Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 2F (Analysis and Topology)

The quotient topology on $\tilde{X}$ isSince inverse images preserve arbitrary unions and finite intersections,Also ${\pi }^{-1}(\varnothing )=\varnothing$ and ${\pi }^{-1}(\tilde{X})=X$. The displayed collection therefore satisfies all the topology axioms.

If $f:\tilde{X}\to Y$ is continuous, then $f\circ \pi$ is continuous because the quotient map $\pi$ is continuous by definition.

Conversely, suppose $f\circ \pi$ is continuous. For every open set $V\subseteq Y$,is open in $X$. The definition of the quotient topology then says that $f^{-1}(V)$ is open in $\tilde{X}$. HenceThis is the universal property of the quotient topology.

No. Take the Hausdorff space $X=\mathbb{R}$ and declare all nonzero points equivalent, leaving $0$ in its own class. The quotient has two points,The nonzero class is open because its inverse image $\mathbb{R}\setminus \{0\}$ is open. However, the only saturated set containing $0$ but not the other class is $\{0\}$, which is not open. Thus every neighbourhood of $[0]$ is the whole quotient, so the two quotient points cannot have disjoint neighbourhoods. Therefore $\tilde{X}$ need not be Hausdorff.