Past exam of the topology and groups course of the University of Oxford

Original solutions to past examinations of the topology and groups course of the University of Oxford.

Solutions to all three questions of the topology and groups course of the University of Oxford Part B examination, sat on 31 May 2023. Source: original released PDF, bundle pages 221–223; paper pages 1–3. The Mathematics and Statistics version is labelled B3.4. All eighteen parts are treated below. Question 3(a)(iii), as printed, needs an additional hypothesis such as local path connectedness; its solution includes both the intended argument and a counterexample to the unrestricted assertion.

Throughout Question 1, $I=[0,1]$ has its usual topology, and the mapping cylinder uses the paper's convention that $(x,1)$ is attached to $f(x)$.

Two continuous maps $f_0,f_1:X\to Y$ are homotopic if there is a continuous mapHere the Cartesian product has the product topology. Such an $H$ is a homotopy, written $f_0\simeq f_1$.

The topological spaces $X,Y$ are homotopy equivalent if there are continuous maps $f:X\to Y$ and $g:Y\to X$ such thatEach is then a homotopy inverse of the other.

For $A\subseteq X$ with its subspace topology with inclusion map $i:A\hookrightarrow X$, a homotopy retract, in the course's convention, is a continuous map $r:X\to A$ withThus $r$ is an actual retraction, and $i,r$ give a homotopy equivalence. Requiring the latter homotopy to fix $A$ pointwise at all times gives the stronger condition of a deformation retraction as used in this wiki. The course's definition of homotopy retract requires no such restriction on that homotopy.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Let $i:A\hookrightarrow X$ be the inclusion map and $r:X\to A$ the retraction. Since $X$ is contractible, choose a contraction of a topological spaceThe function compositionis a continuous map. Its endpoints satisfyConsequently $K$ is a homotopy from the identity map of $A$ to a constant function. This provesThe contracting point of $X$ need not belong to $A$: applying the retraction sends it to the contracting point $r(x_0)$ of $A$. This is the retract of a contractible space principle.

Writefor the quotient map defining the mapping cylinder, and put $j(x)=q(x,0)$. Since the equivalence relation only attaches points at height $1$ to $Y$, distinct points at height $0$ remain distinct. Thus $j$ is an injective continuous map.

Define the homotopyIt is a continuous map, with $H(x,0)=q(x,1)=i(f(x))$ and $H(x,1)=j(x)$. HenceEven if $f$ identifies many points, the free end of the mapping cylinder retains a separate copy of every point of $X$.

DefineThe two formulas agree on the attached end, so the universal property of the quotient topology makes $r$ a continuous map. Also $ri={\mathrm{id}}_Y$.

Slide each cylinder point towards its attached end usingAt $t=1$ these formulas both give $i(f(x))$, so they define a homotopy on the mapping cylinder. For continuity, the product of a quotient map with the unit interval is a quotient map; the formulas before taking the quotient topology are continuous maps and agree on every identified pair. ThusMoreover $Y$ stays fixed throughout. We have obtained the stronger conclusionIn particular $r$ is the requested homotopy retract.

Identify the bottom $j(X)=q(X\times \{0\})$ with $X$; its natural map $j$ is a topological embedding. To verify this directly, an open subset $O\subseteq X$ gives the saturated open subset $O\times [0,1/2)$ of the space before taking the quotient topology; its image is open in $M(f)$ and meets the bottom in $j(O)$.

Suppose first that there is a retraction $R:M(f)\to j(X)$. Define the continuous mapThenis a homotopy with $L(x,0)=x$ and $L(x,1)=g(f(x))$. Hence $gf\simeq {\mathrm{id}}_X$.

Conversely, suppose a continuous map $g:Y\to X$ has $gf\simeq {\mathrm{id}}_X$. Reverse the parameter if necessary to choose a homotopy $L$ with $L(x,0)=x$ and $L(x,1)=gf(x)$. SetAt the glued end, $j(L(x,1))=j(gf(x))$, exactly the value prescribed at $i(f(x))$. The universal property of the quotient topology therefore gives a continuous map $R:M(f)\to j(X)$. At the bottom, $R(j(x))=j(x)$, proving that $R$ is a retraction. We have established the mapping-cylinder bottom retraction criterion:

A group presentation of $G$ consists of a set $S$ of symbols and a subset $R\subseteq F(S)$ of relators, together with a group isomorphismHere $F(S)$ is the free group on $S$, and $⟨⟨R⟩⟩$ is the normal closure of $R$. The images of $S$ form a generating set of a group; the relators prescribe which words become the identity element.

For the canonical presentation of a group, use one symbol $s_g$ for every group element $g\in G$. The universal property of a free group gives a surjective group homomorphismLet $R(G)=\ker \epsilon$, namely all words whose evaluation in $G$ is the identity element. ThenIndeed $R(G)$ is already a normal subgroup, so its normal closure is itself, and the first isomorphism theorem identifies the resulting quotient group with $G$. This is the course's canonical presentation of a group.

An equivalent, smaller collection of relators is $s_g{s}_h{s}_{gh}^{-1}$ for all $g,h\in G$. These encode the multiplication table. They imply $s_1=1$ and $s_g^{-1}=s_{g^{-1}}$, so every word reduces to the single symbol recording its product. This proves directly that they normally generate $R(G)$, although the canonical collection above consists of every word in that group homomorphism kernel.

Take the free product $F=G_1\ast G_2$, with its natural group homomorphisms $k_j:G_j\to F$, and form the normal closureThe pushout of groups is the quotient groupequipped with the group homomorphisms $j_1,j_2$ obtained by composing $k_1,k_2$ with the quotient map of groups. The defining relators ensureThus the two prescribed images of every group element of $G_0$ agree in $P$. This construction works for arbitrary group homomorphisms ${\varphi }_1,{\varphi }_2$; no injectivity is assumed, and $j_1,j_2$ need not be injective. Its universal property is proved in the next part.

The universal property of a pushout of groups says: for every group $H$ and group homomorphisms $u_j:G_j\to H$ satisfying $u_1{\varphi }_1=u_2{\varphi }_2$, there is exactly one group homomorphism $u:P\to H$ such thatTo prove existence, the universal property of a free product supplies a unique group homomorphism $v:G_1\ast G_2\to H$ with $v{k}_j=u_j$. Every defining relator of $N$ maps toSince $\ker v$ is a normal subgroup, it contains the entire normal closure $N$. The universal property of a quotient group therefore gives a group homomorphism $u:P\to H$ through which $v$ factors. Its restrictions are the required $u_j$.

For uniqueness, $j_1(G_1)$ and $j_2(G_2)$ together generate $P$: their preimages already generate the free product. Consequently a group homomorphism out of $P$ is determined by its restrictions to these two subgroups. Any two candidates for $u$ therefore coincide.

This also proves that the pushout of groups is unique up to a unique compatible group isomorphism. Given another such object $P^{\prime }$, apply the universal property in both directions. Both composites restrict to the original maps from $G_1,G_2$, so uniqueness makes them the respective identity maps.

Put $A=T^2\setminus B^2$ and let $E=\partial A=S^1$ denote the attaching circle. The version of the Seifert-van Kampen theorem used here states that if $X=U\cup V$ is an open cover with $U,V,U\cap V$ path-connected and a common base point in $U\cap V$, thenThe expression on the right means the pushout of groups of the two inclusion-induced group homomorphisms; they need not be injective.

Choose a thin collar neighbourhood $C_A$ of $E$ inside $A$, and a thin equatorial band $C_S$ inside $S^2$. For example, before the square's side identifications take $C_A=\{v:1\le |v|<1+ϵ\}$ with $0<ϵ<1/2$, and take $C_S=\{(x,y,z)\in S^2:|z|<ϵ\}$. In the glued topological space setThese are open subsets: their preimages in the two pieces are open and respect the gluing. They cover $X$, and the three deformation retractionsfix $E$. Thus all three spaces are path-connected, and a base point $b\in E$ is preserved.

For completeness, the fundamental group of a once-punctured torus is $F(a,b)$, and its boundary loop of a once-punctured torus represents $[a,b]=ab{a}^{-1}{b}^{-1}$, up to orientation and a change of base point. Here is the geometry behind both facts. In the square model, move each point of $A$ radially outwards to the square boundary:The denominator never vanishes on $A$. The homotopy stays outside the missing open disk, stays inside the square, and fixes the square boundary, so it descends through the side identifications. The resulting retract is a wedge of two circles. Its fundamental group is the free group on the two oriented side based loops $a,b$. The inner boundary loop travels around the square once under this deformation retraction; its successive sides read $a,b,a^{-1},b^{-1}$.

The sphere $S^2$ is simply connected. Hence the group homomorphism from ${\pi }_1(E)\cong \mathbb{Z}$ to ${\pi }_1(V)$ is trivial, while its map to ${\pi }_1(U)$ sends a group generator to $[a,b]$. Applying the Seifert-van Kampen theorem givesThe final group isomorphism follows because the single relator makes $a,b$ commute, leaving the free abelian group on two group generators. The two hemispheres kill the same boundary loop; they do not create another group generator.

Let $n=(0,0,1)$ and use the open coverThe cap $V$ is open in $Y$: it equals $Y\cap \{z>1/2\}$ because the attached chord has $z=0$. The cap is homeomorphic to an open disk, and $U\cap V=V\setminus \{n\}$ is a punctured open disk, hence homotopy equivalent to a circle. Both sets and their intersection are path-connected. Choose the base point $b$ in that intersection.

We first calculate ${\pi }_1(U,b)$. A stereographic projection identifies $S^2\setminus \{n\}$ with ${\mathbb{R}}^2$ and sends the two chord endpoints to $(-1,0),(1,0)$. Thus $U$ becomes the Euclidean plane with an extra closed interval attached at those two points. This description is a homeomorphism: gluing the two closed pieces along their endpoints gives exactly the original subspace topology.

The Euclidean plane has a deformation retraction onto $L=[-1,1]\times \{0\}$ while fixing its endpoints. An explicit homotopy isExtend it by the identity map on the extra chord. The pasting lemma for closed subspaces gives a deformation retraction of $U$ onto two distinct closed intervals sharing just their endpoints, a circle. Consequently ${\pi }_1(U,b)\cong \mathbb{Z}$, with the routine base point transport if the retraction moves $b$.

The inclusion-induced group homomorphism ${\pi }_1(U\cap V,b)\to {\pi }_1(U,b)$ is trivial: it factors through $S^2\setminus \{n\}\cong {\mathbb{R}}^2$, which is contractible. Also ${\pi }_1(V,b)=1$. All hypotheses of the Seifert-van Kampen theorem are now checked, and it gives the fundamental group of a sphere with an attached chord:where both maps from the amalgamating infinite cyclic group are trivial. A group generator follows the chord from one endpoint to the other and returns along any path on the sphere, with a connecting path from the chosen base point.

A covering map is a continuous surjection $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichwhere the $U_{\alpha }$ are pairwise disjoint open subsets of $\tilde{X}$. Such a $U$ is an evenly covered neighbourhood, and the sets $U_{\alpha }$ are its covering sheets. The index set may be infinite. The requirement that every sheet map onto all of $U$ is part of the definition.

For a covering map over a nonempty connected space, the degree of a covering map isthe cardinality of any fibre of a function. To show independence of $x$, choose an evenly covered neighbourhood $U$. Each covering sheet contains exactly one point over every $y\in U$, so $|p^{-1}(y)|$ equals the number of covering sheets over $U$. The fibre of a function cardinality is therefore locally constant. For a fixed cardinal number $\kappa$, the points with that cardinality form an open subset, and their complement is also open. Connectedness forces the value to be constant everywhere.

For a path-connected base there is also an explicit fibre bijection by path lifting: lift a path from $x$ to $y$ starting at each point over $x$ and take its endpoint. The path lifting theorem gives existence and uniqueness; path reversal gives the inverse bijection. The bijection may depend on the path, but the resulting equality of cardinalities does not.

A connected base is needed for one global degree. Without it, use a degree on each connected component on which the cardinality is constant. For example, a covering map of a discrete two-point topological space can have one point over the first point and two over the second. Thus the unqualified printed definition needs the usual connected-base convention, or the explicit assumption that every fibre of a function has the same cardinality.

With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe fundamental group of the circle is $\mathbb{Z}$. Since the image of a finite group under a group homomorphism is finite and $\mathbb{Z}$ has no nontrivial finite subgroups,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns from $f$ to the constant function $f(x_0)$. This proves the claimed null-homotopy under that additional hypothesis, in particular for CW complexes and the topological graphs considered later in the paper.

For arbitrary path-connected spaces, the printed assertion is false. Here is an explicit Warsaw circle counterexample, including the obstruction that the usual lifting argument misses. In the Euclidean plane takeJoin $P=(1,0)$ to $Q=(0,-1)$ by the polygonal arc $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its subspace topology. The arc meets $G\cup L$ only at $P,Q$. The topologist's sine curve $G\cup L$ has two path components, but the added arc joins them, making $W$ path-connected.

Every loop in $W$ is contained infor some $\delta >0$. To justify the assertion, a path cannot reach $L$ along the oscillating tail: the intermediate value theorem for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting continuity. The only alternative route from the tail to $L$ uses $P$ and the joining arc. If a loop visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop lies in $L$. On a sufficiently small parameter interval around that time its image avoids $P$. That would give a forbidden path from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_{\delta }$ is a finite tree, in fact a single arc, so every such loop contracts. ConsequentlyNow define a continuous map $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the pasting lemma for closed subspaces proves continuity.

There is no continuous lift $\tilde{h}:W\to \mathbb{R}$. Any such lift is integer-valued on $G\cup L$, so it is constant on each of the connected spaces $G,L$. Since $G$ accumulates on $L$, continuity forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a null-homotopy, the homotopy lifting theorem for a covering map, starting with a lift of the final constant function and reversing time, would produce a continuous lift of $h$. This theorem needs no local path connectedness of the parameter space. Hence $h$ is not null-homotopic, although $W$ has finite, indeed trivial, fundamental group.

The native PDF explicitly assumes only path connectedness here. The missing local path connectedness cannot be inferred from the finiteness of the fundamental group.

A finite tree is a nonempty connected graph with no graph cycles. We show by mathematical induction on its number $v$ of vertices that it has $v-1$ edges. For $v=1$ there are no edges, so the assertion holds.

If $v>1$, take a path in a graph of maximal length among those without repeated vertices. Each endpoint is a leaf of a tree: an additional incident edge would either extend the path in a graph or create a graph cycle. Delete one such leaf of a tree and its unique incident edge. The remaining graph stays connected, since a path in a graph between remaining vertices never needs that terminal edge, and it has no graph cycles. It is therefore a tree with $v-1$ vertices and, by mathematical induction, $v-2$ edges. Restoring the deleted edge gives $e=v-1$.

Thus its Euler characteristic is

Choose a spanning tree $T$ of the finite connected graph $X$. It contains every vertex of $X$ and has $v(X)-1$ edges by part (i). Collapsing $T$ to a point is a homotopy equivalence: $T$ is a contractible subcomplex, and collapsing a contractible cofibration in a CW complex preserves homotopy type.

Every edge outside $T$ then becomes a circle attached to the common point. Hence the quotient space $X/T$ is a bouquet of circles withpetals. The fundamental group of a bouquet of circles is the free group with one member of a free basis of a group for each petal. Therefore the rank of the fundamental group of a finite connected graph isMore concretely, choose a vertex as base point; follow the unique path in a graph in $T$ to one end of a non-tree edge, traverse it, and return through $T$. These based loops give the free basis of a group. Thus $r$ is the number of free group generators, equivalently the minimum size of a generating set of a group, rather than the size of every possible redundant generating set of a group. Minimality also follows by abelianization, since $F_r$ has abelianization ${\mathbb{Z}}^r$.

Assume $p$ has a common finite degree of a covering map $d$. Declare the lifted vertices to be $p^{-1}(V(X))$, so there are exactly $dv(X)$ of them. Each open edge is homeomorphic to $(0,1)$, a simply connected interval. The restriction of the covering map above it is a disjoint union of $d$ copies of that open edge.

These copies acquire endpoints as follows. Lift the parametrizing path $[0,1]\to X$ of each oriented edge, starting at each of the $d$ points over its initial vertex. The path lifting theorem gives a unique lift all the way to $1$. Its endpoint lies over the terminal vertex, and the interiors of these lifted paths are exactly the $d$ open edges already described. This also handles a graph loop, whose two endpoints may coincide downstairs or upstairs. Local covering sheets above a small star at a vertex show that the lifted incidence relations have precisely the original topology. Thus the graph structure lifted through a covering map is a finite topological graph withSubtracting gives the Euler characteristic under a finite covering formulaThe formula holds even when $\tilde{X}$ is disconnected. If a disconnected base is allowed to have degrees $d_C$ that vary with its connected components $C$, the precise replacement isThe printed formula with a single $d$ uses the constant-degree convention explained in part (a)(ii).

Use a topological graph consisting of the ray $[0,\infty )$ with vertices $v_n=n$ and one oriented graph loop ${\ell }_n$ attached at each $v_n$, for $n=0,1,2,\dots$. Set the base point to $b=v_0$. This is a connected, locally finite graph; it is deliberately infinite.

Define continuous maps $f,g:X\to X$ byHere each graph loop is parametrized by $s\in [0,1]$ with its endpoints at the attaching vertex. The formulas agree at every vertex, and their restrictions to every closed edge are continuous. The weak topology of a CW complex therefore makes $f,g$ continuous maps. Directly on the ray and on each graph loop,

We check the two fundamental group maps at exactly the specified base points. The ray is a spanning tree. At $b=v_0$ a free basis of a group consists of $a_n$, obtained by going along the ray to $v_n$, following ${\ell }_n$, and returning. At $f(b)=v_1$, use the corresponding free basis of a group $c_n$ with connecting paths starting at $v_1$; in particular $c_0$ travels backwards along the first edge before following ${\ell }_0$. The standard spanning tree description applies to this infinite topological graph as well: every loop and every homotopy between based loops has compact image, so lies in a finite subcomplex.

The induced group homomorphisms areThus $f_{\ast }$ misses $c_0$: words in $c_1,c_2,\dots$ cannot equal $c_0$ by the normal form for free groups and free product of groups. Consequently $f_{\ast }$ is not surjective. Meanwhile $c_0$ is a nontrivial free group element killed by $g_{\ast }$, so $g_{\ast }$ is not injective. Neither induced map is a group isomorphism, while functoriality of the fundamental group correctly gives $g_{\ast }{f}_{\ast }=\mathrm{id}$.