Past exam of the topology and groups course of the University of Oxford / 2023 / 3 / a

A covering map is a continuous surjection $p:\tilde{X}\to X$ such that every $x\in X$ has an open neighbourhood $U$ for whichwhere the $U_{\alpha }$ are pairwise disjoint open subsets of $\tilde{X}$. Such a $U$ is an evenly covered neighbourhood, and the sets $U_{\alpha }$ are its covering sheets. The index set may be infinite. The requirement that every sheet map onto all of $U$ is part of the definition.

For a covering map over a nonempty connected space, the degree of a covering map isthe cardinality of any fibre of a function. To show independence of $x$, choose an evenly covered neighbourhood $U$. Each covering sheet contains exactly one point over every $y\in U$, so $|p^{-1}(y)|$ equals the number of covering sheets over $U$. The fibre of a function cardinality is therefore locally constant. For a fixed cardinal number $\kappa$, the points with that cardinality form an open subset, and their complement is also open. Connectedness forces the value to be constant everywhere.

For a path-connected base there is also an explicit fibre bijection by path lifting: lift a path from $x$ to $y$ starting at each point over $x$ and take its endpoint. The path lifting theorem gives existence and uniqueness; path reversal gives the inverse bijection. The bijection may depend on the path, but the resulting equality of cardinalities does not.

A connected base is needed for one global degree. Without it, use a degree on each connected component on which the cardinality is constant. For example, a covering map of a discrete two-point topological space can have one point over the first point and two over the second. Thus the unqualified printed definition needs the usual connected-base convention, or the explicit assumption that every fibre of a function has the same cardinality.

With local path connectedness, the intended conclusion holds. Choose a base point $x_0$ and the standard covering mapThe fundamental group of the circle is $\mathbb{Z}$. Since the image of a finite group under a group homomorphism is finite and $\mathbb{Z}$ has no nontrivial finite subgroups,is trivial. If $X$ is also locally path-connected, the lifting criterion for a covering space supplies a continuous map $\tilde{f}:X\to \mathbb{R}$ with $p\tilde{f}=f$, because $p_{\ast }{\pi }_1(\mathbb{R})=0$. The homotopyruns from $f$ to the constant function $f(x_0)$. This proves the claimed null-homotopy under that additional hypothesis, in particular for CW complexes and the topological graphs considered later in the paper.

For arbitrary path-connected spaces, the printed assertion is false. Here is an explicit Warsaw circle counterexample, including the obstruction that the usual lifting argument misses. In the Euclidean plane takeJoin $P=(1,0)$ to $Q=(0,-1)$ by the polygonal arc $A$ passing successively through $(2,0),(2,-2),(0,-2)$, and put $W=G\cup L\cup A$ with its subspace topology. The arc meets $G\cup L$ only at $P,Q$. The topologist's sine curve $G\cup L$ has two path components, but the added arc joins them, making $W$ path-connected.

Every loop in $W$ is contained infor some $\delta >0$. To justify the assertion, a path cannot reach $L$ along the oscillating tail: the intermediate value theorem for its $x$-coordinate would force its $y$-coordinate to attain both $1$ and $-1$ arbitrarily close to the supposed arrival time, contradicting continuity. The only alternative route from the tail to $L$ uses $P$ and the joining arc. If a loop visited tail points with $x\to 0$, a convergent subsequence of their parameter values would approach a time at which the loop lies in $L$. On a sufficiently small parameter interval around that time its image avoids $P$. That would give a forbidden path from $L$ into the tail without passing through $P$. Thus those tail coordinates have a positive lower bound. The space $T_{\delta }$ is a finite tree, in fact a single arc, so every such loop contracts. ConsequentlyNow define a continuous map $h:W\to S^1$ to be $1$ on $G\cup L$, and to make one full turn on $A$: for a parametrization $a:[0,1]\to A$ from $P$ to $Q$, set $h(a(t))=e^{2\pi it}$. The pieces $G\cup L$ and $A$ are closed in $W$, and the formulas agree at $P,Q$, so the pasting lemma for closed subspaces proves continuity.

There is no continuous lift $\tilde{h}:W\to \mathbb{R}$. Any such lift is integer-valued on $G\cup L$, so it is constant on each of the connected spaces $G,L$. Since $G$ accumulates on $L$, continuity forces those two constants to be the same. But along $A$ a lift of $e^{2\pi it}$ has the form $t+k$, giving values at $Q$ and $P$ differing by $1$, a contradiction. Finally, if $h$ had a null-homotopy, the homotopy lifting theorem for a covering map, starting with a lift of the final constant function and reversing time, would produce a continuous lift of $h$. This theorem needs no local path connectedness of the parameter space. Hence $h$ is not null-homotopic, although $W$ has finite, indeed trivial, fundamental group.

The native PDF explicitly assumes only path connectedness here. The missing local path connectedness cannot be inferred from the finiteness of the fundamental group.