Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2009 / A (Topology and Groups)

An edge loop is a finite sequence of vertices $v_0,\dots ,v_m$, starting and ending at $b$, in which consecutive distinct vertices span an edge. Two such sequences are equivalent when related by finitely many elementary edge homotopies, with endpoints fixed: insert or delete a repeated vertex, cancel or insert a spur $(u,v,u)$, and replace $(u,v,w)$ by $(u,w)$ or conversely when the vertices lie in one simplex. Repetitions are allowed in this last convention; a nondegenerate move uses a triangle.

The edge-loop group $E(K,b)$ consists of these equivalence classes, with product given by concatenating the sequences at $b$. Elementary moves remain valid inside a longer concatenation, so this product is well defined. Concatenation is associative after the harmless repeated base vertex is removed; the constant edge loop is the identity. Reversing a sequence gives its inverse, because a sequence followed by its reverse cancels by successive spurs. Thus these operations do form a group.

Realize an edge loop as a piecewise linear path in the geometric realization $|K|$. A spur contracts in an edge, and a triangle move is a homotopy across that triangle, always fixing endpoints. Hence realization gives a well-defined map $\theta$ to the fundamental group. Realization respects concatenation up to endpoint-preserving reparametrization, so $\theta$ is a homomorphism.

The relative Simplicial approximation theorem says that a continuous map from a finite complex to a geometric realization is homotopic, after sufficiently fine subdivision of its domain, to a simplicial map, with a chosen already simplicial subcomplex kept fixed. Apply its based-circle version to a loop, keeping its marked vertex at $b$. The resulting subdivided simplicial circle traverses edges, so every based homotopy class has an edge-loop representative. Thus $\theta$ is surjective.

For the full two-skeleton, use the triangles through $b$ to replace each edge $(u,v)$ of a loop by $(u,b,v)$. Degenerate cases use repeated-vertex or spur moves. The resulting path contracts by spurs, so its edge-loop group, and therefore its fundamental group, is trivial. This also covers one or two vertices, when only repeated vertices and spurs are needed.

If the deleted triangle has vertices $u,v,w$, choose a fourth vertex $t$. Its boundary is filled by the three present triangles $(t,u,v)$, $(t,v,w)$ and $(t,w,u)$: their union is the cone from $t$ on that boundary. Every elementary homotopy that formerly crossed the missing triangle can therefore be replaced by a homotopy through these three triangles, staying in $|L|$. Since the one-skeleton has not changed, the same contraction of any edge loop is available in $L$. Surjectivity of its realization map gives

A canonical presentation of a group uses a symbol $[g]$ for each $g\in G$ and relations $[g][h]=[gh]$ for every pair. Evaluation to $G$ is surjective, and $g\mapsto [g]$ is its inverse homomorphism; the relation with $g=h=1$ implies $[1]=1$. Thus this presentation recovers any group.

The pushout of the two homomorphisms isHere $\ast$ is the free product, and the denominator is the normal closure. Its canonical maps $j_i:G_i\to G$ satisfy $j_1{\varphi }_1=j_2{\varphi }_2$. The universal property says that maps $u_i:G_i\to H$ agreeing on $G_0$ factor uniquely as $u_i=u{j}_i$ for one homomorphism $u:G\to H$. The free-product universal property first gives the map before quotienting, and agreement makes it kill precisely the imposed relations, proving this property.

Suppose $G_0$ is generated by $s_1,\dots ,s_r$, and $G_i=⟨S_i\mid R_i⟩$ are finite presentations with disjoint generator sets. Choose words $w_{ij}$ representing ${\varphi }_i(s_j)$. ThenAgreement on these generators implies agreement on every word in $G_0$, so these finitely many extra relations suffice by the universal property. No finite presentation for $G_0$ is required.

The Seifert-van Kampen theorem states that for a space covered by path-connected open sets $U,V$ with path-connected nonempty intersection and basepoint in that intersection, ${\pi }_1(U\cup V)$ is the pushout of ${\pi }_1(U)\leftarrow {\pi }_1(U\cap V)\to {\pi }_1(V)$.

To see the two bands in the Klein bottle model, take one region to be $1/4\le y\le 3/4$ and the other to be the outer strips $0\le y\le 1/4$, $3/4\le y\le 1$, joined along the top-bottom identification. Each is an interval bundle over the horizontal circle, and going once around reverses its transverse coordinate. For the middle band use the coordinate $y-1/2$; for the outer band use the signed coordinate about zero modulo one. Both are therefore Möbius bands. Their common boundary is the one circle represented by the lines $y=1/4,3/4$, joined at the identified vertical sides. They cover the whole bottle and are glued along that circle.

A Möbius band deformation retracts onto its core circle, so its fundamental group is infinite cyclic. Going once around its boundary traverses the core twice: one pass along one boundary edge changes sides, and the second pass returns. After choosing core generators and consistent boundary orientation, the two inclusion maps from the common circle send its generator to $x^2$ and $y^2$. Thicken the bands slightly so their overlap is an annular neighbourhood of their common boundary. Seifert-van Kampen theorem then givesReversing a chosen generator absorbs an opposite boundary parametrization, so the presentation is independent of that sign choice.

A lift of $f:Y\to X$ is a continuous map $\tilde{f}:Y\to \tilde{X}$ satisfying $p\tilde{f}=f$. The covering homotopy theorem says that a homotopy $H:Y\times I\to X$ has a unique lift once a continuous lift of its initial map $H(-,0)$ is prescribed. No additional local connectivity assumption on $Y$ is needed for this homotopy-lifting statement.

To prove injectivity on fundamental groups, let a loop $\tilde{\gamma }$ based at $\tilde{b}$ project to a based null-homotopic loop. Lift its null-homotopy, starting with $\tilde{\gamma }$. The two endpoint tracks lift constant paths at $b$ starting at $\tilde{b}$, so remain at $\tilde{b}$ by uniqueness of path lifting. The final path lies in the discrete fibre over $b$ and starts at $\tilde{b}$, so it is constant. Thus the lifted homotopy contracts $\tilde{\gamma }$ relative its basepoint, provingIn the following constructions, take $x$ around the circle factor and $y$ around the projective-plane factor. Above each sheet point, the circle has one incoming and one outgoing lifted $x$ edge. A fixed sheet for $y$ carries an identity cover of ${\mathbb{RP}}^2$; a transposed pair of sheets is joined as the antipodal fibre of a sphere covering ${\mathbb{RP}}^2$. These pieces give genuine covering neighbourhoods at the wedge point as well as away from it.

Use two sheet points $v_0,v_1$. Attach a copy of ${\mathbb{RP}}^2$ at each, each mapping identically onto the projective-plane summand. Put an oriented $x$ edge from $v_0$ to $v_1$ and another from $v_1$ to $v_0$, each mapping once along the cut-open circle. This is a connected two-sheeted cover.

Its underlying two-edge circle contributes one infinite cyclic free factor, and its two attached projective planes contribute two factors of order two. Collapsing one of the connecting edges, or applying Seifert-van Kampen theorem, givesChoose the first $x$ edge as a tree path to $v_1$. The circle loop projects to $x^2$, and the two projective-plane generators project to $y$ and $xy{x}^{-1}$. By injectivity from the preceding proof,The order-two relations come from the respective projective planes; there are no additional relations between these free factors.

Take the sphere double cover $S^2\to {\mathbb{RP}}^2$ and denote the two points above the wedge point by $v_0,v_1$. Attach one $x$ circle at each point, mapping identically onto the circle summand. At both points there is one incoming and one outgoing $x$ branch, so this is a connected two-sheeted cover of the wedge.

The sphere is simply connected, while each attached circle supplies one generator. Join the two attachment points by an arc in the sphere and collapse that contractible arc; the fundamental group is the free product of the two circle groups, hence free of rank two. A lift of $y$ joins $v_0$ to $v_1$. The two loop generators therefore project toInjectivity of the covering map identifies their subgroup with the free group on these two generators.

Put $r=n-1$. Take $r$ spheres, each mapping by the double cover onto ${\mathbb{RP}}^2$, and write their antipodal sheet points as $a_j,b_j$ for $0\le j<r$. Add an oriented $x$ edge from $a_j$ to $a_{j+1}$, with indices modulo $r$, and add an $x$ loop at each $b_j$. Every sheet point has the required two circle branches, so the resulting space is a $2r$-sheeted cover. It is connected because the $a_j$ edges form a cycle and each sphere connects its own two sheet points.

Choose a connecting arc from $a_j$ to $b_j$ in each sphere. Collapsing these arcs leaves the spheres simply connected, and leaves a graph with $r$ vertices, one cyclic edge at each step through the $a_j$, and one additional loop per vertex. This graph has $2r$ edges and rank $2r-r+1=r+1=n$. The sphere summands add no fundamental-group generators. Thus the cover has free fundamental group of rank $n$.

Based at $a_0$, choose the first $r-1$ cyclic $x$ edges as a spanning tree. The closing cycle projects to $x^r$. To traverse the loop at $b_j$, travel along the tree by $x^j$, use the sphere connector projecting to $y$, traverse its $x$ loop, and return. The resulting free basis in the base group isThere are exactly $n$ elements. For $n=2$ this is the previous pair $x,yxy$. Again injectivity of a covering's induced map makes these actual subgroups with the claimed free bases, rather than just quotient presentations.

This realizes Free subgroups of every finite rank in the free product of Z and C2.