Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2007 (C3.1, DMAC 2463, DMPC 2463, DMAB 2463)

Topology, Groups and Algebraic Topology, Mathematics Parts C and B and Mathematics and Philosophy Part C; Wednesday 30 May 2007, 9.30am–12.30pm. The complete five-page source and both sections are privately transcribed from original Part C bundle pages 15–19, supplied through the past-papers disclosure. No exact worked-answer advertisement was found in the bounded availability check. Homology and cohomology use integer coefficients unless another coefficient group is stated.

A homotopy from $g$ to $h$ is a continuous $H:X\times [0,1]\to Y$ with $H(x,0)=g(x)$ and $H(x,1)=h(x)$. Precomposing gives $(w,t)\mapsto H(f(w),t)$, a homotopy from $gf$ to $hf$; postcomposing gives $(x,t)\mapsto k(H(x,t))$, a homotopy from $kg$ to $kh$.

A homotopy equivalence consists of maps $u:X\to Y$ and $v:Y\to X$ with $vu\simeq {\mathrm{id}}_X$ and $uv\simeq {\mathrm{id}}_Y$. Identity maps prove reflexivity, and exchanging $u,v$ proves symmetry. For equivalences $X\leftrightarrows Y$ and $Y\leftrightarrows Z$, compose their forward maps and compose their inverse maps in reverse order. The two compositions with these candidate inverses are homotopic to the respective identities by the pre- and postcomposition result. Concatenating the homotopies proves transitivity.

For the first listed space, radial projection onto the unit sphere gives a strong deformation retraction:The scalar is positive, so a point outside the two coordinate axes never enters either removed axis. Their intersection with the sphere is the four points $\pm e_1,\pm e_2$. The allowed homeomorphism statement therefore identifies this retract with the four-punctured sphere in the list.

Stereographic projection from one puncture identifies that sphere with the plane minus three points. Remove small disjoint open disks around the punctures and choose a larger enclosing disk. Radial motions in the three punctured disks push their remaining points to the small boundary circles, while the exterior pushes to the large boundary circle. This reduces the space to a compact disk with three holes. It has a spine consisting of three loops connected by a tree of arcs; cutting along those arcs gives a disk and collapsing its collar gives the retraction onto the spine. Collapsing a maximal tree in the spine is a homotopy equivalence onto a wedge of three circles.

Finally the three parallel removed lines give the productThe homotopy $(x,y,z)\mapsto (x,y,(1-t)z)$ retracts this onto the punctured plane, hence onto the same rank-three graph. Composing these maps and homotopy inverses proves all four listed spaces have the same homotopy type. The two intersecting axes produce four spherical punctures, which is why three loops appear rather than two.

Use letters $x_i$ and their formal inverses. An elementary contraction deletes an adjacent pair $x_i{x}_i^{-1}$ or $x_i^{-1}{x}_i$. A reduced word has no such pair. Repeated contraction terminates because each step reduces length by two, so every word has a reduced descendant.

For uniqueness, two possible first contractions either occur on disjoint pairs, when performing both yields a common descendant, or overlap in three letters $a{a}^{-1}a$, when either contraction produces the same remaining letter. Induct on word length: the shorter descendants of both first contractions therefore have the same eventual reduced word. This proves every contraction sequence from a given word ends at the same reduced word. An insertion or deletion of an inverse pair preserves that reduced result. Since the word model of the free group identifies words exactly under finite sequences of these moves, every group element has a unique reduced representative.

To define the requested homomorphism, substitute $w_i$ for $x_i$ and $w_i^{-1}$ for $x_i^{-1}$ in any word. Inverse-pair insertions or deletions map to inverse words that cancel, so substitution is well defined on group elements and respects concatenation. This proves the universal property of the free group in this case.

For an automorphism of order four, take $\varphi (x_1)=x_2$ and $\varphi (x_2)=x_1^{-1}$. Its iterates on $x_1$ are $x_2,x_1^{-1},x_2^{-1},x_1$; its fourth iterate also fixes $x_2$. Unique reduced representatives show the first three iterates on $x_1$ differ from $x_1$. Hence its order is exactly four.

For the final map, every iterate of $x_2$ is a positive word and contains an $x_2$. Substituting once replaces each occurrence of $x_2$ by $x_{2}w$ and leaves each $x_1$ unchanged. If $\ell =|w|\ge 1$, its length consequently increases by at least $\ell$ on every iteration. No cancellation occurs in a positive word, so $|{\varphi }^n(x_2)|\ge 1+n\ell >1$. It cannot equal $x_2$, proving no positive iterate of this endomorphism is the identity. Surjectivity of this last endomorphism is not assumed.

This uses positive-word growth under a free-group endomorphism.

One standard five-operation version of the Tietze transformations is: invert a relator; conjugate a relator by a word; multiply a relator on either side by another relator; insert or delete the empty relator; and introduce or eliminate a generator together with a defining relation $y=w$, where $w$ uses the other generators and an eliminated generator occurs nowhere else. The elementary relator operations are reversible. Renaming and reordering the generators and relators is harmless. The Tietze theorem states that two finite presentations define isomorphic groups exactly when they are related by a finite sequence of these transformations.

For a connected cell complex with one vertex, take one generator for each oriented one-cell and one relator from the attaching loop of each two-cell. Higher-dimensional cells do not change the fundamental group. This is the cellular presentation theorem, obtained by applying van Kampen to the two-skeleton. Conversely a finite presentation is realized by a bouquet of its generator circles with one disk attached along each relator word, giving a finite connected complex with the required fundamental group.

Triangulate each relator polygon by adding diagonals from one boundary vertex. Each diagonal is a new one-cell, and every triangular two-cell has an attaching word of length at most three. Algebraically, for a word $s_1\cdots s_{\ell }$ with $\ell \ge 4$, introduce successive partial-product generators $y_2,\dots ,y_{\ell -2}$. Use relatorsEliminating the new generators recovers the original word; words already of length at most three need no change. This proves the triangular-presentation claim geometrically and algebraically.

For the free abelian group of rank two, useIts two relations say $c=ab=ba$. For the requested transformation sequence, replace the second relator by the product of the first and second, freely reducing it to $ab{a}^{-1}{b}^{-1}$. Invert the first relator, giving $c{b}^{-1}{a}^{-1}$, the definition $c=ab$. Eliminate $c$ together with this defining relation. The remaining presentation is exactly the stated two-generator commutator presentation.

The construction is a triangular group presentation.

The based lifting criterion isUnder the stated path-connected and locally path-connected domain hypotheses, it is necessary and sufficient for a continuous based lift. Two based coverings are equivalent if a homeomorphism between their total spaces takes one chosen basepoint to the other and commutes with their projections.

For the usual classification by subgroups, total spaces are taken to be connected. They are then path-connected and locally path-connected because they cover a locally path-connected base. If two such coverings have the same subgroup $H$, apply the lifting criterion to each projection to obtain based maps $F:\tilde{X}\to {\tilde{X}}^{\prime }$ and $G:{\tilde{X}}^{\prime }\to \tilde{X}$ over $X$. Their composites and the identity maps are based lifts of the same projections. Uniqueness of lifts on a connected domain gives $GF=\mathrm{id}$ and $FG=\mathrm{id}$. Thus $F$ is an equivalence, proving at most one connected based covering realizes a given subgroup.

The connected-total-space qualification is necessary. If arbitrary disconnected coverings are allowed, the identity cover of a circle and the disjoint union of two identity covers have the same subgroup at a basepoint in the first component, yet are inequivalent. The subgroup only records the chosen component in that setting.

The degree is the cardinality of a fibre. Over a path-connected base this is independent of the basepoint, by path lifting. For a connected total space it equals the subgroup index $[{\pi }_1(X,b):H]$, and it may be infinite.

On the torus with basepoint $(1,1)$ take $p(z,w)=(z^2,w)$ and $p^{\prime }(z,w)=(z,w^2)$. Both are connected two-sheeted coverings. Their induced subgroups of ${\pi }_1(S^1\times S^1)={\mathbb{Z}}^2$ are $2\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times 2\mathbb{Z}$, which are different. Equivalence over the fixed base would force equal based subgroups. Thus the two degree-two covers are inequivalent.

Write the wedge's fundamental group as the free group on $a,b$. Use the index-two subgroupswhere each map has target $\mathbb{Z}/2$. Explicitly, the first cover has two vertices, two directed lifts of $a$ exchanging the vertices, and a $b$-loop at each vertex. The second interchanges the roles of $a,b$. Each vertex has exactly one incoming and one outgoing lift of each labelled edge, so these are genuine two-sheeted graph coverings. They are connected, and a chosen vertex gives a based cover.

The subgroups differ because $a$ belongs to $H^{\prime }$ but not $H$, while $b$ belongs to $H$ but not $H^{\prime }$. Hence the covers have the same degree and are inequivalent.

Choose an orientation for each simplex. The chain group $C_k$ is the free abelian group on the oriented $k$-simplices, with orientation reversal changing sign. On an ordered simplex setIts face maps respect the identifications in a Delta-complex. In ${\partial }_{k-1}{\partial }_k$, deleting any pair of vertices occurs in the two possible orders with signs $(-1{)}^{i+j-1}$ and $(-1{)}^{i+j}$, which cancel. Hence ${\partial }^2=0$. The groups of homological cycles and boundaries are $Z_k=\ker {\partial }_k$ and $B_k=\mathrm{im}{\partial }_{k+1}$; the identity just proved gives $B_k\subseteq Z_k$, so $H_k=Z_k/B_k$ is defined. Mod-two homology uses the same chain complex with coefficients reduced modulo two.

Here the edge identifications put all six vertices into one vertex class. Let the three boundary edge classes be $a,b,c$ in cyclic order, and orient the internal diagonals as $d=[v_1,v_3]$, $e=[v_3,v_5]$, $f=[v_5,v_1]$. With the four face orientations listed in the source, the boundaries areThus $C_2={\mathbb{Z}}^4$, $C_1={\mathbb{Z}}^6$, $C_0=\mathbb{Z}$, and ${\partial }_1=0$. The Euler characteristic isThe face-boundary map is injective over the integers: coefficients of $a,b,c$ force the first three face coefficients to vanish, after which the diagonal coefficients force the fourth to vanish. Its cokernel eliminates $d,e,f$ and leaves generators $a,b,c$ with the single relation $2(a+b+c)=0$. Since $a+b+c$ is primitive in ${\mathbb{Z}}^3$, this cokernel is ${\mathbb{Z}}^2\oplus \mathbb{Z}/2$. ThereforeModulo two the first three face-boundary columns are $d,e,f$, and the last is their sum. The rank is three, the kernel is one-dimensional, and the edge quotient has dimension three. ConsequentlyThe top mod-two class is the sum of all four faces. Integral homology has no such top class, reflecting nonorientability.

For an open cover $X=U\cup V$, the Mayer–Vietoris theorem gives the long exact sequenceIt ends in $H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$. To prove it, let $C_{\ast }^{U,V}(X)$ be the subcomplex generated by singular simplices lying wholly in $U$ or wholly in $V$. There is a short exact sequence of chain complexesExactness follows from the singular-simplex bases: the intersection of the two chain subgroups consists exactly of simplices lying in both sets. The small-chain locality theorem for singular homology says the inclusion of this small-chain complex into $C_{\ast }(X)$ induces an isomorphism on homology; barycentric subdivision is its standard justification. The long exact sequence of a short exact sequence of chain complexes now proves the theorem. For a homological cycle represented as $a+b$, the connecting map is $[a+b]\mapsto [\partial a]$, since $\partial a=-\partial b$ lies in the intersection. This also identifies the displayed maps and their signs.

For the glued solid tori, use open collar enlargements of the two halves. Each retracts to a circle; the overlap retracts to their boundary torus. Thus the halves have $H_0=H_1=\mathbb{Z}$ and higher homology zero, while the overlap has $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$. In the overlap choose longitude and meridian generators. The meridian bounds a disk in each solid torus and the longitude generates each first homology, so the map on $H_1$ isIts kernel is the meridian copy of $\mathbb{Z}$ and its cokernel is $\mathbb{Z}$. The map on $H_0$ is the injective map $z\mapsto (z,-z)$. Exactness therefore gives $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}$. The preceding segment is $0\to H_3(X)\to H_2(U\cap V)\to 0$, giving $H_3(X)=\mathbb{Z}$. Connectedness gives $H_0(X)=\mathbb{Z}$, and all higher groups vanish. HenceAs a geometric check, the identity boundary gluing identifies the product with $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.

For ordered simplicial cochains, define the Alexander-Whitney cup product byusing the front and back faces of each characteristic simplex. Write $\delta$ for the coboundary, $\delta \varphi =\varphi \partial$. Expanding the face-boundary formula gives the graded Leibniz identityIn this expansion, face omissions on the front and back portions supply the two displayed terms, and their extra terms at the shared vertex cancel with opposite signs. Hence the cup of cocycles is a cocycle. If $\varphi$ changes by $\delta a$, its product with a cocycle changes by $\delta (a⌣\psi )$; changing $\psi$ by $\delta b$ changes it by $(-1{)}^i\delta (\varphi ⌣b)$. The product therefore descends to cohomology. It is associative by the three consecutive-face formula and has the constant degree-zero cochain as unit.

In the wedge, every positive-dimensional simplex belongs to exactly one factor. Restricting positive-degree cochains therefore gives a direct sum of the two cochain groups and splits the coboundaries in degrees above one. In degree one, the same splitting of coboundaries holds: any pair of degree-zero cochains on the factors can have their values at the glued vertices made equal by adding a constant to one factor, which does not change its coboundary. Thus restriction inducesRepresent classes from the two factors by cochains extended by zero on the other factor. If both degrees are positive, a simplex lies wholly in one factor, so one of the front or back evaluations is zero. Their cup product is already zero as a cochain. This proves all the stated mixed products vanish.

For the indicated wedge there is one unit in degree zero, two generators $a,b$ in degree one, and one generator $c$ in degree two; higher groups vanish. Every product of positive-degree elements is zero. Its ring is thereforeThe torus has the same additive groups, but its degree-one generators have nonzero cup product generating degree two. The two graded rings are not isomorphic. Since a homotopy equivalence induces a cohomology-ring isomorphism, the torus and this wedge are not homotopy equivalent.

With the usual second-countability convention, an $n$-manifold is a Hausdorff second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^n$. A local orientation of a manifold at $x$ is a choice of generator ofExcision and a coordinate ball identify this relative group with $H_n({\mathbb{R}}^n,{\mathbb{R}}^n\setminus \{0\})$. An orientation of a topological manifold is a coherent choice of such generators: near each point the choices are induced by one relative class on a neighbourhood, equivalently by an atlas with orientation-preserving transition maps.

For a connected closed oriented manifold, its fundamental class $[M]\in H_n(M;\mathbb{Z})$ is the unique class mapping to every chosen local generator. Here closed means compact without boundary. Poincare duality states that cap product with this class is an isomorphismOver rational coefficients it gives $b_k=b_{n-k}$. ThereforeWhen $n$ is odd this says $\chi =-\chi$, so $\chi (M)=0$. Torsion does not contribute to the Betti numbers or this Euler characteristic.

Yes, for a nonnegative integer $g$: the closed orientable surface of genus $g$ has precisely these groups. A standard polygon cell structure has one vertex, $2g$ one-cells and one two-cell attached by a product of commutators. Its integral cellular boundary maps are zero, because each generator occurs once in each orientation in the attaching word. Thus $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^{2g}$. At genus zero use the sphere.

No. Its top nonzero group would force dimension three. Poincare duality requires $H_2\cong H^1$, and the universal coefficient theorem gives $H^1\cong \mathrm{Hom}(H_1,\mathbb{Z})=\mathbb{Z}$, since $H_0$ is free. This contradicts the proposed vanishing of $H_2$. Equivalently its listed Betti numbers give Euler characteristic $-1$, impossible for a closed oriented odd-dimensional manifold.

Yes: for $n\ge 1$, the sphere $S^n$ is connected, closed and oriented, with $H_0=H_n=\mathbb{Z}$ and no intermediate homology. These groups do not require the manifold to be homeomorphic to a sphere, but the sphere provides the requested example.

No. The dimension would again be three. The universal coefficient theorem gives $H^1=0$, since both $\mathrm{Hom}(H_1,\mathbb{Z})$ and $\mathrm{Ext}(H_0,\mathbb{Z})$ vanish. Poincare duality then forces $H_2=0$, contradicting the proposed torsion group. In general the second integral homology of a connected closed oriented three-manifold is free abelian, being dual to its first homology.

Yes: real projective three-space ${\mathbb{RP}}^3$ has these groups. Its usual cell structure has one cell in each dimension zero through three, with boundary maps ${\partial }_1=0$, ${\partial }_2=2$ and ${\partial }_3=0$. Hence its integral homology is $\mathbb{Z}$ in degrees zero and three, $\mathbb{Z}/2$ in degree one, and zero in degree two. It is a connected closed oriented manifold: the antipodal map on $S^3$ preserves orientation, so the sphere's orientation descends to the quotient.