The Subgroup Condition and Its Basepoints
A lift of a map does more than reproduce its values after projection: it also carries every based loop in the source to a based loop upstairs. Projecting that lifted loop recovers the original loop after applying the map. Passing to homotopy classes gives the necessary subgroup condition for a lift. The argument is short, but the basepoints and the direction of the inclusion matter.
We use based spaces \((X,x_0)\), \((B,b_0)\), and \((E,e_0)\). A loop at \(x_0\) is a continuous map \(\alpha:I\to X\) with \(\alpha(0)=\alpha(1)=x_0\), where \(I=[0,1]\). The fundamental group \(\pi_1(X,x_0)\) consists of based homotopy classes \([\alpha]\) of such loops, with multiplication given by concatenation. If \(g:(X,x_0)\to(Y,y_0)\) is continuous and \(g(x_0)=y_0\), its induced homomorphism is \(g_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)\), defined by \(g_*([\alpha])=[g\circ\alpha]\). This is well-defined because a based homotopy of loops remains a based homotopy after composition with \(g\).
The covering property gives useful local structure, but the necessary subgroup argument uses only that \(p\) is continuous. The condition concerns the induced homomorphisms at the specified basepoints: \(p_*\) has domain \(\pi_1(E,e_0)\), while \(f_*\) has domain \(\pi_1(X,x_0)\). The image of a homomorphism is a subgroup, so both sides of the condition below are subgroups of \(\pi_1(B,b_0)\).
Proof. Assume that \(\widetilde f\) is a based lift. Let \([\alpha]\) be any element of \(\pi_1(X,x_0)\), represented by a loop \(\alpha:I\to X\). The composite \(\widetilde f\circ\alpha:I\to E\) is a loop at \(e_0\), since \(\alpha(0)=\alpha(1)=x_0\) and \(\widetilde f(x_0)=e_0\). Thus \([\widetilde f\circ\alpha]\in\pi_1(E,e_0)\). By the lift identity \(p\circ\widetilde f=f\), $$ p_*\bigl([\widetilde f\circ\alpha]\bigr) =[p\circ\widetilde f\circ\alpha] =[f\circ\alpha] =f_*([\alpha]). $$ Therefore every element \(f_*([\alpha])\) belongs to the image of \(p_*\). Since \([\alpha]\) was arbitrary, the stated subgroup inclusion follows. This argument also covers the case where either fundamental group is trivial: its only element is the identity class, and the same loop calculation applies. \(\square\)
Why Composition Gives the Calculation
The proof can be understood as a compatibility between composing maps and passing to loop classes. We record that compatibility explicitly, so the subgroup argument does not depend on an unspoken rule about induced homomorphisms.
Proof. Let \([\alpha]\in\pi_1(X,x_0)\), represented by a loop \(\alpha\). By the definition of the induced homomorphism, $$ (h\circ g)_*([\alpha])=[(h\circ g)\circ\alpha]. $$ Associativity of composition gives \((h\circ g)\circ\alpha=h\circ(g\circ\alpha)\). Applying the definition first to \(g\) and then to \(h\) gives $$ h_*\bigl(g_*([\alpha])\bigr) =h_*\bigl([g\circ\alpha]\bigr) =[h\circ g\circ\alpha]. $$ The two classes are equal for every \([\alpha]\), so the homomorphisms agree. If one of the groups has only its identity element, this calculation still applies to that element; no nontrivial-loop assumption is needed. \(\square\)
For a based lift, \(p\circ\widetilde f=f\). Applying the proposition yields \(f_*=p_*\circ\widetilde f_*\). Hence the image of \(f_*\) is contained in the image of \(p_*\). The loop-by-loop proof above gives the same conclusion while showing exactly which loop upstairs produces each class downstairs.
Worked Subgroup Calculations
Worked Example: Following a Single Loop Class
Suppose a based map \(f:(X,x_0)\to(B,b_0)\) has a based lift \(\widetilde f:(X,x_0)\to(E,e_0)\). Take any loop \(\alpha\) at \(x_0\), and write \(u=[\alpha]\). The class contributed by \(u\) to the image of \(f_*\) is \(f_*(u)=[f\circ\alpha]\). The loop upstairs obtained from the same \(\alpha\) is \(\widetilde f\circ\alpha\), and it is based at \(e_0\).
Applying \(p_*\) to its class gives $$ p_*\bigl([\widetilde f\circ\alpha]\bigr) =[p\circ\widetilde f\circ\alpha] =[f\circ\alpha] =f_*(u). $$ Thus, for this particular input \(u\), the output \(f_*(u)\) is explicitly an element of \(p_*(\pi_1(E,e_0))\). Repeating this calculation for every \(u\in\pi_1(X,x_0)\) proves the full inclusion, not merely a claim about one chosen loop. As a check, applying the displayed equality to the constant loop gives the identity class on both sides, as it must.
Worked Example: An Image That Fails the Necessary Test
Consider subgroup data in which the image of \(p_*\) is \(4\mathbb Z\), and the image of \(f_*\) is \(6\mathbb Z\), with both regarded as subgroups of the same group \(\mathbb Z\). This is an algebraic test of the necessary condition: it does not assert that every such pair of subgroups comes from a particular covering diagram. We calculate whether the required inclusion holds.
The elements of \(4\mathbb Z\) are the integer multiples of \(4\), while those of \(6\mathbb Z\) are the integer multiples of \(6\). In particular, \(6=6\cdot1\) belongs to \(6\mathbb Z\). If \(6\) belonged to \(4\mathbb Z\), there would be an integer \(k\) with \(6=4k\), so \(k=6/4=3/2\), which is not an integer. Hence \(6\notin4\mathbb Z\), and therefore \(6\mathbb Z\not\subseteq4\mathbb Z\).
The necessary condition fails, so a based lift with these induced-image subgroups is impossible. This conclusion uses only the proved implication: existence of a based lift would force the inclusion. The failure of the inclusion is an obstruction, not a construction of a lift in any other case.
Worked Example: Checking an Image Generated by Two Elements
Suppose the target fundamental group is \(\mathbb Z\), the image of \(p_*\) is \(3\mathbb Z\), and the image of \(f_*\) is generated by \(6\) and \(9\). The subgroup generated by \(6\) and \(9\) consists of all integer combinations \(6a+9b\), where \(a,b\in\mathbb Z\). We check the inclusion by expressing each generator of the image of \(f_*\) as a multiple of \(3\): $$ 6=3\cdot2,\qquad 9=3\cdot3. $$ For arbitrary integers \(a,b\), $$ 6a+9b=3(2a+3b). $$ Since \(2a+3b\) is an integer, every such combination belongs to \(3\mathbb Z\). Thus the subgroup generated by \(6\) and \(9\) is contained in the image of \(p_*\).
In fact it equals \(3\mathbb Z\): the combination \(9-6=3\) belongs to the subgroup generated by \(6\) and \(9\), so every multiple of \(3\) belongs to that subgroup. This proves equality in this example. The necessary condition is satisfied, but that calculation alone does not prove that a lift exists; it verifies only the subgroup requirement.
Basepoints, Scope, and a Common Mistake
The basepoint condition on the lift is essential to the stated form of the theorem. It ensures that when \(\alpha\) starts and ends at \(x_0\), its lifted composite starts and ends at the chosen point \(e_0\), so it represents an element of \(\pi_1(E,e_0)\). Without this condition, the same composite might instead be a loop at a different point of the fiber \(p^{-1}(\{b_0\})\). Its class would then belong to a fundamental group with a different basepoint, not automatically to \(\pi_1(E,e_0)\).
No path-connectedness assumption on \(X\), \(E\), or \(B\) is needed for this necessary implication. The calculation begins with a loop at \(x_0\) and produces a loop at \(e_0\) directly. Nor does the proof use the local homeomorphisms in the definition of a covering: it applies to any continuous based maps \(p\) and \(f\) whenever \(p\circ\widetilde f=f\) and the lift preserves the chosen basepoint. The covering hypothesis becomes important in broader lifting questions, but it is not an extra step in this direction of the argument.
Common mistake: reversing the inclusion. The theorem says that the image of \(f_*\) must lie inside the image of \(p_*\), not the reverse. In the second example, \(6\in6\mathbb Z\) but \(6\notin4\mathbb Z\). So \(6\mathbb Z\not\subseteq4\mathbb Z\). Reversing the inclusion would instead ask whether \(4\mathbb Z\subseteq6\mathbb Z\); that also fails here, since \(4\in4\mathbb Z\) but \(4\notin6\mathbb Z\). The correct direction is determined by the loop calculation: start with a loop class in \(X\), apply \(f_*\), then exhibit that resulting class as the projection under \(p_*\) of a loop class in \(E\).
The subgroup condition is useful because it can rule out a lift before any attempt to define one. Its logical force is one-way: every based lift passes the test, while passing the test is only a necessary condition in this argument. To establish existence in a particular problem requires an additional construction and hypotheses appropriate to that construction.
Check Your Understanding
Use the loop calculation and keep track of the basepoints in each answer.
- Write the loop in \(E\) that proves \(f_*([\alpha])\in p_*(\pi_1(E,e_0))\) when a based lift is given.
- Explain why \(\widetilde f\circ\alpha\) represents an element of \(\pi_1(E,e_0)\), rather than a fundamental group based at another point.
- If \(\operatorname{im}(f_*)=10\mathbb Z\) and \(\operatorname{im}(p_*)=5\mathbb Z\), check the necessary inclusion by expressing an arbitrary element of \(10\mathbb Z\) as a multiple of \(5\).
- If \(f_*(\pi_1(X,x_0))\) is not contained in \(p_*(\pi_1(E,e_0))\), what conclusion about a based lift follows, and why?
- Which part of the proof uses the local homeomorphism condition in the definition of a covering map?