Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / E

A topological space is compact if every open cover has a finite subcover. It is connected if it cannot be written as the union of two disjoint nonempty open subsets. It is Hausdorff if every two distinct points have disjoint open neighborhoods. These definitions apply to the space itself and, with the subspace topology, to its subsets.

True. If open sets cover $Y$, their inverse images under the continuous surjection $f$ cover $X$. Compactness of $X$ gives a finite subcover; surjectivity then makes the corresponding finite family cover $Y$. This proves the compactness of a continuous image of a compact space.

True. A separation of $Y$ into two disjoint nonempty open subsets would pull back under the continuous surjection $f$ to a separation of $X$. The inverse images are nonempty by surjectivity, contradicting connectedness of $X$. Thus a continuous image of a connected space is connected.

True. The Hausdorff assumption makes $\{y\}$ closed in $Y$. Its inverse image under the continuous map $f$ is closed in the compact space $X$, and a closed subset of a compact space is compact. This is the compactness of the fiber with its subspace topology.

False. Take $X=[0,1]$, $Y=S^1$ and the surjection $f(t)=e^{2\pi it}$. The interval is compact and connected, and the circle is Hausdorff. But $f^{-1}(1)=\{0,1\}$ is a disconnected two-point subspace. Thus a continuous surjection need not have connected fibers.

False. Let $X$ be a two-point set with the indiscrete topology, let $Y$ be a one-point space, and let $f$ be the unique map. The space $X$ is finite and hence compact; its only nonempty open subset is all of $X$, so it is connected. It is not Hausdorff, whereas $Y$ is Hausdorff and $f$ is a continuous surjection.

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.

Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.

Thus there are precisely two vertex classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four edge classes and two two-dimensional cells, so the resulting closed connected surface has Euler characteristicIt is a nonorientable surface: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired edge. The vertex links are circles, so these edge identifications produce a closed surface, rather than a singular cell space. The classification of compact surfaces now givesA torus also has Euler characteristic zero, but is excluded by the failed orientation test.