The Fundamental Group Criterion for Lifting Maps

Why Lifting Depends on Loops

A covering map looks locally like a collection of separate copies of its base, but a map into the base may travel around loops in a way that cannot be followed continuously in any one of those copies. The fundamental group records this loop behavior. The lifting criterion makes the obstruction precise: every loop traced by the map must represent a loop class that comes from the chosen point of the covering space.

Let $p:E\to B$ be a covering map, let $X$ be a topological space, and let $f:X\to B$ be continuous. A lift of $f$ through $p$ is a continuous map $\widetilde f:X\to E$ such that $p\circ\widetilde f=f$. Fix $x_0\in X$, put $b_0=f(x_0)$, and choose $e_0\in E$ with $p(e_0)=b_0$. The induced homomorphisms on fundamental groups are based at these points.

Definition: The fundamental group $\pi_1(X,x_0)$ consists of based homotopy classes of loops at $x_0$, with concatenation as its group operation. For a based continuous map $h:(X,x_0)\to(Y,y_0)$, the induced homomorphism $h_*:\pi_1(X,x_0)\to\pi_1(Y,y_0)$ sends the class of a loop $\alpha$ to the class of $h\circ\alpha$. Thus $f_*\pi_1(X,x_0)$ is a subgroup of $\pi_1(B,b_0)$, and $p_*\pi_1(E,e_0)$ is another subgroup of that same group.

Here a covering map means a continuous surjection $p:E\to B$ such that each $b\in B$ has an open neighborhood $V$ whose preimage is a disjoint union of open sets, each mapped homeomorphically onto $V$ by $p$. We use the path lifting theorem, uniqueness of path lifts, and the homotopy lifting property for coverings from the earlier material on covering spaces.

The Fundamental Group Criterion

Theorem (Lifting criterion): Let $p:E\to B$ be a covering map, let $X$ be path-connected and locally path-connected, and let $f:X\to B$ be continuous. Choose $x_0\in X$ and $e_0\in p^{-1}(f(x_0))$. There exists a lift $\widetilde f:X\to E$ with $\widetilde f(x_0)=e_0$ if and only if $$ f_*\pi_1(X,x_0)\subseteq p_*\pi_1(E,e_0). $$

Proof. First suppose a lift $\widetilde f$ exists and $\widetilde f(x_0)=e_0$. For any loop $\alpha$ at $x_0$, the composition $\widetilde f\circ\alpha$ is a loop at $e_0$, and $p\circ\widetilde f\circ\alpha=f\circ\alpha$. Therefore $f_*[\alpha]=p_*[\widetilde f\circ\alpha]$, which belongs to $p_*\pi_1(E,e_0)$. This holds for every loop class, proving the inclusion.

Conversely, suppose $f_*\pi_1(X,x_0)\subseteq p_*\pi_1(E,e_0)$. For each $x\in X$, choose a path $\alpha$ from $x_0$ to $x$. Lift $f\circ\alpha$ to a path in $E$ starting at $e_0$, and provisionally define $\widetilde f(x)$ to be its endpoint. We first check that this value does not depend on the choice of $\alpha$. Suppose $\beta$ is another path from $x_0$ to $x$, and write $\overline{\beta}$ for $\beta$ traversed in reverse. The concatenation $\alpha*\overline{\beta}$ is a loop at $x_0$. By the assumed inclusion, its image under $f$ represents $p_*[\lambda]$ for some loop $\lambda$ at $e_0$. The homotopy lifting property, together with endpoint invariance under path homotopy, says that the lift of $f\circ(\alpha*\overline{\beta})$ starting at $e_0$ ends at the same point as the lift of $p\circ\lambda$ starting there. The latter lift is $\lambda$, so it ends at $e_0$. If the lifts of $f\circ\alpha$ and $f\circ\beta$, both starting at $e_0$, end at $u$ and $v$, respectively, then the lift of the reversed path $f\circ\overline{\beta}$, starting at $u$, ends at $e_0$. Reversing that lifted segment gives a lift of $f\circ\beta$ starting at $e_0$ and ending at $u$. Uniqueness of path lifts makes $u=v$. Thus $\widetilde f(x)$ is well defined, and the constant path at $x_0$ gives $\widetilde f(x_0)=e_0$.

It remains to prove continuity. Fix $x\in X$. Choose an evenly covered open neighborhood $V$ of $f(x)$. By continuity of $f$, the set $f^{-1}(V)$ is an open neighborhood of $x$. Local path-connectedness gives a path-connected open neighborhood $U$ of $x$ contained in $f^{-1}(V)$. Let $W$ be the sheet over $V$ containing $\widetilde f(x)$; thus $p|_W:W\to V$ is a homeomorphism. For every $y\in U$, choose a path $\delta$ in $U$ from $x$ to $y$. The lift used to define $\widetilde f(y)$ can be taken to be a path from $x_0$ to $x$, followed by $\delta$. Along $\delta$, the map $f$ stays in $V$. Its lift starting at $\widetilde f(x)$ therefore stays in $W$, by path lifting and the fact that the sheet $W$ maps homeomorphically to $V$. Consequently $$ \widetilde f(y)=(p|_W)^{-1}(f(y))\qquad (y\in U). $$ The right-hand side is continuous on $U$, so $\widetilde f$ is continuous near $x$. Since $x$ was arbitrary, $\widetilde f$ is continuous on $X$. This proves sufficiency and completes the criterion. $\square$

The two hypotheses on $X$ play different roles. Path-connectedness lets us reach every point by a path from $x_0$, which is needed to define the candidate lift everywhere. Local path-connectedness gives small path-connected neighborhoods, which is used to prove that the candidate is continuous. The subgroup condition alone does not replace either hypothesis.

Worked Applications

Worked Example: Maps Between Circles

Give $S^1$ its subspace topology from $\mathbb C$, and consider the $n$-fold covering $p_n:S^1\to S^1$, $p_n(z)=z^n$, for $n\in\mathbb N$. Let $f_m:S^1\to S^1$ be $f_m(z)=z^m$, where $m\in\mathbb Z$. Determine exactly when $f_m$ lifts through $p_n$ to a lift taking $1$ to $1$.

Use the standard identification $\pi_1(S^1,1)\cong\mathbb Z$, in which the integer is the winding number. The map $f_m$ induces multiplication by $m$, so $f_{m*}\pi_1(S^1,1)=m\mathbb Z$. The covering $p_n$ induces multiplication by $n$, so $p_{n*}\pi_1(S^1,1)=n\mathbb Z$. The criterion is therefore $m\mathbb Z\subseteq n\mathbb Z$, which holds exactly when $n$ divides $m$. This includes $m=0$: then $m\mathbb Z=\{0\}$, and the constant map has the constant lift $1$.

For a positive instance, take $m=6$ and $n=3$. Since $6\mathbb Z\subseteq3\mathbb Z$, a lift is $\widetilde f(z)=z^2$. Direct substitution gives $p_3(\widetilde f(z))=(z^2)^3=z^6=f_6(z)$, and $\widetilde f(1)=1$.

For a negative instance, take $m=2$ and $n=3$. The class $2\in2\mathbb Z$ does not belong to $3\mathbb Z$, so the inclusion fails and there is no lift taking $1$ to $1$. This agrees with the direct obstruction: a lift $z\mapsto z^k$ would require $3k=2$, which has no integer solution. The subgroup test rules out any continuous based lift, not only lifts of this displayed form.

Worked Example: A Map from the Torus

Give $T^2=S^1\times S^1$ the product topology. Consider $f:T^2\to S^1$ defined by $f(z,w)=z^4w^8$, and the covering $p_4:S^1\to S^1$, $p_4(u)=u^4$. Use the basepoint $(1,1)$ and choose $1$ above its image.

The fundamental group of the torus is $\mathbb Z^2$, with the two coordinates recording winding in the two circle factors. A loop with winding numbers $(a,b)$ is sent by $f$ to a loop with winding number $4a+8b$. Thus $f_*\pi_1(T^2,(1,1))=4\mathbb Z+8\mathbb Z=4\mathbb Z$. The covering sends winding number $k$ to $4k$, so $p_{4*}\pi_1(S^1,1)=4\mathbb Z$. The required inclusion holds.

Indeed, define $\widetilde f(z,w)=zw^2$. Then $$ p_4(\widetilde f(z,w))=(zw^2)^4=z^4w^8=f(z,w), $$ and $\widetilde f(1,1)=1$. As a check on the induced-group calculation, the loop with winding numbers $(1,0)$ maps to winding number $4$, and the loop with winding numbers $(0,1)$ maps to winding number $8$; both are in $4\mathbb Z$, as are all their integer combinations.

Worked Example: A Map from the Real Line

Give $\mathbb R$ its standard topology and $S^1$ its subspace topology from $\mathbb C$. Let $p:\mathbb R\to S^1$ be $p(s)=e^{2\pi i s}$, the standard covering. Set $x_0=0$, $e_0=0$, and define $f:\mathbb R\to S^1$ by $f(t)=e^{2\pi i(t^2+3t)}$.

Every loop in $\mathbb R$ is null-homotopic: a loop $\alpha$ based at $0$ contracts through the based homotopy $H(r,s)=(1-r)\alpha(s)$, where $r,s\in[0,1]$. Hence $\pi_1(\mathbb R,0)$ is trivial and $f_*\pi_1(\mathbb R,0)=\{0\}$. The subgroup inclusion in the criterion holds for any subgroup $p_*\pi_1(\mathbb R,0)$, in particular for this covering. The predicted lift is $\widetilde f(t)=t^2+3t$. Substitution gives $p(\widetilde f(t))=e^{2\pi i(t^2+3t)}=f(t)$, and $\widetilde f(0)=0$, as required.

A Common Pitfall: Reversing the Inclusion

The criterion requires that every loop class produced by $f$ belongs to the subgroup produced by $p$. In symbols, the image under $f_*$ must be the smaller subgroup or equal to the subgroup under $p_*$; the reverse inclusion is not enough.

Wrong calculation. For $f_2:S^1\to S^1$, $f_2(z)=z^2$, and the three-fold covering $p_3(z)=z^3$, one might claim that a lift exists because $3\mathbb Z\subseteq2\mathbb Z$. That inclusion points in the wrong direction: it compares the covering subgroup with the image subgroup, not the required image subgroup with the covering subgroup.

Correction. The actual test is $2\mathbb Z\subseteq3\mathbb Z$. Since $2\in2\mathbb Z$ but $2\notin3\mathbb Z$, the test fails and no lift with the chosen basepoint exists. This example also shows why it is important to name both induced subgroups and keep their basepoints fixed: the criterion is about the specific groups $f_*\pi_1(X,x_0)$ and $p_*\pi_1(E,e_0)$.

Check Your Understanding

For each question, specify the basepoints and state which subgroup inclusion you are testing.

For the $8$-fold circle covering $z\mapsto z^8$, which maps $z\mapsto z^m$ lift as based maps when $m=12$, and when $m=16$?
For the covering $z\mapsto z^5$, explain why the map $z\mapsto z^0=1$ passes the subgroup test.
For $f:T^2\to S^1$ given by $f(z,w)=z^3w^6$, determine whether a based lift through $z\mapsto z^3$ exists.
In the proof of sufficiency, where are path-connectedness and local path-connectedness of $X$ used?
Explain why the path-defined value at $x$ must be checked for independence of the chosen path from $x_0$ to $x$.

The Fundamental Group Criterion for Lifting Maps

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