$h:\mathbb{R}_{/\sim}\rightarrow \mathbb{R}^2$: A Bijection from a Quotient Space to the Unit Circle (Geometrically Considered)

Question

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NOTE: This is not a duplicate.

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Define a relation $\sim$ on $\mathbb{R}$ as follows: for any $a,b \in \mathbb{R}$, $$a\sim b \iff a-b\in \mathbb{Z}.$$

Let $S=\mathbb{R}_{/\sim}$. That is, $S$ is the set of equivalence classes of elements of $\mathbb{R}$ under the equivalence relation $\sim$.

Let $C=\{(x,y)\in\mathbb{R^2}:x^2+y^2=1\}$, and define $h:S\rightarrow C$ by $$h([t])=(\cos(2\pi t),\sin(2\pi t)).$$

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TASK: Prove that $h$ is a bijection.

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I know that I have to show $$h([t])=h([t'])\implies [t]=[t'],$$ and $$\forall x\in C \exists x'\in S : h([x'])=x,$$ but I'm not really sure how to go about this.

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Further question: What is the geometric interpretation of this?

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WORK: I get that $\mathbb{R}_{/\sim}=\{y\in\mathbb{R} : y-x\in \mathbb{Z}\}$, right?

Answer 1

Hint

• $h$ is injective:

  $h([t]=h([t'])\Rightarrow (\cos(2\pi t),\sin(2\pi t))=(\cos(2\pi t'),\sin(2\pi t'))\Rightarrow e^{2\pi it}=e^{2\pi it'}$ so $t-t'\in \mathbb Z$ and then $[t]=[t']$.

• $h$ is surjective:

Let $(\cos(2\pi t),\sin(2\pi t))\in C$ then take $[t]\in S$ and we have $h([t])=(\cos(2\pi t),\sin(2\pi t))$

Geometrically $C$ is the unit circle and we have $h(S)=C$ with $h$ is bijective so we can identify geometrically $S$ to the unit circle.

Answer 2

Algebraically, $S$ is the quotient group $\mathbb{R}/\mathbb{Z}$, and the map $\mathbb{R}/\mathbb{Z} \to C$ is a group isomorphism. To see this, note that the map $\mathbb{R} \to C$, $t \to e^{2\pi i t}$ is a group homomorphism (the circle in the complex plane is a group under multiplication), and the kernel of this map is precisely $\mathbb{Z}$. Hence, by the first isomorphism theorem $\mathbb{R}/\mathbb{Z} \cong C$.

Topologically, the resulting quotient space $\mathbb{R}/\sim$ is homeomorphic to $C$ via the same map.