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A quotient topological space is a quotient set equipped with the quotient topology. For a concrete gluing, give the closed interval $I=[0,1]$ its usual subspace topology, identify $0\sim 1$, and leave every interior point in its own singleton equivalence class. The resulting quotient space is a circle: the two endpoints become one point, while interior points remain distinct.

The exact continuous map to the unit circle in the plane isHere sine and cosine give $p(0)=p(1)=(1,0)$, and $p(s)=p(t)$ holds exactly when $s=t$ or $\{s,t\}=\{0,1\}$. Thus $p$ induces a bijection $\bar{p}:I/\sim \to S^1$. ^[ref] The universal property of the quotient topology makes $\bar{p}$ continuous. The quotient of a compact space is compact, and the unit circle is Hausdorff, so the compact-to-Hausdorff continuous bijection theorem makes $\bar{p}$ a homeomorphism. This verifies the topology, not merely the point identifications.

To inspect an open neighbourhood of the seam, choose $0<\epsilon <1/2$ and let $U_{\epsilon }=p([0,\epsilon )\cup (1-\epsilon ,1])$. Its full preimage isBoth pieces are open in the subspace topology on $I$, so $U_{\epsilon }$ is open in the circle. In contrast, $A=[0,\epsilon )$ is open in $I$, butis not open in $I$: it contains $1$ without any neighbourhood of $1$. Hence $p(A)$ is not open in the circle. A quotient map can therefore fail to be an open map.

For a continuous function $f:I\to \mathbb{R}$, the only gluing condition is $f(0)=f(1)$. For example, $f(t)=\cos (2\pi t)$ defines a continuous function on the circle, whereas $f(t)=t$ does not. This lets us use an interval as a coordinate for a periodic function while making the seam invisible. More general quotient spaces encode glued geometric pieces in the same way: describe continuous maps before gluing and check that their values agree on every equivalence class.

The Hausdorff property need not survive a quotient topology. In the non-Hausdorff quotient of the real line by rational translation, $x\sim y$ means $x-y\in \mathbb{Q}$. A nonempty open preimage $W$ must contain all translates by rational numbers of any open interval inside it. The density of the rational numbers then forces $W=\mathbb{R}$: for any $y$, some rational number $r$ has $y-r$ in that interval. Thus this quotient space has the indiscrete topology. The irrationality of the square root of two gives distinct equivalence classes for $0$ and $\sqrt{2}$, so this quotient space is not Hausdorff. Unlike the endpoint-gluing circle, this gluing identifies a dense subset with each point.