Past exam of the options II course of the University of Oxford / 2007 / E / E3

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.