Past exam of the options II course of the University of Oxford / 2006 / E (Topology)

A subset is compact if every cover of it by open sets of the ambient space has a finite subcover; equivalently use open sets of its subspace topology. A cover of $A\cup B$ separately covers $A$ and $B$. Compactness of each supplies two finite collections, whose union is finite and covers $A\cup B$. If $A$ is closed in a compact space, add the open complement of $A$ to any open cover of $A$. Compactness supplies a finite subcover of the whole space, and discarding the complement leaves a finite cover of $A$.

For the enlarged space, every old open set is open in $C$ and has the same trace on $X$. The trace on $X$ of a new open set $V\cup \{\infty \}$ is $V$, which is open in $X$ because its complement is closed. Thus the induced topology on $X$ is exactly the original one.

Given an open cover of $C$, choose a member containing infinity. It has the form $V\cup \{\infty \}$ with $K=X\setminus V$ compact. The remaining cover restricted to $K$ has a finite subcover. Together with the chosen infinity-neighborhood, these finitely many sets cover $C$. Hence $C$ is compact, with no Hausdorff assumption needed.

Finally put $a=f(\infty )$. Continuity makes $U=f^{-1}((a-ϵ,a+ϵ))$ open and containing infinity. Therefore $U=V\cup \{\infty \}$ with $W=X\setminus V$ compact and closed in $X$. For every $x\in X\setminus W=V$ we haveThis expresses convergence to the value at infinity outside compact closed subsets.

A quotient map $\pi :X\to Y$ is a surjection such that $U\subseteq Y$ is open if and only if ${\pi }^{-1}(U)$ is open in $X$. Thus the topology on $Y$ is exactly the final topology induced by $\pi$.

For an open continuous surjection, continuity gives one implication. For the other, if ${\pi }^{-1}(U)$ is open then $U=\pi ({\pi }^{-1}(U))$ is open by openness and surjectivity. Hence it is a quotient map.

If $\pi :X\to Y$ and $\sigma :Y\to Z$ are quotient maps, they are surjective andThis set is open exactly when ${\sigma }^{-1}(U)$ is open, exactly when $U$ is open. Thus the composition is a quotient map.

For the factorization, continuity of $g$ implies continuity of $f=g\circ \pi$. Conversely, if $f$ is continuous and $U$ is open in $Z$, then ${\pi }^{-1}(g^{-1}(U))=f^{-1}(U)$ is open. The quotient property makes $g^{-1}(U)$ open. This holds for every open $U$, so $g$ is continuous.

The two singly marked sides of the first triangle are glued with their endpoints at $A$ matched. A cross-section parallel to the remaining side has its two endpoints identified, becoming a circle; the apex becomes its cone point. Thus this quotient triangle is a disc, whose boundary is the doubly marked edge.

In the second triangle, the two triply marked sides run consecutively along the boundary from $E$ through $D$ to $F$. Pairing these sides in the indicated directions introduces one reversal; its boundary word is $aa{b}^{-1}$, where $b$ is the remaining doubly marked edge. This is a Möbius band with boundary $b$. One can obtain the usual rectangle model by cutting from the identified triple-edge seam to the midpoint of $b$ and opening the triangle.

The double markings glue the disc boundary to the boundary of that Möbius band. Capping a Möbius band by a disc givesThis description also verifies that the quotient is a closed surface, including the vertices. As a check, the vertex classes are $\{A\}$ and $\{B,C,D,E,F\}$; there are three edge classes and two faces. The Euler characteristic is $2-3+2=1$, consistent with the projective plane. The Möbius band exhibits nonorientability directly.