Past exam of the options II course of the University of Oxford / 2007 / E (Topology)

Compactness means every open cover has a finite subcover. Connectedness means the space cannot be written as the union of two disjoint nonempty open sets. The Hausdorff property means every two distinct points have disjoint open neighbourhoods.

True. Pull an open cover of $Y$ back through $f$. Surjectivity makes those inverse images cover $X$. A finite subcover of the compact space $X$ maps back to a finite subcover of $Y$. Thus a continuous image of a compact space is compact.

True. A separation of $Y$ into two disjoint nonempty open sets would pull back to a separation of $X$: continuity gives openness, and surjectivity gives nonemptiness. This contradicts connectedness of $X$.

True. Every singleton in a Hausdorff space is closed. Thus $f^{-1}(\{y\})$ is closed in $X$ by continuity. A closed subset of a compact space is compact, whether or not that compact space is Hausdorff. Hence every fibre is compact.

False. Take $X=[-1,1]$, $Y=[0,1]$ and $f(x)=x^2$, with their usual topologies. The domain is compact and connected, the codomain is Hausdorff, and the map is continuous and onto. But the fibre over one is $\{-1,1\}$, a disconnected two-point subspace.

False. Let $X$ be an indiscrete two-point space and $Y$ a one-point space. The constant map is continuous and onto; $X$ is compact because it is finite and connected because its only nonempty open set is the whole space. It is not Hausdorff, since its two points have no disjoint nonempty neighbourhoods. The one-point codomain is Hausdorff.

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 surface is a Klein bottle. Here is a cell-count justification that keeps all the arrow directions. Name the corners of the first square $a,b,c,d$, beginning at its upper left and proceeding clockwise; name those of the second $e,f,g,h$ in the same way. The single-arrow gluing identifies $a$ with $e$ and $b$ with $h$. The triple-arrow gluing identifies $d$ with $h$ and $c$ with $g$. The first square's quadruple-arrow gluing identifies $b$ with $d$ and $c$ with $a$. The second square's double-arrow gluing identifies $e,f,g$ with one another. There are exactly two resulting vertex classes:There are four edge classes and two faces, so the Euler characteristic is $2-4+2=0$. Every edge is paired and the corner links close into circles, giving a compact surface without boundary; the cross-square identifications make it connected. It is nonorientable: the quadruple-arrow identification glues two boundary edges of the same square in the same boundary-traversal direction, so no orientation of that face can extend across the gluing. The classification of compact connected surfaces now identifies the nonorientable surface of Euler characteristic zero as the Klein bottle.

Geometrically, the internal identification in each square produces a Möbius band. The remaining single- and triple-arrow edges comprise their two boundary circles, which are glued together. The double of a Möbius band is another description of the same Klein bottle.