Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / E / E4

A subset is connected if it has no decomposition into two disjoint nonempty sets both relatively open in it. Equivalently its only relatively clopen subsets are itself and the empty set. Connectedness here refers to the subset's induced topology.

Suppose $f(X)=U\cup V$ were a separation into two disjoint nonempty relatively open sets. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$, and they are disjoint, nonempty and cover $X$. This contradicts connectedness of $X$. Hence $\boxed{f(X)\text{ is connected}}$.

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.

Let $A$ be clopen and $C$ an equivalence class. The intersections $C\cap A$ and $C\setminus A$ are complementary relatively open subsets of $C$. Connectedness of $C$ means one must be empty. Thus every equivalence class is entirely in $A$ or entirely outside it, proving $\boxed{q^{-1}(q(A))=A}$. This is precisely saturation.

If $X$ is connected, its continuous surjective image under $q$ is connected by (a)(ii). Conversely suppose $X/R$ is connected and $A\subseteq X$ is clopen. By (i) both $A$ and its complement are saturated. The quotient definition then makes $q(A)$ open, and its complement $q(X\setminus A)$ open too. Thus $q(A)$ is clopen in the connected quotient, so is empty or the whole quotient. Pulling back gives $A=\varnothing$ or $X$. By the clopen criterion $X$ is connected. Hence $\boxed{X\text{ connected}⟺X/R\text{ connected}}$.

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 $\boxed{X\text{ is connected}}$.