§ Covering spaces

§ Covering spaces: Intuition

§ Covering spaces: Definition

Slogan: Covering space is locally disjoint copies of the original space.

§ Path lifting and Monodromy

§ Path lifting lemma

Theorem :Suppose p:yXp: y \rightarrow X is a covering map. Let δ:[0,1]X\delta: [0, 1] \rightarrow X be a path such that δ(0)=x\delta(0) = x, and let yp1(x)y \in p^{-1}(x) [ yy is in the fiber of xx]. Then there is a unique path γ:[0,1]Y\gamma: [0,1] \rightarrow Y which "lifts" δ\delta. That is, δ(p(y))=γ(y)\delta(p(y)) = \gamma(y), such that γ(0)=Y\gamma(0) = Y.
Slogan: Paths can be lifted. Given how to begin the lift, can be extended all the way.
  • Let NN be a collection of elementary neighbourhoods of XX.
  • {δ1(U):UN}\{ \delta^{-1}(U) : U \in N \} is an open cover (in the compactness sense) of [0,1][0, 1].
  • By compactness, find a finite subcover. Divide interval into subintervals 0=t0<t1<tn=10 = t_0 < t_1 < \dots t_n = 1such that δk=δ[tk,tk+1]\delta|k = \delta|_{[t_k, t_{k+1}]} lands in UkU_k, an elementary neighbourhood.
  • Build γ\gamma by induction on kk.
  • We know that γ(0)\gamma(0) should be yy.
  • Since we have an elementary neighbourhood, it means that there are a elementary sheets living over U0U_0, indexed by some discrete set FF. yy lives in one of thse sheets. We have local inverses qmq_m. One of them lands on the sheet of yy, call it qq. So we get a map q:U0Yq: U_0 \rightarrow Y such that q(x)=yq(x) = y.
  • Define γ(0)q(δ(0))=q(x)=y\gamma(0) \equiv q(\delta(0)) = q(x) = y.
  • Extend γ\gamma upto t1t_1.
  • Continue all the way upto tkt_k.
  • To get γ\gamma from (tk,tk+1(t_k, t_{k+1}, there exists a qk:UkYq_k: U_k \rightarrow Ysuch that qk(δ(tk))=γ(tk)q_k(\delta(t_k)) = \gamma(t_k). Define γ(tkttk1)qk(δ(tk))\gamma(t_k \leq t \leq t_{k_1}) \equiv q_k(\delta(t_k)).
  • This is continuous because δ\delta continuous by definition, qkq_k continuous by neighbourhood, γ\gamma is pieced together such that endpoints fit, and is thus continuous.
  • Can check this is a lift! We get pγ=pqkδkp \circ \gamma = p \circ q_k \circ \delta_k. Since qkq_k is a local inverse of pp, we get pγ=δkp \circ \gamma = \delta_kin the region.

§ 7.03: Path lifting: uniqueness

If we have a space XX and a covering space YY, for a path gammagamma that starts at xx, we can find a path γ\gamma' which starts at yp1(x)y \in p^{-1}(x) and projects down to γ\gamma: γ(t)=p(γ(t))\gamma(t) = p(\gamma'(t)). We want to show that this path lift is unique

§ Lemma

Let p:YXp: Y \rightarrow X be a covering space. Let TT be a connected space Let F:TXF: T \rightarrow X be a continuous map (for us, T[0,1]T \simeq [0, 1]). Let F1,F2:TYF_1, F_2: T \rightarrow Y be lifts of FF ( pF1=Fp \circ F_1 = F, pF2=Fp \circ F_2 = F). We will show that F1=F2F_1 = F_2 iff the lifts are equal for some tt \in T.
Slogan: Lifts of paths are unique: if they agree at one point, they agree at all points!
  • We just need to show that if F1F_1 and F2F_2 agree somewhere in YY, they agree everywhere. It is clear that if they agree everywhere, they must agree somewhere.
  • To show this, pick the set SS where F1,F2F_1, F_2 agree in YY: S{tT:F1(t)=F2(t)}S \equiv \{ t \in T : F_1(t) = F_2(t) \}.
  • We will show that SS is open and closed. Since TT is connected, SS must be either the full space or the empty set. Since SS is assumed to be non-empty, S=TS = T and the two functions agree everywhere.
  • (Intuition: if both SS and ScS^c are open, then we can build a function that colors T=SScT = S \cup S^cin two colors continuously; ie, we can partition it continuously; ie the spaces must be disconnected. Since TT is connected, we cannot allow that to happen, hence S=S = \emptyset or S=TS = T.)
  • Let t inTt \ in T. Let UU be an evenly covered neighbourhood/elementary neighbourhood of F(t)F(t) downstairs (in XX). Then we have p1(U)=αVαp^{-1}(U) = \sqcup_\alpha V_\alpha such that pVα:VαUp|_V{\alpha}: V_\alpha \rightarrow Uis a local homeomorphism.
  • Since F1,F2F_1, F_2 are continuous, we will have opens V1,V2V_1, V_2 in VαV_\alpha, which contain F1(t),F2(t)F_1(t), F_2(t) upstairs (mirrroring UU containing F(t)F(t) downstairs).
  • The pre-images of V1V_1, V2V_2 along F1,F2F_1, F_2 give us open sets tT1,T2Tt \in T_1, T_2 \subseteq T.
  • Define T=T1T2T* = T_1 \cap T_2. If F1(t)F2(t)F_1(t) \neq F_2(t), then V1V2V_1 \neq V_2and thus F1F2F_1 \neq F_2 on all of TT*. So, Sc=TS^c = T* is open.
  • If F1(t)=F2(t)F_1(t) = F_2(t), then V1=V2V_1 = V_2 and thus F1=F2F_1 = F_2 on TT*(since pF1=F=pF2p \circ F_1 = F = p \circ F_2, and ppis injective within UU, ie within V1,V2V_1, V_2). So SS is open.
  • Hence we are done, as SS is non-empty and clopen and is thus equal to TT. Thus, the two functions agree on all of TT.

§ Homotopy lifting, Monodromy

  • Given a loop γ\gamma in XX based at xx , the monodromy around γ\gamma is a permutation σγ:p1(x)p1(x)\sigma_\gamma : p^{-1}(x) \rightarrow p^{-1}(x), where σγ(y)γy(1)\sigma_{\gamma}(y) \equiv \gamma^y(1)where γy\gamma^y is the unique lift of γ\gamma staring at yy. We have that σγPerm(p1(x))\sigma_{\gamma} \in Perm(p^{-1}(x)).
  • Claim: if γ1γ2\gamma_1 \simeq \gamma_2 then σγ1=σγ2\sigma_{\gamma_1} = \sigma_{\gamma_2}.
  • We need a tool: homotopy lifting lemma.
Slogan: permutation of monodromy depends only on homotopy type

§ Homotopy lifting lemma/property of covering spaces

Suppose p:YXp: Y \rightarrow X is a covering map and γs\gamma_s is a homotopy of paths rel. endpoints ( γs(0)\gamma_s(0) and γs(1)\gamma_s(1) are independent of ss / endpoints are fixed throughout the homotopy). Then there exists for each lift γ0:[0,1]Y\gamma'_0 : [0, 1] \rightarrow Y of γ0:[0,1]X\gamma_0:[0,1] \rightarrow X (ie, pγ0=gamma0p \circ \gamma'_0 = gamma_0), a completion of the lifted homotopy γs:[0,1]Y\gamma'_s: [0, 1] \rightarrow Y (ie, pgammas=gammasp \circ gamma'_s = gamma_s). Moreover, this lifted homotopy is rel endpoints: ie, the endpoints of gammagamma' are independent of ss.
Slogan: homotopy lifted at 0 can be lifted for all time
  • Let H:[0,1]×[0,1]XH: [0, 1] \times [0, 1] \rightarrow X be the homotopy in XX such that H(s,t)=γs(t)H(s, t) = \gamma_s(t). Subdivide the square into rectangles RijR_{ij} such that H(Rij)H(R_{ij}) is contained in UijU_{ij} for some elementary neighbourhood UijU_{ij}. We build H:[0,1]×[0,1]YH': [0, 1] \times [0, 1] \rightarrow Y by building local inverses qij:UijYq_{ij} : U_{ij} \rightarrow Y such that pqij=Rijp \circ q_{ij} = R_{ij}. We then set HRij=qijHH'|_{R_{ij}} = q_{ij} \circ H.