TITLE: How does the quotient $\mathbb{R}/\mathbb{Z}$ become the circle $S^1$?
QUESTION [27 upvotes]: I came to know that the circle $S^1$ is actually the quotient space $\mathbb{R}/\mathbb{Z}$. But I don't understand how. To my knowledge elements of the quotient space $X/Y$ are of the form $xY$, i.e. the cosets. Right? But how $\mathbb{R}/\mathbb{Z}$ becomes a circle! Can anyone explain?

REPLY [11 votes]: It suffices to show the map $p:\mathbb{R}\to S^1$ defined by $p(x)=e^{2\pi ix}$ is a quotient map. Then you can apply the following theorem (see  Munkres' Topology Theorem 22.1) to conclude that $\mathbb{R}/\mathbb{Z}$ and $S^1$ are homeomorphic.

Theorem. Let $f:X\to Y$ be a quotient map. Let $\sim$ be a equivalence relation on $X$ by declaring $x\sim x'$ iff $f(x)=f(x')$. Let $X^*$ be the collection of equivalence class of $X$ (i.e., $X^*=\{[x]:x\in X\}$). We give $X^*$ the quotient topology induced by the map $p:X\to X^*$ that sends each point of $X$ to the equivalence class containing it (i.e., $p(x)=[x]$). Then the map $f$ induces a homeomorphism $\widetilde{f}:X^*\to Y$ characterized by $f=\widetilde{f}\circ p$.