Past exam of the options II course of the University of Oxford / 2009 / E / E4

A subset is connected when its subspace topology admits no separation into two disjoint nonempty open sets. Equivalently, its only subsets that are both open and closed in that subspace are the empty set and the entire subset.

A separation of $f(X)$ would pull back under the continuous map to two disjoint nonempty open sets covering $X$: nonemptiness follows because their points belong to the image. This contradicts connectedness of $X$. Hence a continuous image of a connected space is connected.

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.

Let $C$ be any equivalence class. Since $A$ is both open and closed, $C\cap A$ is both open and closed in the subspace topology on $C$. Its connectedness makes $C\cap A$ either empty or all of $C$. Thus an equivalence class meeting $A$ lies wholly in $A$. Hence $A$ is saturated, a union of whole equivalence classes.

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.