Past exam of the topology and groups course of the University of Oxford / 2023 / 2

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.