Past exam of the options II course of the University of Oxford / 2012 / E / E4 / a

The quotient topology is defined byPreimages commute with arbitrary unions and finite intersections and take the empty and whole sets to the corresponding sets, so this definition indeed gives a topology.

A map is continuous exactly when the preimage of every open set is open. By the definition in part (i), every open $U\subseteq X/\mathcal{R}$ has open preimage $p^{-1}(U)$. Thus $p$ is continuous, directly from the defining property of the quotient topology.

If $g$ is continuous, so is the composition $g\circ p$. Conversely, assume the composition is continuous. For every open set $V\subseteq Y$,is open in $X$. The quotient definition then says that $g^{-1}(V)$ is open in $X/\mathcal{R}$. Hence $g$ is continuous andThis is the universal property of the quotient topology.

Let two distinct quotient points correspond to two equivalence classes. By hypothesis those classes lie in disjoint open saturated sets $U,V\subseteq X$. Saturation gives $p^{-1}(p(U))=U$, so the quotient definition makes $p(U)$ open, and likewise $p(V)$.

The image sets contain the two quotient points and are disjoint. Indeed, a class meeting both $U$ and $V$ would be contained in each because both sets are saturated, contradicting their disjointness. Thus the quotient is Hausdorff.

Define $\tilde{f}:X/\mathcal{F}\to Y$ by $\tilde{f}([x])=f(x)$. Its definition is independent of the representative because the classes are precisely the fibres of $f$. It is onto since $f$ is onto, and injective because two points with equal $f$-values lie in the same fibre.

The relation $\tilde{f}\circ p=f$ and part (iii) imply continuity of $\tilde{f}$. Since $X$ is compact and $p$ is continuous, $X/\mathcal{F}$ is compact. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Therefore