Why Covering-Space Arguments Need Care
Many lifting arguments have a short outline: choose a point over a base point, lift paths, and use the lifted endpoints to define a map. Each step has a condition attached to it. Omitting the chosen point in the fiber can lose control of which lift is being constructed; assuming path independence can make the proposed definition contradictory; and writing \(p\circ g=f\) does not by itself prove that \(g\) is continuous. This tutorial examines these errors through explicit examples.
A map \(p:E\to Y\) is a covering map if it is surjective and every \(y\in Y\) has an open neighborhood \(V\) such that \(p^{-1}(V)\) is a disjoint union of open sets, each mapped homeomorphically onto \(V\) by \(p\). Such a \(V\) is evenly covered. A lift of \(f:X\to Y\) is a map \(g:X\to E\) satisfying \(p\circ g=f\). A path is a continuous map \(\alpha:I\to Y\), where \(I=[0,1]\); a lift of \(\alpha\) is a path \(\widetilde{\alpha}:I\to E\) with \(p\circ\widetilde{\alpha}=\alpha\).
The conclusion is not merely “there is a lift.” It is “there is a lift taking the specified value \(e_0\) at \(x_0\).” The subgroup on the right also depends on that choice. When applying the criterion, record the full based data \((x_0,e_0)\).
Worked Example: What the Chosen Fiber Point Specifies
Let \(Y=S^1\) with its usual topology, let \(E=S^1\bigsqcup S^1\) have the disjoint-union topology, and let \(p:E\to Y\) be the identity map on each component. This is a covering: all of \(Y\) is evenly covered by the two components. Set \(X=Y\) and \(f=\operatorname{id}_Y\). There are two lifts, \(g_1\) mapping \(Y\) identically to the first component and \(g_2\) mapping it identically to the second.
Choose \(x_0\in Y\). If \(e_0\) is the point corresponding to \(x_0\) in the first component, then \(g_1(x_0)=e_0\), whereas \(g_2(x_0)\ne e_0\). Choosing the corresponding point in the second component instead selects \(g_2\). In both cases the subgroup condition holds: on the selected component, \(p\) is a homeomorphism, so \(p_*\) maps its fundamental group onto \(\pi_1(Y,x_0)\). The condition therefore produces a lift with the prescribed value; it does not say that there is only one lift if no value is prescribed.
Check: for each selected \(e_0\), the corresponding component map is a lift and has value \(e_0\) at \(x_0\). The two lifts differ at every point of \(Y\), so omitting the fiber point loses genuine information.
Error 1: Assuming Path Independence
The path construction defines a candidate value at \(x\in X\) by choosing a path \(\gamma\) from \(x_0\) to \(x\), lifting \(f\circ\gamma\) from \(e_0\), and taking its endpoint. For this to define a function, different choices of \(\gamma\) must give the same endpoint. The lifting criterion is what guarantees this independence under its hypotheses; it cannot simply be assumed.
Worked Example: Two Paths Give Different Endpoints
Write \([u]_1\) for the class of \(u\in\mathbb{R}\) in \(Y=\mathbb{R}/\mathbb{Z}\), and \([u]_2\) for its class in \(E=\mathbb{R}/2\mathbb{Z}\). Give both spaces their quotient topologies and define \(p:E\to Y\) by \(p([u]_2)=[u]_1\). This is a covering: an arc in \(Y\) of length less than \(1\) has two disjoint inverse-image arcs in \(E\), and \(p\) maps each homeomorphically onto the chosen arc. Take \(f=\operatorname{id}_Y\), \(x_0=[0]_1\), and \(e_0=[0]_2\).
Consider the constant path \(\gamma_0(t)=[0]_1\). Its lift starting at \(e_0\) is \(\widetilde{\gamma}_0(t)=[0]_2\), so its endpoint is \([0]_2\). Now consider the loop \(\gamma_1(t)=[t]_1\), which has the same initial and final points. Its lift starting at \(e_0\) is \(\widetilde{\gamma}_1(t)=[t]_2\), because \(p([t]_2)=[t]_1\) at every \(t\). Its endpoint is \([1]_2\), which differs from \([0]_2\): their difference, \(1\), is not an integer multiple of \(2\).
Thus the proposed path-defined value at \(x_0\) would have to be both \([0]_2\) and \([1]_2\). There is no lift of \(\operatorname{id}_Y\) taking \([0]_1\) to \([0]_2\): if such a lift existed, composing it with either loop would give a lift of that loop starting at \([0]_2\), and uniqueness of path lifts would force the two loop endpoints to agree. They do not.
The precise repair is to compare any two paths with the same endpoints. Following the first and then the reverse of the second gives a loop. The subgroup condition in the lifting criterion ensures that the image of this loop lifts to a loop at the chosen \(e_0\); consequently the two lifted path endpoints agree. The example shows why that step cannot be replaced by an unsupported assertion of independence.
Error 2: Reversing the Order of Loop Actions
For paths \(\alpha\) and \(\beta\) with \(\alpha(1)=\beta(0)\), let \(\alpha*\beta\) mean “traverse \(\alpha\), then traverse \(\beta\).” For a loop \(\alpha\) based at \(y_0\), define \(T_\alpha\) on the fiber \(p^{-1}(y_0)\) by lifting \(\alpha\) from a point of the fiber and taking the endpoint. Path lifting and its uniqueness are standard results from the earlier material on covering spaces.
Proof. Fix \(e\in p^{-1}(y_0)\). Let \(\widetilde{\alpha}\) be the lift of \(\alpha\) starting at \(e\), and let \(e_1=T_\alpha(e)\). Let \(\widetilde{\beta}\) be the lift of \(\beta\) starting at \(e_1\). The path formed by traversing \(\widetilde{\alpha}\) and then \(\widetilde{\beta}\) projects to \(\alpha*\beta\) and starts at \(e\). By uniqueness of the path lift starting at \(e\), it is the lift of \(\alpha*\beta\). Its endpoint is \(T_\beta(e_1)=T_\beta(T_\alpha(e))\). This holds for every \(e\) in the fiber, proving the formula. \(\square\)
Worked Example: Checking the Order with Permutations
Consider the bouquet of two circles, with loops labelled \(a\) and \(b\), and a three-sheeted graph covering over it. Such a covering can be built with three vertices over the bouquet vertex: the lifted \(a\)-edges connect sheet \(i\) to sheet \(\sigma_a(i)\), and the lifted \(b\)-edges connect sheet \(i\) to sheet \(\sigma_b(i)\), where \(\sigma_a=(12)\) and \(\sigma_b=(23)\). Each edge maps homeomorphically to its labelled circle edge; at each vertex there is one lift of each incident edge direction, so small neighborhoods of the bouquet vertex are evenly covered.
Starting on sheet \(1\), traverse \(a\) and then \(b\). The first loop sends \(1\) to \(\sigma_a(1)=2\); the second sends \(2\) to \(\sigma_b(2)=3\). Thus \(T_{a*b}(1)=3\). The correct function composition gives \(\bigl(T_b\circ T_a\bigr)(1)=T_b(2)=3\). Reversing the composition order gives \(\bigl(T_a\circ T_b\bigr)(1)=T_a(\sigma_b(1))=T_a(1)=2\), which is not the endpoint. Therefore the reversed formula fails in this example.
Check: \(\sigma_b(1)=1\), since the transposition \((23)\) fixes \(1\); and \(\sigma_a(1)=2\), since \((12)\) exchanges \(1\) and \(2\). These values account for the incorrect endpoint \(2\).
Error 3: Treating the Lift Equation as a Continuity Proof
The equation \(p\circ g=f\) is necessary for a lift, but it does not imply that \(g\) is continuous. Continuity of a path-defined lift is proved using evenly covered neighborhoods and the local inverse branches of \(p\), together with the hypotheses used in the path construction. It is not a consequence of the equation alone.
Worked Example: A Discontinuous Lift of a Continuous Map
Give \(I=[0,1]\) its usual topology, give \(\{0,1\}\) the discrete topology, and give \(E=I\times\{0,1\}\) the product topology. Let \(p:E\to I\) be \(p(t,i)=t\). This is a covering: for every open \(U\subseteq I\), \(p^{-1}(U)=U\times\{0\}\,\cup\,U\times\{1\}\), and each sheet maps homeomorphically to \(U\). Let \(f=\operatorname{id}_I\), and define \(g(t)=(t,0)\) when \(t\in\mathbb{Q}\), and \(g(t)=(t,1)\) when \(t\notin\mathbb{Q}\).
For every \(t\in I\), \(p(g(t))=t=f(t)\). Nevertheless \(g\) is not continuous. The set \(I\times\{0\}\) is open in \(E\), while its preimage under \(g\) is \(\mathbb{Q}\cap I\). This preimage is not open in \(I\): every relative neighborhood of any rational point in \(I\) contains irrational points. Hence the inverse-image test for continuity fails.
The lift equation has been checked pointwise, but continuity has not followed. In contrast, the path-defined lift theorem proves continuity by restricting to suitable neighborhoods on which the covering map has a continuous inverse branch.
Error 4: Confusing Path-Lift and Map-Lift Uniqueness
Uniqueness of path lifts says that two lifts of the same path with the same starting point agree. Uniqueness of map lifts has a different form: connectedness of the domain and agreement at one base point are essential. The following proposition gives the exact condition needed for this uniqueness argument.
Proof. Let \(A=\{x\in X:g(x)=h(x)\}\). We show that both \(A\) and \(X\setminus A\) are open. Fix \(x\in X\), and choose an evenly covered open neighborhood \(V\) of \(f(x)\). By continuity of \(f,g,h\), there is an open neighborhood \(W\) of \(x\) such that \(f(W)\subseteq V\), and such that \(g(W)\) and \(h(W)\) lie in sheets over \(V\). If \(g(x)=h(x)\), the two values lie in the same sheet. On that sheet \(p\) is one-to-one, so for every \(z\in W\), the equalities \(p(g(z))=f(z)=p(h(z))\) imply \(g(z)=h(z)\). Thus \(W\subseteq A\). If \(g(x)\ne h(x)\), they lie in distinct sheets; the sheets are disjoint, so \(g(z)\ne h(z)\) for every \(z\in W\). Thus \(W\subseteq X\setminus A\). Both sets are open. Since \(x_0\in A\), the set \(A\) is nonempty. Connectedness of \(X\) prevents \(A\) and \(X\setminus A\) from being two disjoint nonempty open sets covering \(X\), so \(A=X\), as required. \(\square\)
Worked Example: Why a Base Point and Connectedness Matter
Use again \(X=Y\) with its usual topology and the covering \(p:Y\bigsqcup Y\to Y\) that is the identity on each component. For \(f=\operatorname{id}_Y\), the two component inclusions are distinct lifts. They do not agree at any point of \(X\), so the uniqueness proposition does not apply to that pair. Prescribing a value at \(x_0\) selects one of them.
Now take the disconnected domain \(X=\{u,v\}\) with the discrete topology, let \(Y=\{*\}\) have the one-point topology, and let \(E=\{e_0,e_1\}\) have the discrete topology. The unique map \(p:E\to Y\) is a covering. The unique map \(f:X\to Y\) has four lifts, since each of \(u\) and \(v\) can independently be sent to either point of \(E\). For example, the lifts \(g(u)=h(u)=e_0\), \(g(v)=e_0\), and \(h(v)=e_1\) agree at \(u\) but differ at \(v\). All these maps are continuous because their domains are discrete. Thus agreement at one point does not force equality when the domain is disconnected.
Path-lift uniqueness is still applicable to a fixed path with a fixed initial lift. It cannot replace the connected-domain argument for map lifts: a map on a disconnected domain can make independent choices on different components.
A Reliable Check Before Concluding
When auditing a covering-space argument, write down the data and the claim before using a lifting result:
- For a lifting criterion, identify \(x_0\) and the chosen \(e_0\in p^{-1}(f(x_0))\).
- For a path-defined map, compare endpoints from any two paths with the same endpoints; do not presume independence.
- For concatenated paths, state whether the first path is traversed first. With \(\alpha*\beta\) meaning \(\alpha\) then \(\beta\), the endpoint operators compose as \(T_\beta\circ T_\alpha\).
- For a proposed lift, check both \(p\circ g=f\) and continuity of \(g\).
- For uniqueness of map lifts, check that the lifts agree at a point and that the domain is connected.
Common pitfall. A frequent incorrect argument says, “The lifts agree at the base point, so uniqueness of path lifts implies the maps agree everywhere.” Path-lift uniqueness applies only after one has selected a particular path in the domain and compared the two lifts along that path. To reach an arbitrary point, the maps must be restricted along a path from the base point; that reasoning needs the domain to be path-connected. The map-lift proposition above gives the broader connected-domain result directly, using evenly covered neighborhoods. Neither conclusion permits dropping the base-point agreement or the connectedness hypothesis.
Check Your Understanding
For each question, name the hypothesis that prevents the corresponding error.
- In the lifting criterion, what point in the fiber must be specified, and what value must the resulting lift take at the domain base point?
- In the quotient covering \(\mathbb{R}/2\mathbb{Z}\to\mathbb{R}/\mathbb{Z}\), compare the lifted endpoints of the constant loop and \(t\mapsto[t]_1\), both starting at \([0]_2\).
- If \(\alpha\) is traversed before \(\beta\), write the endpoint-operator formula for \(T_{\alpha*\beta}\).
- Why does \(p\circ g=f\) not prove that a proposed lift \(g\) is continuous? Give the inverse-image test used in the discontinuous example.
- State the connectedness and base-point hypotheses that allow uniqueness of lifts of maps.