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Let $W=(\mathbb{N}\times S^1)/A$, and choose in each circle the point opposite $o$:In the subspace topology from the Euclidean plane, the sequence $z_n$ converges to $o$, which does not belong to $B$. Thus $B$ is not closed in $X$.

On the other hand, let $D={\bar{f}}^{-1}(B)\subseteq W$. With the quotient map $q$ from part (i),This is closed in $\mathbb{N}\times S^1$: its complement is the open set $\mathbb{N}\times (S^1\setminus \{(0,1)\})$. The definition of the quotient topology, applied to complements, says that $D$ is closed in $W$. A homeomorphism would carry this closed set to the closed set $B$, a contradiction. Therefore $\bar{f}$ is not a homeomorphism.

The obstruction occurs at the common point. The open set $W\setminus D$ contains the collapsed point, but its image $X\setminus B$ contains no neighbourhood of $o$, since $z_n\to o$. The wedge sum topology allows a separate small neighbourhood on each circle; a neighbourhood of $o$ in the Hawaiian earring must contain all sufficiently small whole circles.

Use positive indices $n\ge 1$ and write $p=(0,-1)\in S^1$. SetFor fixed $n$, this function is a homeomorphism from the unit circle to $C_n$: it scales its radius to $1/n$ and translates its centre to $(0,1/n)$. Thus it takes values in the Hawaiian earring $X$ and is surjective onto $X$. It is continuous on each open slice $\{n\}\times S^1$; since these slices form an open cover of $\mathbb{N}\times S^1$, $f$ is continuous on the whole product topology.

Distinct circles $C_n,C_m$ meet only at $o=(0,0)$. To see this directly, their equations areFor $m\ne n$, subtraction gives $y=0$, then $x=0$. Each slice contains exactly one preimage of $o$, namely $(n,p)$. Therefore the only nonsingleton fibre of $f$ is $A$.

Let $q:\mathbb{N}\times S^1\to (\mathbb{N}\times S^1)/A$ be the quotient map collapsing $A$. The universal property of the quotient topology supplies a unique continuous map $\bar{f}$ with $f=\bar{f}\circ q$. The fibre calculation proves injectivity, and surjectivity follows from that of $f$. Hence $\bar{f}$ is the required continuous bijection.

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is finite and cannot be all of $\mathbb{R}$.

Suppose the quotient space had distinct points with disjoint open neighbourhoods $U,V$. Since the quotient map $p$ is surjective, both preimages $p^{-1}(U),p^{-1}(V)$ would be nonempty. Since $p$ is continuous, these would be disjoint open subsets of $\mathbb{R}$ with its cofinite topology, contradicting the observation above. Thus any such quotient space with at least two points is not Hausdorff.

False. Give $X=\mathbb{R}$ its usual topology, identify all points of $(0,1)$ with one another, and leave every other point in a singleton equivalence class. Take the closed set $K=\{1/2\}$.

Its image $p(K)$ is a singleton in the quotient space, but its full preimage, also called the saturation of a subset, isThis is not closed in $\mathbb{R}$, since $0$ belongs to its closure. In the quotient topology, a subset is closed exactly when its preimage is closed, by applying the defining open-set rule to complements. Thus $p(K)$ is not closed, and this quotient map is not a closed map.

If $q:E\to B$ is a quotient map, then $q\times {\mathrm{id}}_{[0,1]}$ is a quotient map. More generally the second factor may be any locally compact space that is Hausdorff. To prove the assertion, let $W\subseteq B\times [0,1]$ have open preimage $O$, and take $(b,t)\in W$. Choose $e\in q^{-1}(b)$ and a compact neighbourhood $K$ of $t$ with $\{e\}\times K\subseteq O$. The set $E_K=\{u:\{u\}\times K\subseteq O\}$ is open by compactness of $K$, and is saturated under $q$ because membership in $O$ depends only on $q(u)$ and the second coordinate. Hence $q(E_K)$ is open by the quotient topology. The rectangle $q(E_K)\times \mathrm{int}K$ is a neighbourhood of $(b,t)$ contained in $W$, proving that $W$ is open. This justifies parameterized homotopies defined on quotient spaces.

For a bounded linear operator into a Banach space, compactness is equivalent to the displayed condition, where ${\pi }_Y$ is the quotient map by a finite-dimensional subspace of the target. A finite net for the image of the unit ball gives such a subspace by taking its span. Conversely, if the image lies within $\epsilon$ of a finite-dimensional subspace, choose approximants there; they remain bounded because the original image is bounded. A finite net for these approximants gives a slightly larger finite net for the image. This proves total boundedness and hence compactness. This condition does not require choosing the approximants by a linear map, and therefore does not assume an approximation property for the Banach space.

The quotient norm measures the smallest norm among representatives of a coset. Its infimum need not be attained. Nevertheless, for every $\epsilon >0$, an element $z\in X/Y$ has a representative $v\in X$ with $\Vert v\Vert <\Vert z\Vert +\epsilon$. The quotient map is contractive.

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.

The quotient topology turns an instruction to glue points together into a precise topological space. Start with a topological space $(X,{\tau }_X)$ and an equivalence relation $\sim$. The quotient set $X/\sim$ consists of equivalence classes, and its canonical projection of a quotient set isThe quotient topology is defined by the rule using preimages of open sets:Thus a subset of the quotient space is open if and only if its entire preimage is open in $X$. Each fiber $q^{-1}(\{[x]\})$ is the whole equivalence class $[x]$, so the preimage includes every point glued to any of its members.^[ref]

This is the finest topology making $q$ a continuous map: every topology with that property can contain only subsets whose preimages are open. It really is a topology, because preimages preserve the empty set, the whole set, arbitrary unions, and finite intersections. The gluing example below shows why checking a complete preimage matters, especially at the seam.

The quotient topology also makes continuous maps out of a glued space easy to construct. A continuous map $f:X\to Z$ descends to a unique continuous map $\bar{f}:X/\sim \to Z$ exactly when it is constant on every fiber of $q$, meaning $x\sim y\Rightarrow f(x)=f(y)$. Define $\bar{f}([x])=f(x)$; the equivalence class representative cannot affect the answer. For every open set $V\subseteq Z$,The right side is open, so the quotient topology makes ${\bar{f}}^{-1}(V)$ open. Conversely, if $f=\bar{f}\circ q$ for a continuous map $\bar{f}$, then composition of continuous functions makes $f$ continuous, and equal equivalence classes force equal values. This is the universal property of the quotient topology. ^[ref]

A surjective continuous map is not automatically a quotient map. On the same two-point set $D=\{a,b\}$, take the identity map from the discrete topology to the indiscrete topology. It is continuous and surjective, but $\{a\}$ is not open in the target even though its preimage is open in the domain. The quotient topology for this identity map would be the discrete topology. Thus continuity supplies only one direction of the rule using preimages of open sets; a quotient map requires both.

A quotient space of a space carrying the cofinite topology has a topology contained in the cofinite topology on its underlying quotient set. If $q:X\to Y$ is the quotient map and $U$ is a nonempty open subset of $Y$, its preimage has finite complement. Since $q$ is surjective, the complement of $U$ is the image of that finite complement, hence is finite. Equality need not hold: partition $\mathbb{Z}$ into its even and odd integers. Both classes are infinite, so their two-point quotient space has the indiscrete topology, whereas the cofinite topology on a finite set is the discrete topology.

For the canonical projection of a quotient set $p:X\to X/\sim$, the saturation of $A\subseteq X$ is the union of all equivalence classes meeting $A$, namely $p^{-1}(p(A))$. A subset is called saturated when it equals its saturation. The quotient topology says that $p(A)$ is closed exactly when the saturation of $A$ is closed, and similarly for openness. A closed set need not have closed saturation: collapsing $(0,1)\subset \mathbb{R}$ to one point makes the saturation of $\{1/2\}$ equal to $(0,1)$. Hence a quotient map need not be a closed map.