Worked Applications of the Lifting Criterion

From a General Criterion to Concrete Decisions

The lifting criterion turns a question about whether a continuous map can be lifted into a question about fundamental groups. Two useful applications are especially direct. A contractible domain has no nontrivial loops to obstruct lifting, while for a circle domain the obstruction can be tested by comparing subgroups of the integers. In each case, the basepoint and the specified covering matter: a lift must project to the given map, and its value at the basepoint must be chosen in the appropriate fiber.

A covering map $p:E\to Y$ is a continuous surjection such that every $y\in Y$ has an open neighborhood $V$ for which $p^{-1}(V)$ is a union of disjoint open sets, each mapped homeomorphically onto $V$ by $p$. A lift of $f:X\to Y$ through $p$ is a continuous map $\widetilde f:X\to E$ satisfying $p\circ\widetilde f=f$. The fundamental group $\pi_1(X,x_0)$ consists of based loops at $x_0$, up to based homotopy, with concatenation as its operation.

Theorem (Fundamental Group Criterion for Lifting Maps): Let $X$ be path-connected and locally path-connected, let $p:E\to Y$ be a covering map, and let $f:X\to Y$ be continuous. Choose $x_0\in X$ and $e_0\in E$ with $p(e_0)=f(x_0)$. There is a lift $\widetilde f:X\to E$ with $\widetilde f(x_0)=e_0$ if and only if $$ f_*\bigl(\pi_1(X,x_0)\bigr)\subseteq p_*\bigl(\pi_1(E,e_0)\bigr). $$ Here $f_*$ and $p_*$ are the homomorphisms induced by the maps. This criterion is the standard lifting criterion established in the earlier material on lifting maps.

The inclusion is in the fundamental group $\pi_1(Y,f(x_0))$. It depends on the chosen point $e_0$ in the fiber, so that point should be recorded when applying the criterion. The hypotheses on $X$ are part of the result; in the applications below the domains are disks or circles, which are path-connected and locally path-connected.

Why Contractible Domains Have No Loop Obstruction

Corollary: Let $X$ be contractible, path-connected, and locally path-connected. If $p:E\to Y$ is a covering map and $f:X\to Y$ is continuous, then for every $x_0\in X$ and every $e_0\in p^{-1}(f(x_0))$, there is a lift $\widetilde f:X\to E$ with $\widetilde f(x_0)=e_0$.

Proof. Contractibility means that the identity map on $X$ is homotopic to a constant map. Consequently every loop in $X$ is freely homotopic to a constant loop. A freely null-homotopic loop represents the identity in the based fundamental group: changing the point at which a loop is based conjugates its class, and a conjugate of the identity is still the identity. Thus $\pi_1(X,x_0)$ is trivial. It follows that $f_*(\pi_1(X,x_0))$ is the trivial subgroup of $\pi_1(Y,f(x_0))$, which is contained in every subgroup, including $p_*(\pi_1(E,e_0))$. The fundamental group criterion gives the required lift. $\square$

This conclusion is an existence statement for each permitted choice of $e_0$; it does not say that lifts based at different points of the fiber are the same. Uniqueness, when a basepoint value is fixed, is a separate property of lifts.

Worked Example: A Disk Map Lifted Through the Circle Covering

Let $D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}$ have the subspace topology from $\mathbb{R}^2$, and give $S^1=\{z\in\mathbb{C}:|z|=1\}$ its subspace topology. Use the covering $p_3:S^1\to S^1$, $p_3(w)=w^3$, and define $$ f(x,y)=e^{2\pi i(x^2+y^2)}. $$

The disk is contractible, so the corollary gives a lift after any compatible point in the fiber is chosen. An explicit lift is $$ \widetilde f(x,y)=e^{2\pi i(x^2+y^2)/3}. $$

To check it, cube the proposed value: $$ p_3(\widetilde f(x,y)) =\left(e^{2\pi i(x^2+y^2)/3}\right)^3 =e^{2\pi i(x^2+y^2)} =f(x,y). $$ At the basepoint $(0,0)$, the proposed lift has value $1$, and $p_3(1)=1=f(0,0)$. Thus it also has the stated basepoint compatibility. The formula is continuous because $x^2+y^2$ and the exponential function are continuous.

Check: at $(1,0)$, the map has value $e^{2\pi i}=1$, while the lift has value $e^{2\pi i/3}$. Cubing that lift gives $e^{2\pi i}=1$, as required. The lift need not equal the original map; it must project to it.

Worked Example: A Disk Map Lifted Through a Torus Covering

Give $T^2=S^1\times S^1$ the product topology and $\mathbb{R}^2$ its standard topology. Consider the standard covering $$ q:\mathbb{R}^2\to T^2,\qquad q(u,v)=(e^{2\pi i u},e^{2\pi i v}). $$ For $D^2$ as above, define $$ F(x,y)=\bigl(e^{2\pi i(x+y)},e^{2\pi i(x-y)}\bigr). $$

A lift is $\widetilde F:D^2\to\mathbb{R}^2$ given by $\widetilde F(x,y)=(x+y,x-y)$. Each coordinate is continuous. Substitution into the covering gives $$ q(\widetilde F(x,y)) =q(x+y,x-y) =\bigl(e^{2\pi i(x+y)},e^{2\pi i(x-y)}\bigr) =F(x,y). $$

For the basepoint $(0,0)$, the lift takes the value $(0,0)$, and $q(0,0)=(1,1)=F(0,0)$. As a numerical check at $(1,0)$, $\widetilde F(1,0)=(1,1)$, so $q(1,1)=(e^{2\pi i},e^{2\pi i})=(1,1)$; also $F(1,0)=(e^{2\pi i},e^{2\pi i})=(1,1)$.

The Circle Test: Subgroup Inclusion Becomes Divisibility

For a circle domain, the fundamental group is identified with $\mathbb{Z}$. In the applications here, take as input the induced homomorphism and the subgroup associated with the chosen covering point. In particular, for the standard $n$-fold covering $p_n:S^1\to S^1$, $p_n(z)=z^n$, the subgroup in the target is $n\mathbb{Z}$ under the standard identification. The test below uses these subgroup data; it does not require computing them from a formula for a particular circle map.

Proposition (Circle Lifting Test): Suppose $f:S^1\to S^1$ induces multiplication by $m\in\mathbb{Z}$ on fundamental groups, and suppose the subgroup associated with a chosen covering point is $d\mathbb{Z}$, where $d$ is a nonnegative integer and $0\mathbb{Z}=\{0\}$. Then $f$ lifts with that basepoint value if and only if $m\mathbb{Z}\subseteq d\mathbb{Z}$. If $d>0$, this is equivalent to $d$ dividing $m$. If $d=0$, it is equivalent to $m=0$.

Proof. Since $f_*$ is multiplication by $m$, its image is $\{mk:k\in\mathbb{Z}\}=m\mathbb{Z}$. The fundamental group criterion says exactly that this image must be contained in $d\mathbb{Z}$. If $d>0$, containment implies $m\in d\mathbb{Z}$, because $m=m\cdot 1$ belongs to $m\mathbb{Z}$; hence $m=dr$ for some $r\in\mathbb{Z}$, so $d$ divides $m$. Conversely, if $m=dr$, then every element $mk$ equals $d(rk)$, so $m\mathbb{Z}\subseteq d\mathbb{Z}$. If $d=0$, the target subgroup is $\{0\}$. The inclusion holds exactly when every $mk$ is zero, which holds if $m=0$; if $m\ne0$, the element $m=m\cdot1$ is nonzero and the inclusion fails. Applying the lifting criterion proves both directions. $\square$

Worked Example: A Circle Map That Lifts to a Three-Fold Cover

Suppose the supplied subgroup data for a map $f:S^1\to S^1$ are $f_*(\mathbb{Z})=12\mathbb{Z}$, and the covering is $p_3$ with subgroup $3\mathbb{Z}$. The test compares $12\mathbb{Z}$ with $3\mathbb{Z}$. Since $12=3\cdot4$, every element of the first subgroup has the form $$ 12k=3(4k),\qquad k\in\mathbb{Z}. $$ Thus $12\mathbb{Z}\subseteq3\mathbb{Z}$, and the criterion guarantees a lift for the chosen compatible point in the fiber.

Check: the generator $12$ of the image subgroup is in $3\mathbb{Z}$, since $12=3\cdot4$. More generally, the calculation above checks every element $12k$, not just the generator.

Worked Example: A Circle Map That Does Not Lift to a Four-Fold Cover

Suppose the induced image is $6\mathbb{Z}$, while the chosen covering subgroup is $4\mathbb{Z}$. The element $6=6\cdot1$ belongs to $6\mathbb{Z}$. It does not belong to $4\mathbb{Z}$, because there is no integer $r$ with $6=4r$. Therefore $6\mathbb{Z}$ is not contained in $4\mathbb{Z}$, and the map has no lift with the specified basepoint value.

Check: the divisibility test gives the same answer: $4$ does not divide $6$, since $6/4=3/2$ is not an integer. The failed element $6$ explicitly witnesses the failed subgroup inclusion.

Worked Example: The Trivial Subgroup Case

Suppose the covering subgroup is $0\mathbb{Z}=\{0\}$, and the map induces multiplication by $0$. Its image is $$ 0\mathbb{Z}=\{0\}. $$ Hence the image is contained in the covering subgroup, and a lift exists for the chosen compatible point.

If instead the induced multiplier is $2$, then the image contains $2=2\cdot1$, which is not in $\{0\}$. The inclusion fails, so no lift exists with that basepoint value. This confirms why the case $d=0$ must be treated separately rather than using a positive-divisor test.

A Common Divisibility Error

Common pitfall. Reversing the divisibility condition leads to an incorrect decision. For a covering subgroup $3\mathbb{Z}$ and induced image $6\mathbb{Z}$, the correct condition is $3\mid6$, so the lift exists. The inclusion can be checked element by element: every element of $6\mathbb{Z}$ is $6k=3(2k)$, hence belongs to $3\mathbb{Z}$. The incorrect calculation $6\mid3$ asks the opposite question and would wrongly reject the lift.

The subgroup test also keeps the zero case honest: every subgroup contains $0$, so a zero induced image always passes; a nonzero induced image cannot be contained in the trivial subgroup. For any application, first identify the image subgroup, then the covering subgroup, and test inclusion in that order.

Check Your Understanding

For each question, state the subgroup inclusion being tested before deciding whether a lift exists.

Why does the lifting criterion guarantee a lift for a map from a contractible, locally path-connected space, once a point in the appropriate fiber is selected?
A circle map has induced image $15\mathbb{Z}$, and the covering subgroup is $5\mathbb{Z}$. Does it lift with the specified basepoint value?
A circle map has induced image $8\mathbb{Z}$, and the covering subgroup is $6\mathbb{Z}$. Find an element witnessing failure of inclusion, or prove inclusion.
For covering subgroup $\{0\}$, which integer multipliers give a lift?
Explain why the condition for $m\mathbb{Z}\subseteq d\mathbb{Z}$, when $d>0$, is $d\mid m$, not $m\mid d$.