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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.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.

A continuous map injective map $f:X\to Y$ that is a homeomorphism onto its image with the subspace topology. A continuous map injective map from a compact space into a Hausdorff space is a topological embedding.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

For $n\ge 1$, a topological $n$-manifold with boundary is a Hausdorff second-countable space locally homeomorphic to relatively open sets in the half-space ${\mathbb{H}}^n$. The openness here is in the subspace topology of ${\mathbb{H}}^n$, not necessarily in ${\mathbb{R}}^n$. Its intrinsic boundary $\partial M$ consists of points sent to $x_n=0$; the remaining points have ordinary coordinate charts with images open in Euclidean space. The intrinsic boundary is independent of the chosen chart.^[ref]

The closed interval $[0,1]$ illustrates the distinction: its interior points have open interval charts, while $0$ and $1$ have charts modeled on $[0,\epsilon )\subseteq {\mathbb{H}}^1$, using $x$ and $1-x$. Its intrinsic boundary is $\{0,1\}$. The circle has empty intrinsic boundary even though its boundary of a set as a subset of the Euclidean plane is the whole circle. A topological manifold without boundary is therefore a special case with $\partial M=\varnothing$, and a smooth atlas on the half-space charts gives a manifold with boundary in the smooth sense.

A topological subspace of a topological space $X$ is a subset $A\subseteq X$ equipped with the subspace topology inherited from $X$.

A topological tripod is three copies of the closed interval $[0,1]$ with their zero endpoints identified to a single junction $o$. Equivalently, embed three line segments with distinct directions in the Euclidean plane and use the subspace topology. It is a compact, connected metric space, and is Hausdorff, second countable, and paracompact. Every interior arm point has a coordinate chart to an open interval, and each outer endpoint has a half-interval chart as in a topological manifold with boundary. The junction has neither kind of chart.

Here is a precise local obstruction. Let $S$ be a small open neighbourhood of $o$ consisting of initial segments of the three arms, excluding their outer endpoints. Every connected open neighbourhood $N$ of $o$ contained in $S$ contains an initial segment on every arm, because $N$ is open at $o$. Its intersection with each arm has no gaps: if a point between $o$ and a point of $N$ were missing, cutting there would separate the part beyond it from the rest of $N$, contradicting connectedness. Thus $N\setminus \{o\}$ has exactly three connected components, one interval per arm.

Suppose a coordinate chart $h$ to the real line existed at $o$. Choose a small open interval $I$ about $h(o)$ lying inside $h(S\cap \mathrm{domain}(h))$, and restrict the chart to $N=h^{-1}(I)$. This is a connected open neighbourhood of $o$ contained in $S$. Removing $o$ leaves three connected components in $N$, while removing $h(o)$ leaves two in $I$. A homeomorphism preserves connected components, so this is impossible. This restriction step is essential: the count is proved for any sufficiently small connected chart neighborhood, not merely one chosen drawing.

A higher dimension of a manifold cannot repair the junction. For $n\ge 2$, repeat the argument with an open ball about $h(o)$: a punctured open ball in ${\mathbb{R}}^n$ is path-connected, whereas $N\setminus \{o\}$ still has three connected components. A zero-dimensional chart would make $o$ an isolated point, which it is not. A chart at a boundary point of a topological manifold with boundary similarly gives a punctured relative half-ball: one connected component when $n=1$, and a path-connected space when $n\ge 2$. Consequently the topological tripod fails the local definition even when intrinsic boundary is allowed.

The circle provides a useful warning about scale: deleting a point from the whole circle leaves a connected space, but deleting it from a sufficiently small connected arc leaves two connected components. Local chart tests use the latter. The circle and interval are not homeomorphic article uses deletion from the whole spaces for a different, global obstruction.

Consider the graph of a function $x\mapsto \sin (1/x)$, using the sine of the multiplicative inverse of $x$, together with a vertical closed interval:Give $T$ the subspace topology from the Euclidean plane. The symbol $\times$ denotes a Cartesian product. The right endpoint $(1,\sin 1)\approx (1,0.841471)$ belongs to $G$. The continuous map $x\mapsto (x,\sin (1/x))$ sends the connected real interval $(0,1]$ onto $G$. Hence $G$ is connected, and its closure $\overline{G}=T$ is connected by closure of a connected set. Every height in $[-1,1]$ is approached as $x$ tends to zero; for example,For $n=1$, these inputs are $p_1=2/(5\pi )\approx 0.127324$ and $q_1=2/(7\pi )\approx 0.090946$. For any other height $y\in [-1,1]$, choose $\theta \ge 1$ with $\sin \theta =y$; then the sequence $(1/(\theta +2\pi n),y)$ lies in $G$ and has limit of a sequence $(0,y)$.

Nevertheless, $T$ has exactly two path components, $G$ and $L$. Within $G$, join inputs $a,b\in (0,1]$ by the continuous path $t\mapsto ((1-t)a+tb,\sin (1/((1-t)a+tb)))$. Within $L$, use $t\mapsto (0,(1-t)c+td)$ for heights $c,d\in [-1,1]$.

The obstruction to a continuous path between $L$ and $G$ is persistent oscillation, not a visible gap. Proof outline: suppose such a path exists, directed from $L$ to $G$, and take its last time in $L$. Immediately afterwards its first coordinate $x(t)>0$ approaches zero and its second coordinate is $\sin (1/x(t))$. The intermediate value theorem forces the first coordinate to pass through inputs of both displayed types arbitrarily close to that departure time. The second coordinate therefore takes both values $1$ and $-1$ arbitrarily close to that time, contradicting continuity. ^[ref]

The sampled graph of a function illustrates the formulas; it does not prove connectedness or the failure of path connectedness. Taking a closure preserves connectedness, but this example shows that it can destroy path connectedness.

If $f:M\to N$ satisfies transversality of a map to a submanifold $S$, its nonempty preimage is an embedded submanifold of the displayed codimension. Choose a slice chart for an embedded submanifold around $f(p)$ in which $S$ is the zero set of the first $k$ coordinates. Their composition with $f$ has surjective derivative on its zero set, so the regular level set theorem gives a local embedded slice of dimension $\dim M-k$. These slices give the subspace topology and a compatible smooth atlas on the preimage. Its tangent space is

If $D=\mathrm{diag}(a_1,\dots ,a_{n+1})$ is invertible, $x\mapsto Dx$ sends the ellipsoid ${\sum }_i{a}_i^2{x}_i^2=1$ to the unit sphere. On $u=(v,s)\in S^n$, the two stereographic projections $v/(1-s)$ and $v/(1+s)$ omit opposite poles and have inverses $(2w/(1+|w{|}^2),\pm (|w{|}^2-1)/(1+|w{|}^2))$, with the sign chosen for the appropriate pole. Their overlap transition is $w\mapsto w/|w{|}^2$, smooth and involutive on ${\mathbb{R}}^n\setminus \{0\}$. Composing their inverses with $D^{-1}$ gives a two-chart smooth atlas of the ellipsoid. Its subspace topology is Hausdorff and has a second-countable space topology, so it is a smooth manifold. For $n=0$ the overlap is empty and each chart is a point.

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.