose point pre-images are identical to the equivalence classes of $\mathcal R$, i.e. for all $s,t \in [0,2]$ we have $s \mathcal R t$ if and only if $f(s)=f(t)$ (and $f$ is a quotient map, meaning, a surjective map such that a subset $A \subset $ (double circle) is open in the double circle if and only if $f^{-1}(A) \subset [0,2]$ is open in $[0,2]$).

It follows that the quotient space $X = [0,1] / \mathcal R$ is homeomorphic to the double circle. In fact even more follows: the formula $$F[x]=f(x) $$ is a well-defined formula for a homeomorphism $$F : X \to \text{(double circle)} $$ where the input to $F$ is an equivalence class $[x]$ represented by a point $x \in [0,2]$. These facts are all immediate consequences of the universality theorem for quotient maps.

Lee Mosher
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The map $f : [0,2] \to \mathbb R^2 = \mathbb C, f(t) = - 1 + e^{2\pi i t}$ for $t \in [0,1]$, $f(t) = 1 + e^{2\pi i t}$ for $t \in [1,2]$ (which is the same as in your comment) is continuous and its image $D = f([0,2])$ is an embedded double circle. Since $[0,2]$ is compact and $D$ is Hausdorff the map $q : [0,2] \stackrel{f}{\to} D$ is a closed map, therefore a quotient map. It is easy to see that $q(x) = q(y)$ iff $x \mathcal R y$. This means that there exists a unique homeomorphims $h : [0,2]/\mathcal R \to D$ such that $h \circ p = q$. See for example When is a space homeomorphic to a quotient space?

Paul Frost
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