Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 5 / 1

Use the homotopy lifting theorem for a covering map: a homotopy of paths has a unique lift once its initial path is lifted. Also use uniqueness of lifts to a covering space: a lift of a constant path with a specified starting point is constant.

Let $\tilde{\gamma }$ be a loop based at ${\tilde{x}}_0$ whose image under $p_{\ast }$ is trivial. There is a homotopy $F:I\times I\to X$ from $p\tilde{\gamma }$ to the constant loop at $x_0$, fixing both endpoints. Lift this homotopy with initial path $\tilde{\gamma }$. Both vertical boundary paths are constant lifts at ${\tilde{x}}_0$. Its final path is a lift of the constant loop and hence is also constant at ${\tilde{x}}_0$. Thus the lifted homotopy contracts $\tilde{\gamma }$ relative to its endpoints. The kernel of the induced group homomorphism is trivial, provingNo universal-cover assumption is needed for this injection; it holds for any actual covering map.

For the classical subgroup-covering construction, assume $X$ is path-connected, locally path-connected, and semilocally simply connected. These are the usual hypotheses underlying the classification of connected covering spaces; an arbitrary space, as literally allowed by the wording, need not admit every subgroup covering.

Construct the path-class construction of a subgroup covering directly. Consider paths $\alpha$ starting at $x_0$, and identify $\alpha ,\beta$ when they have the same endpoint and the loop class $[\alpha {\beta }^{-1}]$ belongs to $H$. The subgroup properties make this an equivalence relation. Let $Y$ be its set of classes and define $p([\alpha ])=\alpha (1)$.

Choose a path-connected open neighbourhood $U$ of $\alpha (1)$ whose loops become null-homotopic in $X$. Such neighbourhoods form a local basis under the stated local hypotheses. Define a sheetFor each point of $U$ the resulting class is independent of the chosen path $\eta$, since the difference is a loop in $U$, conjugated by $\alpha$, and is trivial in ${\pi }_1(X,x_0)$. Thus $p$ maps this sheet bijectively onto $U$. Give $Y$ the topology generated by these sheets. Restrictions to smaller such neighbourhoods show that the sheets form a basis and that their endpoint maps are homeomorphisms. Two sheets over $U$ that intersect coincide, by concatenating and reversing paths within $U$; otherwise they are disjoint. Therefore $p^{-1}(U)$ is a disjoint union of copies of $U$, and $p$ is a covering map.

The basepoint is the class of the constant path. The path $\alpha$ lifts to its successive prefix classes, giving a path from this basepoint to $[\alpha ]$; hence $Y$ is path-connected. A loop $\gamma$ lifts closed exactly when its final prefix class agrees with the constant class, namely when $[\gamma ]\in H$. Consequently $p_{\ast }({\pi }_1(Y))=H$. Part (i) supplies injectivity, so

The missing local hypothesis is substantive. In the Hawaiian earring, every neighbourhood of the common basepoint contains a whole sufficiently small circle. Its loop is nontrivial in the whole space, since retracting onto that circle preserves it. An evenly covered neighbourhood would lift that loop to a closed loop in one sheet, so the image subgroup of any covering contains a nontrivial element. Therefore the trivial subgroup cannot be realized by a simply connected covering of this space. The intended result is valid under the local covering-space hypotheses above, not for every topological space.

Let $F$ be the free group on a set $A$. Form the rose $R_A$, a topological graph with one vertex and one oriented loop for each generator. Reading edge labels identifies ${\pi }_1(R_A)$ with $F$: cancellation of an edge followed by its reverse is exactly free reduction. This also applies to an infinite generating set, using the graph's CW topology rather than an accumulation topology.

Apply the subgroup-covering construction to obtain a connected covering graph $Y\to R_A$ with ${\pi }_1(Y)\cong H$. Choose a spanning tree $T\subseteq Y$ and a base vertex $v_0$. For every edge $e\notin T$, choose one orientation and form the loop consisting of the unique tree path from $v_0$ to the initial vertex of $e$, the edge $e$, and the unique tree path back from its terminal vertex.

These loops generate ${\pi }_1(Y)$. Indeed, in any edge loop insert the tree paths to and from the base vertex between consecutive edges. A tree edge contributes a contractible loop; a non-tree edge contributes the chosen loop or its inverse. They are freely independent as well: reducing an edge path cancels backtracking, and, after suppressing the tree portions, a nonempty freely reduced word in the oriented non-tree edges cannot disappear. Equivalently, collapsing the tree gives a rose whose petals are precisely the non-tree edges, and the same edge-path reduction identifies its fundamental group with the free group on those petals.

Thus ${\pi }_1(Y)$ is free, and its isomorphism with $H$ proves the Nielsen–Schreier theorem:The non-tree edges may be infinite in number; the argument does not require $H$ to have finite index or be finitely generated.