Past exam of the topology and Groups and Algebraic Topology course of the University of Oxford / 2009 / A / 3

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.