Lift of a map to a covering space

Let $𝑋$ and $𝑌$ be a topological spaces, $𝑝 : ˜ 𝑋 ↠ 𝑋$ be a covering of $𝑋$ with $˜ 𝑋$ , and $𝑓 : 𝑌 \to 𝑋$ be a continuous map. A lift $˜ 𝑓$ of $𝑓$ is any function so that $𝑝 ˜ 𝑓 = 𝑓$ , #m/def/topology i.e. the following diagram commutes in $𝖳 𝗈 𝗉$ :

Lifts fill a fundamental role in Homotopy theory MOC, in particular they allow for the computation of the Fundamental group. Their usefulness follows from the main theorem below.

Main theorem

Let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ be a connected covering, $(𝑌, 𝑦 0)$ be a connected and locally path-connected¹ space, and $𝑓 : (𝑌, 𝑦 0) \to (𝑋, 𝑥 0)$ be a morphism in $𝖳 𝗈 𝗉 •$ . Then there exists a lift $˜ 𝑓 : (𝑌, 𝑦 0) \to (˜ 𝑋, ˜ 𝑥 0)$ of $𝑓$ iff #m/thm/topology

𝜋 1 𝑓 (𝜋 1 (𝑌, 𝑦 0)) \subseteq 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))

i.e. the image of $𝜋 1 𝑓$ is a subset of the image² of $𝜋 1 𝑝$ , where $𝜋 1$ is the Fundamental group functor. Furthermore if $˜ 𝑓$ exists it is unique.

Construction of lift

A lift $˜ 𝑓 : (𝑌, 𝑦) \to (˜ 𝑋, ˜ 𝑥 0)$ of $𝑓 : (𝑌, 𝑦) \to (𝑋, 𝑥 0)$ is constructed as follows: For each $𝑦 \in 𝑌$ , by path-connectedness there exists a path $𝜔 : 𝕀 \to 𝑌$ from $𝑦 0$ to $𝑦$ . Then define a path $𝛼 = 𝑓 𝜔$ in $𝑋$ , which by Second lemma Lifts of paths has a unique lift $˜ 𝛼$ with $˜ 𝛼 (0) = ˜ 𝑥 0$ . Then let $˜ 𝑓 (𝑦) = ˜ 𝛼 (1)$ .

The proof involves four lemmas, each relying on the previous: Uniqueness may be proven immediately, then we prove the special cases of lifts of paths and lifts of homotopies of paths, and then the requirement given for the fundamental group.

First lemma: Uniqueness

Let $𝑝 : ˜ 𝑋 ↠ 𝑋$ be a connected covering, $𝑌$ be a connected space, $𝑓 : 𝑌 \to 𝑋$ be a continuous function, and $˜ 𝑓 1, ˜ 𝑓 2 : 𝑌 \to ˜ 𝑋$ be lifts of $𝑥$ . Then $˜ 𝑓 1 = ˜ 𝑓 2$ iff $˜ 𝑓 1 (𝑦 0) = ˜ 𝑓 2 (𝑦 0)$ for some $𝑦 0 \in 𝑌$ . #m/thm/topology

Proof

The reverse direction is obvious. For the forward direction, consider the set

𝑀 = {𝑦 \in 𝑌 : ˜ 𝑓 1 (𝑦) = ˜ 𝑓 2 (𝑦)}

Let $𝑦 \in 𝑌$ , $𝑥 = 𝑓 (𝑦) \in 𝑋$ , and $𝑈 \subseteq 𝑋$ be an evenly covered open neighbourhood of $𝑥$ . Let $˜ 𝑈 1, ˜ 𝑈 2 \subseteq ˜ 𝑋$ be the sheets over $𝑈$ containing $˜ 𝑓 1 (𝑦)$ and $˜ 𝑓 2 (𝑦)$ respectively. Then $𝑉 = ˜ 𝑓 - 1 1 (˜ 𝑈 1) \cap ˜ 𝑓 - 1 2 (˜ 𝑈 2)$ is an open neighbourhood of $𝑦$ , and

˜ 𝑓 1 ↾ 𝑉 = (𝑝 ↾ ˜ 𝑈 1) - 1 \circ (𝑓 ↾ 𝑉) ˜ 𝑓 2 ↾ 𝑉 = (𝑝 ↾ ˜ 𝑈 2) - 1 \circ (𝑓 ↾ 𝑉)

Now if $˜ 𝑓 1 (𝑦) = ˜ 𝑓 2 (𝑦)$ , then $˜ 𝑈 1 = ˜ 𝑈 2$ and consequently

˜ 𝑓 1 ↾ 𝑉 = (𝑝 ↾ ˜ 𝑈 1) - 1 \circ (𝑓 ↾ 𝑉) = (𝑝 ↾ ˜ 𝑈 2) - 1 \circ (𝑓 ↾ 𝑉) = ˜ 𝑓 2 ↾ 𝑉

Thus $𝑀$ is the union of all $𝑉$ obtained from $𝑦$ with $˜ 𝑓 1 (𝑦) = ˜ 𝑓 2 (𝑦)$ and is thus open. Now if $˜ 𝑓 1 (𝑦) \neq ˜ 𝑓 2 (𝑦)$ , then $˜ 𝑈 1 \neq ˜ 𝑈 2$ and consequently $˜ 𝑓 1 (𝑧) \neq ˜ 𝑓 2 (𝑧)$ for all $𝑧 \in 𝑉$ . Thus $𝑌 ∖ 𝑀$ is the union of all $𝑉$ obtained from $𝑦$ with $˜ 𝑓 1 (𝑦) \neq ˜ 𝑓 2 (𝑦)$ and is thus open. Therefore $𝑀$ is clopen and inhabited ( $𝑦 0 \in 𝑀$ ), so since $𝑌$ is connected, $𝑀 = 𝑌$ . Hence $˜ 𝑓 1 = ˜ 𝑓 2$ .

Second lemma: Lifts of paths

Let $𝑝 : ˜ 𝑋 ↠ 𝑋$ be a connected covering and $𝛼 : 𝕀 \to 𝑋$ be a continuous path from $𝑥 0 = 𝛼 (0)$ . For each $˜ 𝑥 0 \in 𝑝 - 1 {𝑥 0}$ there exists exactly one lifted path $˜ 𝛼 : 𝕀 \to ˜ 𝑋$ from $˜ 𝛼 (0) = ˜ 𝑥 0$ . #m/thm/topology

Proof

Uniqueness follows from First lemma Uniqueness, but is also self-evident in the following argument. For each $𝑥 \in 𝑋$ , let $𝑈 𝑥$ be an evenly covered open neighbourhood of $𝑥$ . Then ${𝑈 𝑥} 𝑥 \in 𝑋$ is an open cover of $𝑋$ , and ${𝛼 - 1 (𝑈 𝑥)} 𝑥 \in 𝑋$ is an open cover of $𝕀$ . Using a Lebesgue number $𝕀$ may be evenly subdivided with

0 = 𝑡 0 < \dots < 𝑡 𝑘 = 1

so that $𝛼 [𝑡 𝑖 - 1, 𝑡 𝑖] \subseteq 𝑈 𝑥 𝑖$ where $𝑥 𝑖 \in 𝑋$ for all $1 \leq 𝑖 \leq 𝑘$ . Now consider a lift $˜ 𝛼 : 𝕀 \to ˜ 𝑋$ . Clearly $˜ 𝛼 ↾ [𝑡 𝑖 - 1, 𝑡 𝑖] = (𝑝 ↾ ˜ 𝑈 𝑖) - 1 \circ (𝛼 ↾ [𝑡 𝑖 - 1, 𝑡 𝑖])$ , where $˜ 𝑈 𝑖$ is the sheet over $𝑈 𝑥 𝑖$ containing $𝛼 (𝑡 𝑖 - 1)$ . Thus if $˜ 𝛼 (0) = ˜ 𝑥 0$ is set, $˜ 𝛼$ is unique and well-defined.

Third lemma: Lifts of homotopies of paths

Let $𝑝 : ˜ 𝑋 ↠ 𝑋$ be a connected covering and $𝛼 0, 𝛼 1 : 𝕀 \to 𝑋$ be continuous paths with the same endpoints $𝑥 0, 𝑥 1$ homotopic to one another via $𝐴 : 𝕀 2 \to 𝑋 : (𝑡, 𝑠) \mapsto 𝛼 𝑠 (𝑡)$ . Let $˜ 𝑥 0 \in 𝑝 - 1 {𝑥 0}$ and $˜ 𝛼 0, ˜ 𝛼 1 : 𝕀 \to ˜ 𝑋$ be the unique lifts of $𝛼 0, 𝛼 1$ respectively with $˜ 𝛼 0 (0) = ˜ 𝛼 1 (0) = ˜ 𝑥 0$ . Then there exists a unique lift $˜ 𝐴 : ˜ 𝛼 0 ≃ ˜ 𝛼 1$ of the homotopy $𝐴$ , and in particular $˜ 𝛼 0 (1) = ˜ 𝛼 1 (1)$ . #m/thm/homotopy

Proof

First, notice that if a lift $˜ 𝐴 : 𝕀 2 \to ˜ 𝑋$ of $𝐴 : 𝕀 2 \to 𝑋$ with $𝐴 (0, 0) = ˜ 𝑥 0$ exists, it is necessarily unique (by First lemma Uniqueness) and a homotopy from $˜ 𝛼 0$ to $˜ 𝛼 1$ : Clearly $𝑝 ˜ 𝐴 (0, 𝑠) = 𝑥 0$ and $𝑝 ˜ 𝐴 (1, 𝑠) = 𝑥 1$ for all $𝑠 \in 𝕀$ , and since $𝑝 - 1 {𝑥 0}$ and $𝑝 - 1 {𝑥 1}$ are discrete, both $˜ 𝐴 (0, 𝑠)$ and $˜ 𝐴 (1, 𝑠)$ must be constant for all $𝑠 \in 𝕀$ , so $˜ 𝐴 (0, 𝑠) = ˜ 𝑥 0$ and we let $˜ 𝐴 (1, 𝑠) = ˜ 𝑥 1$ . By construction $˜ 𝐴$ is a homotopy from $˜ 𝐴 (-, 0)$ to $˜ 𝐴 (-, 1)$ , but $˜ 𝐴 (-, 𝑠)$ is a lift of $𝛼 𝑠$ with $˜ 𝐴 (0, 𝑠) = 0$ for each $𝑠 \in 𝕀$ , thus by uniqueness $˜ 𝐴 (-, 𝑠) = ˜ 𝛼 𝑠$ in particular for $𝑠 = 0, 1$ , and hence $˜ 𝐴$ is the desired homotopy.

For existence we use a similar construction to above: Using a Lebesgue number argument $𝕀 2$ may be subdivided into a grid with

0 = 𝑡 0 < \dots < 𝑡 𝑘 = 1 0 = 𝑠 0 < \dots < 𝑠 𝑘 = 1

so that for each square $𝑄 𝑖 𝑗 = [𝑡 𝑖, 𝑡 𝑖 + 1] \times [𝑠 𝑗, 𝑠 𝑗 + 1]$ with $0 \leq 𝑖 \leq 𝑘 - 1$ , its image $𝐴 (𝑄 𝑖 𝑗)$ is contained entirely within an evenly covered open set $𝑈 𝑖 𝑗$ in $𝑋$ , i.e. $𝐴 (𝑄 𝑖 𝑗) \subseteq 𝑈 𝑖 𝑗$ . Now consider a lift $˜ 𝐴 : 𝕀 2 \to ˜ 𝑋$ . If the bottom left corner $˜ 𝐴 (𝑡 𝑖, 𝑠 𝑗)$ is set, then clearly $˜ 𝐴 ↾ 𝑄 𝑖 𝑗 = (𝑝 ↾ ˜ 𝑈 𝑖 𝑗) - 1 \circ (𝐴 ↾ 𝑄 𝑖 𝑗)$ , where $˜ 𝑈 𝑖 𝑗$ is the sheet over $𝑈 𝑖 𝑗$ containing $˜ 𝐴 (𝑡 𝑖, 𝑠 𝑗)$ . Then by Second lemma Lifts of paths the edges automatically agree, thus by starting with $˜ 𝐴 (0, 0) = ˜ 𝑥 0$ we obtain a well-defined, unique lift $˜ 𝐴$ of $𝐴$ .

Fourth lemma: Condition for the existence of a lift

Let $𝑝 : (˜ 𝑋, ˜ 𝑥 0) ↠ (𝑋, 𝑥 0)$ be a connected covering, $(𝑌, 𝑦 0)$ be a path-connected space, and $𝑓 : (𝑌, 𝑦 0) \to (𝑋, 𝑥 0)$ be a morphism in $𝖳 𝗈 𝗉 •$ . If a lift $˜ 𝑓 : (𝑌, 𝑦 0) \to (˜ 𝑋, 𝑥 0)$ exists, then $𝜋 1 𝑓 (𝜋 1 (𝑌, 𝑦 0)) \subseteq 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ . #m/thm/homotopy

Proof

Since $𝑝 ˜ 𝑓 = 𝑓$ , it follows from functor properties of the Fundamental group that $(𝜋 1 𝑝) (𝜋 1 ˜ 𝑓) = 𝜋 1 𝑓$ , and thus the image of $𝜋 1 𝑓$ must be contained within the image of $𝜋 1 𝑝$ .

Proof of main theorem

The forward direction follows from Fourth lemma Condition for the existence of a lift.

Proof the construction is well-definined

It remains to show that $˜ 𝛼 (1)$ is independent from the choice of path $𝜔$ . To this end let $𝜈 : 𝕀 \to 𝑌$ be a path from $𝑦 0$ to $𝑦$ . Then $―― 𝜔 ⊙ 𝜈$ is a continuous loop with basepoint $𝑦 0$ . Since $𝑓 \circ (―― 𝜔 ⊙ 𝜈)$ is guaranteed a lift by Second lemma Lifts of paths, it follows $𝜋 1 𝑓 [―― 𝜔 ⊙ 𝜈] \in 𝜋 1 𝑝 (𝜋 1 (˜ 𝑋, ˜ 𝑥 0))$ , and thus there exists a continuous loop $˜ 𝜇 : 𝕀 \to ˜ 𝑋$ with basepoint $˜ 𝑥 0$ such that $𝜋 1 𝑓 [―― 𝜔 ⊙ 𝜈] = 𝜋 1 𝑝 [˜ 𝜇]$ . Let $𝛼 = 𝑓 \circ 𝜔$ , $𝛽 = 𝑓 \circ 𝜈$ , and $𝜇 = 𝑝 \circ ˜ 𝜇$ , so $[𝛼] - 1 ⊙ [𝛽] = [𝜇]$ and thus $𝛽 ≃ 𝛼 ⊙ 𝜇$ . Then if $˜ 𝛼$ and $˜ 𝛽$ are the lifts of $𝛼$ and $𝛽$ respectively with $˜ 𝛼 (0) = ˜ 𝛽 (0) = ˜ 𝑦 0$ , then $˜ 𝛼 ⊙ ˜ 𝜇$ is the lift of $𝛼 ⊙ 𝜇$ . Hence by Third lemma Lifts of homotopies of paths, $˜ 𝛽 ≃ ˜ 𝑎 ⊙ ˜ 𝜇$ , and in particular $˜ 𝛽 (1) = ˜ 𝛼 ⊙ ˜ 𝜇 (1) = ˜ 𝛼 (1)$ . Hence $˜ 𝛼 (1)$ is the same regardless of the selected path $𝜔$ .

#state/tidy | #lang/en | #SemBr

And thus path-connected, since A locally path-connected space is path-connected iff it is connected ↩
In Algebraische Topologie wird dies als die charakteristische Untergruppe bezeichnet (p. 91). ↩

Lift of a map to a covering space

Main theorem

First lemma: Uniqueness

Second lemma: Lifts of paths

Third lemma: Lifts of homotopies of paths

Fourth lemma: Condition for the existence of a lift

Proof of main theorem

Footnotes