Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / E / 3

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.