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A compact space is a topological space in which every open cover has a finite subcover. A Hausdorff space is one in which every two distinct points have disjoint open neighbourhoods. A homeomorphism is a bijection that is continuous and whose inverse is continuous.

For an equivalence relation $R$ on $X$, let $X/R$ be the set of equivalence classes and letThe quotient topology declares $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. It follows directly from the definition that $q$ is continuous.

Suppose that the continuous map $f:X\to Y$ is constant on equivalence classes. The only possible factorisation iswhich is well-defined by the hypothesis and satisfies $F\circ q=f$. For every open $V\subseteq Y$,is open in $X$. The definition of the quotient topology therefore makes $F^{-1}(V)$ open, so $F$ is continuous. This proves the universal property of the quotient topology.

Let $p:X\to X/R$ send a point to its equivalence class. The quotient topology isIt is the finest topology on the quotient making $p$ continuous.

Take distinct classes $p(x)\ne p(y)$. The points are not related by $R$, so by the defining negation there is a Hausdorff space $Z$ and a continuous map $f:X\to Z$ with $f(x)\ne f(y)$. Choose disjoint open neighborhoods $U$ and $V$ of those two values. Their inverse images $A=f^{-1}(U)$ and $C=f^{-1}(V)$ are disjoint open subsets of $X$ and are saturated by part (ii).

Hence $p^{-1}(p(A))=A$ and $p^{-1}(p(C))=C$, making $p(A)$ and $p(C)$ open in the quotient topology. They contain the chosen classes and are disjoint: an equivalence class meeting both would contradict saturation and $A\cap C=\varnothing$. Thus $X/R$ is Hausdorff. This quotient is the Hausdorff reflection by continuous maps; every continuous map from $X$ to a Hausdorff space factors uniquely through it.

For the quotient map $q:X\to X/\sim$, the quotient topology is defined byTake the Hausdorff space $X=\mathbb{R}$ and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, neither of which is open in $\mathbb{R}$, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.

Use the equivalence relation of belonging to the same fibre:The map $\bar{f}:X/\mathcal{R}\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the quotient topology: for open $V\subseteq Y$, $q^{-1}({\bar{f}}^{-1}(V))=f^{-1}(V)$ is open.

The quotient space is compact as the continuous image of compact $X$, and $Y$ is Hausdorff. The compact-to-Hausdorff continuous bijection theorem makes $\bar{f}$ a homeomorphism. Indeed, images of closed subsets of a compact space are compact and hence closed in a Hausdorff space, so the inverse is continuous.

Let $q:X\to X/\mathcal{R}$ send each point to its equivalence class. The quotient topology isIt is the finest topology on the quotient set making the canonical surjection $q$ continuous. The preimage test, rather than a choice of representatives, defines its open sets.

Use the equivalence relation of belonging to the same fibre:The map $\bar{f}:X/\mathcal{R}\to Y$, $[x]\mapsto f(x)$, is well defined, injective by the definition of the relation, and surjective because $f$ is surjective. It is continuous by the quotient topology: for open $V\subseteq Y$, $q^{-1}({\bar{f}}^{-1}(V))=f^{-1}(V)$ is open.

The quotient space is compact as the continuous image of compact $X$, and $Y$ is Hausdorff. The compact-to-Hausdorff continuous bijection theorem makes $\bar{f}$ a homeomorphism. Indeed, images of closed subsets of a compact space are compact and hence closed in a Hausdorff space, so the inverse is continuous.

Let $q:X\to X/\mathcal{R}$ send each point to its equivalence class. The quotient topology isIt is the finest topology on the quotient set making the canonical surjection $q$ continuous. The preimage test, rather than a choice of representatives, defines its open sets.

For the surjection $p$ to the set of equivalence classes, define a subset $V$ of the quotient to be open exactly when $p^{-1}(V)$ is open in $X$. Inverse images preserve the empty and full sets, arbitrary unions and finite intersections. Applying the topology axioms in $X$ proves this is a topology on the quotient, and it makes $p$ continuous.

The map need not be open. On $X=\mathbb{R}$, identify zero with one and leave every other point in its own class. The set $U=(-1/4,1/4)$ is open, but its saturation iswhich is not open in $\mathbb{R}$. Thus $p(U)$ is not open in the quotient topology. Continuity of a quotient map does not imply openness.

The underlying set of the quotient space is the set of equivalence classes $X/R$. With quotient map $q(x)=[x]$, its quotient topology is defined by declaring $U\subseteq X/R$ open exactly when $q^{-1}(U)$ is open in $X$. This makes $q$ a continuous surjection and is the finest topology with that property.

If $X$ is connected, its continuous quotient image is connected by part (a). Conversely, suppose $X$ is disconnected. It has a nonempty proper subset $A$ both open and closed. Part (i) makes $A$ and its complement saturated, so $q^{-1}(q(A))=A$ and $q^{-1}(q(X\setminus A))=X\setminus A$. The quotient topology therefore makes these disjoint nonempty images open, and they cover $X/R$. Thus the quotient is disconnected. This proves the two spaces are connected simultaneously.

The difference map $(z,w)\mapsto z-w$ is continuous, and $X$ is its preimage of the open set $\mathbb{C}\setminus \{0\}$. Thus $X$ is open in the product.

The class of $(z,w)$ is $C_z=\{z\}\times (\mathbb{C}\setminus \{z\})$. A punctured plane is path-connected: two points can be joined by a polygonal path avoiding the deleted point. Consequently every class is connected.

The projection $\pi$ is onto, since $(z,z+1)\in X$ for every $z$. It is continuous as a restricted coordinate projection. If $U$ is open in $X$, it is open in the product because $X$ itself is open. For a point $(z,w)\in U$, the product-topology basis gives open sets $V,W$ with $(z,w)\in V\times W\subseteq U$. Then $V\subseteq \pi (U)$. Hence $\pi (U)$ is open.

The map is constant exactly on the equivalence classes, so it induces a bijection $\overline{\pi }:X/R\to \mathbb{C}$. It is continuous by the defining property of the quotient topology. It is also open: for quotient-open $V$, $\overline{\pi }(V)=\pi (q^{-1}(V))$ is open. ThusThe plane is connected, and the classes are connected, so part (b) now yields $X$ is connected.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

The quotient topology is defined byPreimages commute with arbitrary unions and finite intersections and take the empty and whole sets to the corresponding sets, so this definition indeed gives a topology.

For the natural projection $p:X\to X/R$, the quotient topology declares $U\subseteq X/R$ open exactly when $p^{-1}(U)$ is open in $X$. It is the finest topology on the quotient making $p$ continuous.

Take distinct classes $p(x)\ne p(y)$. The failure of $x\mathcal{R}y$ supplies a Hausdorff space $Y$ and a continuous map $g:X\to Y$ with $g(x)\ne g(y)$. Choose disjoint open neighborhoods $V,W$ of these two images. Their inverse images $A=g^{-1}(V)$ and $B=g^{-1}(W)$ are disjoint open subsets of $X$, and part (ii) says both are saturated.

The quotient topology therefore makes $p(A)$ and $p(B)$ open, since their inverse images are precisely $A,B$. They are disjoint as well: an equivalence class meeting both would, by saturation, be contained in both. They contain $p(x)$ and $p(y)$ respectively. Thus $X/\mathcal{R}$ is Hausdorff. This construction is the Hausdorff reflection by continuous maps.

For the quotient map $q:X\to X/\sim$, the quotient topology is defined byTake the Hausdorff space $X=\mathbb{R}$ and identify two numbers precisely when they are both rational or both irrational. The quotient has two points. The inverse images of its singleton subsets are $\mathbb{Q}$ and $\mathbb{R}\setminus \mathbb{Q}$, neither of which is open in $\mathbb{R}$, since each is dense and has dense complement. Hence the quotient has the indiscrete topology and is not Hausdorff.

For the canonical surjection $p:X\to X/\sim$, define a subset $W$ of $X/\sim$ to be open exactly when $p^{-1}(W)$ is open in $X$. This is the quotient topology. The empty set and the whole quotient have open inverse images. Moreover inverse images commute with arbitrary unions and finite intersections, so the corresponding axioms of a topology follow directly from those on $X$.

The map $p$ need not be an open map. For example, on $X=\mathbb{R}$ identify every point of $[0,1]$ to one point and leave all other classes singletons. The set $U=(-1/2,1/2)$ is open, butis not open. The definition of the quotient topology therefore shows that $p(U)$ is not open. The obstruction is that saturating an open set by equivalence classes need not preserve openness.

The set $X/R$ consists of the equivalence classes, with canonical surjection $q:x\mapsto [x]$. The quotient topology declares $V\subseteq X/R$ open exactly when $q^{-1}(V)$ is open in $X$. This makes $q$ continuous and is the finest topology on the quotient set doing so.

The difference map $(z,w)\mapsto z-w$ is continuous and $X$ is its inverse image of the open set $\mathbb{C}\setminus \{0\}$, so $X$ is open in ${\mathbb{C}}^2$. For fixed $z$, the equivalence class is $\{z\}\times (\mathbb{C}\setminus \{z\})$, homeomorphic to a punctured plane. It is path-connected: two points can be joined by a polygonal path avoiding the missing point. Thus all equivalence classes are connected.

The projection $\pi$ is continuous and onto, since $(z,z+1)\in X$. If $U$ is open in $X$, it is open in ${\mathbb{C}}^2$ because $X$ is open. Product-coordinate projection is open: around any $(z,w)\in U$ there is a product of open disks contained in $U$, whose first disk is a neighborhood inside $\pi (U)$. Hence $\pi (U)$ is open.

Its fibers are precisely the equivalence classes, so it induces a bijection $\bar{\pi }:X/R\to \mathbb{C}$. Continuity follows from the quotient topology and $\pi =\bar{\pi }\circ q$. If $V$ is open in $X/R$, then $q^{-1}(V)$ is open in $X$ and $\bar{\pi }(V)=\pi (q^{-1}(V))$ is open. Thus the inverse is continuous too, proving $\boxed{X/R\cong \mathbb{C}}$ as topological spaces. The plane is connected, so the connected-fiber equivalence in (b)(ii) gives $\boxed{X\text{ is connected}}$.