Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 133 / 1

Use the Schreier coset graph of the subgroup, with one directed edge labelled $s$ from $Hg$ to $Hgs$ for each $s\in \{a,b\}$. The homomorphismidentifies the cosets of $\ker \varphi$ with the integers. The covering graph therefore has vertices $v_n$, $n\in \mathbb{Z}$, an $a$-edgeand a $b$-loop at every $v_n$. Thus it is a doubly infinite $a$-line with one $b$-circle attached at each integer vertex.

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Forthe map $\varphi :F_2\to \mathbb{Z}$ is surjective because $\gcd (2,3)=1$. Again identify the cosets of $\ker \varphi$ with $\mathbb{Z}$. The covering has vertices $v_n$ and directed edgesThis labelled graph is connected: integer combinations of $2$ and $3$ reach every vertex. Each vertex has one incoming and one outgoing edge of each label, as required for a covering graph of the two-petalled rose.

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The based Schreier coset graph has vertices $⟨a⟩g$ and an $s$-edge from $⟨a⟩g$ to $⟨a⟩gs$. At the base vertex $⟨a⟩$ there is an $a$-loop. There are no other cycles: the fundamental group of the covering is the subgroup $⟨a⟩\cong \mathbb{Z}$, and its displayed loop already generates it.

Equivalently, start with one $a$-circle at the base vertex and attach labelled trees so that every vertex has exactly one incoming and one outgoing $a$-edge and exactly one incoming and one outgoing $b$-edge. After deleting the base $a$-loop, the underlying graph is a tree. This describes the complete infinite covering and distinguishes it from its finite core, which is just the $a$-loop.

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Whenever a combinatorial loop $\alpha$ traverses an oriented edge and immediately traverses the same edge backwards, delete that backtracking pair. Each deletion is a homotopy relative to endpoints inside $Z$ and reduces the edge length by two, so the process terminates at a reduced, hence locally injective, combinatorial loop $\beta$.

The universal cover $\tilde{Y}$ of a connected graph is a tree. The lift $\tilde{\beta }$ is also locally injective because a covering map is a local graph isomorphism. A locally injective edge path in a tree cannot repeat a vertex: the segment between two successive visits would be a nonempty reduced closed path, whereas every closed path in a tree backtracks. Thus $\tilde{\beta }$ is injective unless $\beta$ is constant.

Now let a loop in $Z$ become null-homotopic in $Y$. Its reduced representative $\beta$ lifts to a closed path in $\tilde{Y}$. The preceding injectivity forces that lift, and hence $\beta$, to be constant. The original loop is null-homotopic in $Z$, proving thatis an injective group homomorphism.

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Choose a finite generating set $h_1,\dots ,h_r$ for $H$. Under the standard classification of connected covering spaces, each $h_i$ is represented by a based combinatorial loop ${\gamma }_i$ in $Y$. Let $Z$ be the union of the images of these finitely many finite edge paths. Then $Z$ is a finite connected subgraph containing $y_0$.

Part (b) makes the inclusion-induced mapinjective. Its image contains every $h_i$, because every ${\gamma }_i$ lies in $Z$, and therefore contains the subgroup they generate, namely all of $H$. The map is consequently an isomorphism.

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