Past exam of the topology course of the University of Oxford / 2023 (A5, A13757W1)

Original solutions to all three questions of the University of Oxford Part A topology examination, sat on 16 June 2023. Source: officially released examination PDF, parent PDF pages 101–103; printed paper pages 1–3. The wording in 1(a)(v) requires an interpretation explained in its solution. The classification reference in 3(b)(iii) should point to 3(a)(iv).

Give the nonempty subset $A\subseteq X$ its subspace topology. It is compact if every open cover admits a finite subcover. Equivalently, whenever $A\subseteq {\bigcup }_{j\in J}{U}_j$ with each $U_j$ open in $X$, there are $j_1,\dots ,j_m$ such that $A\subseteq U_{j_1}\cup \cdots \cup U_{j_m}$.

It is connected if it has no separation of a topological space: there are no disjoint nonempty open subsets $B,C$ of $A$ with $A=B\cup C$. Here openness is relative to the subspace topology; equivalently, the only clopen sets of $A$ are $\varnothing$ and $A$.

Let $(V_j{)}_{j\in J}$ be an open cover of $f(K)$ by open subsets of $Y$. The continuity of $f$ makes each preimage $f^{-1}(V_j)$ open in $X$, and these preimages cover $K$. By compactness, choose $j_1,\dots ,j_m$ withApplying $f$ gives $f(K)\subseteq {\bigcup }_{r=1}^m{V}_{j_r}$. Thus every open cover of $f(K)$ has a finite subcover, proving the continuous image of a compact space theorem. A cover by sets open in $f(K)$ gives the same argument after expressing them as intersections with ambient open sets.

Let $K\subseteq X$ be compact, with $X$ a Hausdorff space. The empty set is closed, so suppose $K\ne \varnothing$. Fix $x\in X\setminus K$. For each $y\in K$, the Hausdorff property supplies disjoint open neighbourhoods $U_y$ of $y$ and $V_y$ of $x$.

The $U_y$ form an open cover of $K$. Choose a finite subcover $U_{y_1},\dots ,U_{y_m}$ and put $V={\bigcap }_{r=1}^m{V}_{y_r}$. This is an open neighbourhood of $x$. If a point of $V$ belonged to $K$, it would belong to some $U_{y_r}$ as well as $V_{y_r}$, a contradiction. Hence $V\subseteq X\setminus K$. Every point of the complement has such an open neighbourhood, so $X\setminus K$ is open and $K$ is closed.

Use the Sierpiński space $X=\{0,1\}$ with topology $\{\varnothing ,\{1\},X\}$. The singleton $A=\{1\}$ is compact: from any open cover, choose one member containing $1$. But $X\setminus A=\{0\}$ is not open, so $A$ is not closed. This also exhibits why the Hausdorff hypothesis in the preceding proof matters.

The intended hypothesis is that $A$ is not connected. The original PDF has the same grammatical ambiguity as the transcription. If only $X$ were assumed not connected, the assertion would fail for the discrete space $X=\{0,1\}$ with its usual metric and the closed subset $A=\{0\}$: two disjoint sets cannot both meet that singleton.

Under the intended reading, choose a separation of a topological space $A=B\bigsqcup C$. Both $B$ and $C$ are nonempty and closed in $A$, because each is the complement of the other relatively open part. Since $A$ is closed in $X$, both are closed in $X$.

For a nonempty subset $F$ of the metric space $(X,d)$, the distance to a set satisfiesIndeed, the triangle inequality gives $d(x,F)\le d(x,z)+d(z,F)$, and swapping $x,z$ proves the bound. Thus the two distance to a set functions below are continuous. SetThese are disjoint open sets. For $b\in B$, we have $d(b,B)=0$ and $d(b,C)>0$, since $C$ is closed and $b\notin C$. Hence $B\subseteq U$. Similarly $C\subseteq V$. Therefore $A\subseteq U\cup V$ and both $U$ and $V$ meet $A$, as required. The construction is a concrete instance of the normality of every metric space.

Suppose $K\setminus \{a\}$ were compact. By the compact subset of a Hausdorff space theorem, it would be closed in $X$, hence closed in $K$. The singleton $\{a\}$ is also closed in the Hausdorff space $X$: each other point has an open neighbourhood avoiding $a$.

Consequently both $\{a\}$ and $K\setminus \{a\}$ are clopen in $K$. They are disjoint, nonempty, and cover $K$, giving a separation of a topological space in contradiction to the connectedness of $K$. Thus $K\setminus \{a\}$ is not compact.

Removing $1/2$ from the closed interval $I=[0,1]$ leaves the separation of a topological spaceThus $1/2$ is a cut point of $I$.

In contrast, removing any point $p$ from the square $Q=[0,1{]}^2$ leaves a path-connected space. To prove this, take distinct $u,v\in Q\setminus \{p\}$. Choose an interior point $w$ outside the two lines through $(u,p)$ and $(v,p)$. Such a choice exists: a small circle in the interior meets each of those lines in at most two points. Since $Q$ is a convex set, the line segments $[u,w]$ and $[w,v]$ stay in $Q$; our choice of $w$ ensures that both avoid $p$. Their concatenation is a continuous path from $u$ to $v$ in $Q\setminus \{p\}$.

The product topology on $I\times I$ agrees with its usual subspace topology in the Euclidean plane. If $h:Q\to I$ were a homeomorphism, its restriction would give a homeomorphismThe domain is connected, since path connectedness implies connectedness, while the target is not connected. This contradicts the preservation of connectedness by a homeomorphism.

The answer is no, because the number of non-isolated points differs. In $X$, every $1/n$ is an isolated point: a sufficiently small open interval around it contains no other member of $X$. The only non-isolated point is $0$, since every neighbourhood of $0$ contains some $1/n$.

In the product topology, $(x,y)\in X\times X$ is an isolated point exactly when both coordinates are isolated points of $X$. One direction follows because the product of two open singletons is open. For the other, any basic open rectangle containing $(x,y)$ contains another point if either coordinate has no open singleton. Hence the set of non-isolated points of $X\times X$ iswhich is infinite. A homeomorphism takes open singletons to open singletons, so it induces a bijection between the sets of non-isolated points. A space with just one such point cannot be homeomorphic to a space with infinitely many.

The quotient set $X/\sim$ consists of the equivalence classes $[x]=\{y\in X:y\sim x\}$. Its collapsing map is the canonical projection of a quotient setEquip this set with the quotient topology:This is a topology because preimages preserve arbitrary unions, finite intersections, the empty set, and the whole set. The resulting topological space is the quotient space. The displayed rule immediately implies that the preimage of every open set is open, so $p$ is continuous; it is also surjective.

Use the equivalence relation induced by a function:Equality in $Y$ supplies reflexivity, symmetry, and transitivity, so this is an equivalence relation. Define $\bar{f}:X/\sim \to Y$ by $\bar{f}([x])=f(x)$. It is well-defined; it is injective by the definition of $\sim$ and surjective because $f$ is surjective.

For an open set $W\subseteq Y$,is open by continuity of $f$. The quotient topology therefore makes ${\bar{f}}^{-1}(W)$ open, proving that $\bar{f}$ is continuous.

The continuous image of a compact space theorem applied to $p$ makes $X/\sim$ compact. If $C$ is closed in $X/\sim$, it is compact, and $\bar{f}(C)$ is compact in the Hausdorff space $Y$, hence closed. Thus $\bar{f}$ is a closed map. Its inverse function is continuous, since its preimage of $C$ is precisely $\bar{f}(C)$. We have proved that $\bar{f}$ is a homeomorphism, including the compact-to-Hausdorff continuous bijection theorem used in the argument.

Here $D^2$ must denote the closed disc $\{(x,y):x^2+y^2\le 1\}$, as its boundary points participate in the identification. For $r=\sqrt{x^2+y^2}$, defineThe Pythagorean trigonometric identity gives $\Vert F(x,y){\Vert }^2=1$, so the image lies in the sphere $S^2$. Away from the origin the formula is continuous. At the origin, the first two coordinates have absolute value at most $|\sin (\pi r)|\to 0$ and the third tends to $1$, proving continuity there as well.

Every boundary point maps to $(0,0,-1)$, while the origin maps to $(0,0,1)$. For a point $(u,v,w)\in S^2$ with $-1<w<1$, there is a unique $r\in (0,1)$ with $\cos (\pi r)=w$, and its unique preimage isThis point has radius $r$, because $u^2+v^2=1-w^2={\sin }^2(\pi r)$. The two poles have the preimages already described. Thus $F$ is surjective and its fibers are exactly the prescribed equivalence classes.

The closed disc is compact by the Heine-Borel theorem, and $S^2$ is Hausdorff with the subspace topology from Euclidean space. The preceding result gives the homeomorphism

Any two nonempty open sets of the infinite cofinite topology intersect: their complements are finite sets, so the complement of their intersection is finite and cannot be all of $\mathbb{R}$.

Suppose the quotient space had distinct points with disjoint open neighbourhoods $U,V$. Since the quotient map $p$ is surjective, both preimages $p^{-1}(U),p^{-1}(V)$ would be nonempty. Since $p$ is continuous, these would be disjoint open subsets of $\mathbb{R}$ with its cofinite topology, contradicting the observation above. Thus any such quotient space with at least two points is not Hausdorff.

True. The canonical projection of a quotient set $p:X\to X/\sim$ is a continuous surjection. Its image is compact by the continuous image of a compact space theorem.

For connectedness, if $U,V$ formed a separation of a topological space of $X/\sim$, then the preimages $p^{-1}(U),p^{-1}(V)$ would be disjoint nonempty open sets covering $X$, contradicting its connectedness. This proves the continuous image of a connected space theorem in this case. Hence the quotient space is both compact and connected.

False. Give $X=\mathbb{R}$ its usual topology, identify all points of $(0,1)$ with one another, and leave every other point in a singleton equivalence class. Take the closed set $K=\{1/2\}$.

Its image $p(K)$ is a singleton in the quotient space, but its full preimage, also called the saturation of a subset, isThis is not closed in $\mathbb{R}$, since $0$ belongs to its closure. In the quotient topology, a subset is closed exactly when its preimage is closed, by applying the defining open-set rule to complements. Thus $p(K)$ is not closed, and this quotient map is not a closed map.

True. Choose the irrational number $\alpha =\sqrt{2}$, and partition the rational numbers intoNo endpoint is rational, so these disjoint sets cover $\mathbb{Q}$. Each is nonempty by the density of the rational numbers, and each is clopen in the subspace topology on $\mathbb{Q}$.

Let two rational numbers be equivalent exactly when they belong to the same $C_n$. The bijectionis well-defined. For every subset $B\subseteq \mathbb{Z}$, the preimage of $h^{-1}(B)$ under the canonical projection of a quotient set is ${\bigcup }_{n\in B}{C}_n$, which is open in $\mathbb{Q}$. Thus every subset of $\mathbb{Q}/\sim$ is open: its quotient topology is the discrete topology. The usual subspace topology on the integers is also the discrete topology, because $\{n\}=\mathbb{Z}\cap (n-1/2,n+1/2)$. Consequently $h$ is a homeomorphism.

False. Take the metric space $X=\mathbb{R}\times \{0,1\}$ with the Euclidean distance and subspace topology inherited from ${\mathbb{R}}^2$. For every $t\ne 0$, identify $(t,0)$ with $(t,1)$; leave the two points $(0,0)$ and $(0,1)$ in separate equivalence classes. All equivalence classes have one or two points, so all are closed in this metric space.

Let $o_0,o_1$ be the two distinct images of the origins. If $U_0,U_1$ are any open neighbourhoods of $o_0,o_1$ in the quotient space, their preimages contain respectivelyfor some $\epsilon ,\delta >0$. Choose $0<t<\min (\epsilon ,\delta )$. The common image $p(t,0)=p(t,1)$ belongs to both $U_0$ and $U_1$. Therefore the origins have no disjoint open neighbourhoods and the quotient space is not Hausdorff. This is the line with two origins.

An abstract simplicial complex consists of a set $V$ of vertices of a simplicial complex and a family $K$ of finite nonempty subsets of $V$, called abstract simplexes, such that every singleton $\{v\}$ belongs to $K$ and every nonempty subset of a member of $K$ also belongs to $K$. One may include the empty set as an additional face by convention. A member with $k+1$ vertices is a $k$-simplex; its nonempty subsets are its faces of a simplex.

The geometric realization has the following explicit description using a barycentric coordinate system:Here the support is $\mathrm{supp}(t)=\{v:t(v)>0\}$. Each $\rho \in K$ gives a standard geometric simplexwith its usual Euclidean topology. Common faces of a simplex are identified using the same barycentric coordinate system.

Precisely, the topology on $|K|$ is defined by declaring $A\subseteq |K|$ closed if and only if $A\cap |\rho |$ is closed in $|\rho |$ for every $\rho \in K$. Equivalently, use the quotient topology on the disjoint union of all the standard geometric simplexes, identifying identical points of common faces of a simplex. For a finite simplicial complex, this agrees with the subspace topology in ${\mathbb{R}}^V$. For an infinite simplicial complex, the stated rule specifies the realization topology without presuming a finite-dimensional ambient space.

Fix a vertex of a simplicial complex $v$, and let $R$ be the set of vertices reachable from $v$ by an edge path. If a simplex contains a vertex of $R$, every vertex of that simplex lies in $R$: append its edge from the reachable vertex to the existing edge path. Thus every simplex has either all its vertices in $R$ or all outside $R$.

Partition the geometric realization as $|K|=A\bigsqcup B$, according to whether the support in the barycentric coordinate system lies in $R$ or outside $R$. On each geometric simplex, the intersection with $A$ is either the whole simplex or the empty set, and the same holds for $B$. The defining realization topology therefore makes both $A$ and $B$ closed, hence also open because they are complements.

The vertex $v$ belongs to $A$. If any vertex were unreachable, it would belong to the nonempty set $B$, giving a separation of a topological space of $|K|$. This contradicts connectedness. Therefore every vertex is reachable from $v$, and any two vertices can be joined by an edge path.

A topological surface is a Hausdorff space with a countable basis of a topology in which every point has an open neighbourhood homeomorphic to an open subset of the Euclidean plane. We use the convention that a surface has no boundary unless a surface with boundary is explicitly specified. A closed surface is compact and has no boundary. Some conventions also include connectedness in the definition; we impose it on the closed combinatorial surface below.

A closed combinatorial surface is a finite connected simplicial complex $K$ with the following properties: every simplex is a face of a simplex of dimension two, and the link of a simplex at every vertex is a cycle graph. Explicitly, for a vertex $v$ its link is the simplicial subcomplexTo say this is a cycle graph means that there are distinct vertices $w_1,\dots ,w_m$, with $m\ge 3$, and exactly the link edges $\{w_i,w_{i+1}\}$, with indices modulo $m$. In particular, every edge of $K$ belongs to exactly two triangles. The vertex-link condition also requires that the triangles around each vertex form one cyclic fan; edge incidence alone would allow singular examples.

These conditions make $|K|$ a closed surface. A point inside a triangle has a disc neighbourhood, the two triangles at an edge give a disc neighbourhood, and the cyclic fan at a vertex gives a disc neighbourhood. Finiteness gives compactness, the Hausdorff property, and a countable basis of a topology from a realization in a finite-dimensional Euclidean space. If disconnected closed combinatorial surfaces are allowed, each connected component satisfies this definition separately.

The classification theorem for closed combinatorial surfaces states that the geometric realization of every closed combinatorial surface is homeomorphic to exactly one of the following:

With ${\Sigma }_0=S^2$, the complete distinguishing data are orientability and the Euler characteristic:Here $\chi (K)=V-E+F$, where $V,E,F$ count the vertices of a simplicial complex, edges, and triangles of a finite triangulation. Thus two such connected closed surfaces are homeomorphic if and only if they have the same orientability and Euler characteristic. In particular $N_1$ is the real projective plane and $N_2$ is the Klein bottle. Every listed type admits a finite triangulation. For disconnected finite closed combinatorial surfaces, apply the theorem to each of the finitely many connected components; the resulting unordered collection of types determines the homeomorphism type.

We construct both the required triangulation and its polygonal schema. Subdivide the square $Q=[0,1{]}^2$ into a $4\times 4$ grid. Insert the centre of each small square and join it by line segments to its four corners, making four triangles per small square. Identify opposite sides of $Q$ by translations: $(0,t)\sim (1,t)$ and $(t,0)\sim (t,1)$. This is the standard quotient space presentation of the torus.

The grid and centre vertices, with these identifications, define an abstract simplicial complex $K_1$. Every triangle contains its own small-square centre and two consecutive grid vertices, so no triangle has identified vertices and no distinct triangles acquire the same vertex set. Only intended common sides or vertices are shared. Thus this is a genuine finite triangulation, whose geometric realization is homeomorphic to $T^2$. There are $32$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic zero.

Choose the simplex $\tau$ represented in the interior of $Q$ by the triangle with verticesRemove its relative interior. Before identifying the square's sides, the remainder is a closed disc with a triangular hole, hence an annulus. Cut this annulus along the line segment from $(0,0)$ to $(1/4,1/4)$, treating the two banks as distinct boundary arcs. The cut space is a closed disc; traversing its boundary follows the four outer edges, one bank, the three inner edges, and the other bank. It therefore has nine marked boundary sides.

A homeomorphism from this cut disc to a convex nine-sided polygon can be chosen to carry its nine marked sides to the nine polygon sides. One justification is to choose a boundary homeomorphism with the prescribed marked vertices and extend it over the disc by radial extension after disc parametrizations. Transport the identifications to this convex polygon. With labels recording traversal direction, its boundary word isIdentify equally labelled sides in their arrow directions: the two $a$ sides and two $b$ sides restore the torus pairings, while the two $c$ sides undo the cut. Leave $d,e,f$ unpaired. Their union is the triangular boundary of the removed simplex. The resulting quotient space is exactlyThis is the polygonal presentation of a punctured surface. The word supplies all nine sides and every gluing; inverse exponents reverse the boundary traversal direction.

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

Use compatible edge identifications: choose a bijection between the three boundary vertices of $\tau$ and those of $\sigma$, and extend it linearly on each edge. This defines a simplicial homeomorphism of the two boundary circles. This compatibility is the usual meaning of gluing the two boundary triangles; independently chosen edge reversals need not agree at their endpoints. The classification needed here is 3(a)(iv); the printed reference to 3(a)(iii) is a source error.

Delete the two triangular simplexes but retain all their proper faces of a simplex, and identify the two resulting boundary simplicial subcomplexes by this map. The result is a finite abstract simplicial complex $L$: vertices from different sides are identified only in the boundary, and the only shared edges are boundary edges. There is no triangle consisting of all three boundary vertices after the two faces have been removed. Consequently no other simplexes become duplicates. Its geometric realization has the specified quotient topology and is homeomorphic to $Z$.

We check the closed combinatorial surface conditions. Away from the seam, vertex links of a simplex remain cycle graphs. At a boundary vertex of $X$ or $Y$, removing the face changes its link from a cycle graph to a graph path, with endpoints corresponding to its two boundary edges. Gluing the boundaries identifies the two paths' respective endpoints, making one cycle graph. Each seam edge now meets one triangle from $X$ and one from $Y$, and all other edges retain their original two incident triangles. Thus every vertex link is a cycle graph and every simplex lies in a triangle. The simplicial complex is connected, since the two connected punctured surfaces meet along their boundary circle. Hence $L$ is a closed combinatorial surface.

This construction is the connected sum of surfaces $T^2\#{\mathbb{RP}}^2$. For completeness, compute its Euler characteristic directly. Removing a triangular face while retaining its edges and vertices subtracts one, soGluing identifies three vertex pairs and three edge pairs, givingFor the explicit triangulations above, this is also $62-189+126=-1$.

Finally $Z$ is a nonorientable surface. If $Z$ were orientable, its restriction would make $Y$ orientable. The consistent choices of orientation of a simplex on its triangles extend across a capping disc: choose the cap's orientation of a simplex so that the induced directions on the common boundary oppose. Capping $Y$ by $\sigma$ would therefore make the real projective plane orientable, a contradiction. The obstruction to orientability in the real projective plane can also be seen from a Möbius band neighbourhood of a projective line.

By the classification theorem for closed combinatorial surfaces, a nonorientable surface with Euler characteristic $-1$ has $2-k=-1$, hence $k=3$. Therefore