MAS61015 Algebraic Topology

22. Covering maps

Consider a continuous map p:XY. For each point yY, we have a subset p-1{y}X, which we call the fibre of p over y. We next define what it means for p to be a covering map. The key points are that

  • All the fibres p-1{y} must be discrete subsets of X;

  • The fibre p-1{y} must depend continuously on y, in an appropriate sense.

It is not easy to formulate the second condition directly, so the formal definition looks rather different from this informal discussion.

Example 22.1.

Consider the exponential map exp:{0}. The fibres are


Each fibre is a discrete set, which suggests that the map should be a covering. We will check that this is true once we have given the proper definition.

Definition 22.2.

Let p:XY be a continuous map of spaces. Consider an open subset VY. We say that V is trivially covered by p if there is a discrete space F and a map f:p-1(V)F such that the combined map p,f:p-1(V)V×F is a homeomorphism. We say that p is a covering map (or that X is a covering space of Y) if for each point yY there is an open set V that contains y and is trivially covered.

Remark 22.3.

Suppose we have open subsets VVY and that V is trivially covered, as witnessed by a map f:p-1(V)F. We then note that p-1(V)p-1(V), se we can restrict f to get a map f:p-1(V)F. It is not hard to check that the combined map p,f:p-1(V)V×F is a homeomorphism, so V is also trivially covered.

Example 22.4.

Take X=× and Y=, and let p:XY be the projection map, given by p(x,n)=x. We claim that the whole space Y is trivially covered. Indeed, we can take F= and define f:XF by f(x,n)=n. Then the combined map p,f:XY×F is just the identity map ××, which is certainly a homeomorphism. From this it is clear that p is a covering map.

Example 22.5.

Take X and Y to be the unit circle in , or in other words X=Y={eiθ|θ}. Define p:XY by p(z)=z2, or equivalently p(eiθ)=e2iθ. Each fibre p-1{y} is a discrete set consisting of the two square roots of y, which are negatives of each other. This suggests that p should be a covering, which we can prove as follows. We first take V0=Y{-1}, so each element yV0 can be expressed in a unique way as y=eiθ with -π<θ<π. The two square roots of y are then x=eiθ/2 (which has Re(x)>0) and -x=-eiθ/2=ei(θ/2+π) (which has Re(-x)<0). We therefore have


and we can define f0:p-1(V0){1,-1} by

f0(z)={1 if Re(z)>0-1 if Re(z)<0.

We then find that the map p,f0:p-1(V0)V0×{1,-1} is a homeomorphism, showing that V0 is trivially covered. A similar approach can be used to check that the set V1=Y{1} is also trivially covered. Any element yV1 can be expressed uniquely as y=eiθ with 0<θ<2π, and the two square roots are then x=eiθ/2 (which has Im(x)>0) and -x (which has Im(-x)<0). We therefore have


and we can define f1:p-1(V1){1,-1} by

f1(z)={1 if Im(z)>0-1 if Im(z)<0.

We then find that the map p,f1:p-1(V1)V1×{1,-1} is a homeomorphism, as required. As Y=V0V1, this proves that p is a covering.

Recall that the real projective space Pn is defined to be the quotient space Sn/, where xy iff y=±x. We therefore have a quotient map π:SnPn, for which π(x)=π(y) iff x=±y. We give Pn the quotient topology, which means that a subset VPn is open iff π-1(V) is open in Sn.

Proposition 22.6.

The map π:SnPn is a covering map.


For each aSn we put Ua={xSn|x.a>0} and Va=π(Ua)Pn. We then find that

π-1(Va)={x|π(x)π(Ua)}={x|xUa or -xUa}={x|x.a0}.

This first shows that π-1(Va) is an open subset of Sn and so (by the definition of the quotient topology) that Va is an open subset of Pn. It also allows us to define a continuous map fa:π-1(Va){1,-1} by

fa(x)={1 if x.a>0-1 if x.a<0.

It is easy to see that the combined map π,fa:π-1(Va)Va×{1,-1} is a homeomorphism, so Va is trivially covered. The sets Va cover all of Pn (because π(a)Va) so π is a covering map as claimed. ∎

Proposition 22.7.

The map exp:{0} is a covering map, as is the map exp:iS1 (where we identify S1 with {z:|z|=1}).


This is closely related to the proof in Example 22.5. We put

V0 =(-,0] V1 =[0,)
U0 ={x+iy|x,-π<y<π} U1 ={x+iy|x,0<y<2π}
W0 ={x+iy|y is not an odd multiple of π} W1 ={x+iy|y is not an even multiple of π}.

If yV0 then there is a unique choice of r and θ with -π<θ<π and r>0 and y=reiθ. It follows that the number x=log(r)+iθ lies in U0 and has exp(x)=y. This means that the restricted map exp:U0V0 is bijective. Standard complex analysis shows that the inverse is also continuous, so we have a homeomorphism exp:U0V0. Similarly, the restricted map exp:U1V1 is also a homeomorphism. From this we see that

exp-1(V0) ={x0+2nπi|x0U0,n}=W0
exp-1(V1) ={x1+2nπi|x1U1,n}=W1.

We can thus define continuous maps fi:Wi (for i=0,1) by fi(xi+2nπ)=n. Equivalently, f0(x+iy) is the closest integer to y/(2π); this is well-defined and continuous on W0 because points where y is an odd multiple of π have been removed from W0. Similarly, f1(x+iy) is the closest integer to (y-π)/(2π). We find that the maps exp,p0:W0V0× and exp,p1:W1V1× are homeomorphisms, so V0 and V1 are trivially covered. We also have {0}=V0V1, so exp:{0} is a covering map. The proof for the restricted map exp:iS1 is essentially the same.

Definition 22.8.

Let p:XY be a covering map.

  • (a)

    Consider a point yY. A lift of y means a point y~X with p(y~)=y.

  • (b)

    Consider a continuous path u:[0,1]Y. A lift of u means a continuous path u~:[0,1]X such that pu~=u. This means in particular that u~(t) is a lift of u(t) for all t.

  • (c)

    More generally, let T be any space and let u:TY be a continuous map. A lift of u is a continuous map u~:TX with pu~=u.

  • (d)

    For a map u:TY as in (c), we say that u is small if there is a trivially covered open set VY such that u(T)V.

Lemma 22.9.

Suppose we have a path-connected space T and a small continuous map u:TY. Suppose we also have points t0T and x0X with u(t0)=p(x0). Then there is a unique lift u~:KX such that u~(t0)=x0.


By assumption, we can choose an open subset VY containing u(K), and a homeomorphism p,f:p-1(V)V×F as in Definition 22.2. If u~:TX is a lift of u, then we have p(u~(t))=u(t)V, so u~(t)p-1(V) for all tT, so we have a well-defined and continuous composite fu~:TF. As T is path connected and F is discrete, this must be constant. Thus, if u~(t0)=x0, then f(u~(t))=f(x0) for all t. It follows that the only possibility is


Proposition 22.10.

Let p:XY be a covering map. Let u be a path from a to b in Y, and let a~X be a lift of a. Then there is a unique lift u~ of u such that u~(0)=a~.


Because p is a covering map, we can find a family of trivially covered open sets ViY such that Y=iVi. The preimages u-1(Vi) then form an open covering of [0,1]. Because [0,1] is a compact metric space, this covering has a Lebesgue number ϵ>0 (by Proposition 8.31). Choose n>1/ϵ and divide [0,1] into subintervals Tk=[(k-1)/n,k/n] for k=0,,n. By the Lebesgue number property, we can choose an index ik such that Tku-1(Vik), so u(Tk)Vik. This means that the restriction of u to Tk is small, so Lemma 22.9 is applicable. We are given a~X with p(a~)=a=u(0). We put x0=a~, and use Lemma 22.9 to show that there is a unique map u~1:T1X with p(u~1(t))=u(t) and u~1(0)=x0. We now define x1=u~1(1/n), so p(x1)=u(1/n). Applying Lemma 22.9 again, we see that there is a unique map u~2:T2X with p(u~2(t))=u(t) and u~2(1/n)=x1. We put x2=u~2(2/n)X, so p(x2)=u(2/n). We then repeat the process in the obvious way, to get a family of maps u~k:[(k-1)/n,k/n]X and points xkX with u~k(k/n)=xk=u~k+1(k/n). It follows that the maps u~k can be patched together to give a continuous map u~:[0,1]X with pu~=u and u~(0)=x0=a~. The same kind of induction shows that this is unique. ∎

Remark 22.11.

Note that Proposition 22.10 does not say anything about u~(1). We know that p(u~(t))=u(t) for all t, so in particular p(u~(1))=u(1)=b, so u~(1)p-1{b}. However, if we have two different paths u,v:ab in Y and we use the same starting point a~p-1{a} in both cases, then it can easily happen that the endpoints u~(1),v~(1) are different elements of p-1{b}. However, this cannot happen if there is a pinned homotopy between u and v, as we will show later.

Corollary 22.12.

Let p:XY be a covering map. Suppose we have a path-connected space T and a continuous map u:TY. Suppose that m,n:TX are continuous lifts of u, and that there is at least one point t0T with m(t0)=n(t0); then m=n.


Consider a point tT; we must show that m(t)=n(t). As T is path connected, we can choose a path v from t0 to t in T. Now mv and nv are both lifts of the path uv:[0,1]Y, and they satisfy (mv)(0)=m(t0)=n(t0)=(nv)(0). Thus, the uniqueness clause in Proposition 22.10 tells us that mv=nv. In particular, we have (mv)(1)=(nv)(1), or in other words m(t)=n(t) as required. ∎

Proposition 22.13.

Let T be a compact convex subset of N, and suppose that t0T. Let p:XY be a covering map, and let u:TY be continuous. Suppose that x0X with p(x0)=u(t0). Then there is a unique continuous lift u~:TX with pu~=u and u~(t0)=x0.


Roughly speaking, the idea is as follows: to define u~(t), we move a short distance from t towards t0 to reach a point t, then u~(t) will already be defined and we define u~(t) to be the unique lift of u(t) that is close to u~(t). The rest of this proof should be seen as a more complete and rigorous version of this idea.

We first claim that there exists ϵ>0 such that for all tT, the restricted map u:OB(t,ϵ)Y is small. (Here and elsewhere in this proof, notation for balls should be interpreted relative to T, so OB(t,ϵ)={tT|t-t<ϵ}.) Indeed, for each tT we can choose a trivially covered open set VtY containing u(t). The set u-1(Vt) is then open in T and contains t. This means that the sets u-1(Vt) form an open cover of the compact metric space T, so there is a Lebesgue number ϵ>0. This has the required property.

Next, for j>0 we put Tj={tT|t-t0jϵ/2}. We will prove by induction on j that there is a unique continuous map u~j:TjX with u~j(t0)=x0 and p(u~j(t))=u(t) for all tTj. To start with, the map u:T1Y is small by our choice of ϵ, so Lemma 22.9 gives u~1. Suppose we have already constructed u~j. For each aTj, we note that the map u:OB(a,ϵ)Y is small, so there is a unique continuous va:OB(a,ϵ)X lifting u with va(a)=u~j(a). Both u~j and va restrict to give lifts of u over the convex set OB(a,ϵ)Tj, and they agree at the point a, and the restricted map u:OB(a,ϵ)TjY is small; it follows that va agrees with u~j on OB(a,ϵ)Tj.

Now suppose that a,bTj and that the set U=OB(a,ϵ)OB(b,ϵ) is nonempty. We then find that the point c=(a+b)/2 must lie in U and it also lies in Tj because Tj is convex. The maps va and vb both agree with u~j at c, so they agree with each other. The restricted map u:UY is small, so we conclude that va|U=vb|U. Because of this consistency property, we see that the maps va can be combined to give a map v:aTjOB(a,ϵ)X. As the sets OB(a,ϵ) are all open, an open patching argument shows that v is continuous. As each map va is a lift of u, we see that v is a lift of u. As va agrees with u~j on Tj, we see that v agrees with u~j on Tj.

It is also easy to see that Tj+1 is contained in the domain of v, so we can define u~j+1 to be the restriction of v to Tj+1. This is a continuous lift of u extending u~j and therefore satisfying u~j+1(t0)=x0 as required.

As T is assumed to be compact, it must be bounded. We therefore have Tj=T for sufficiently large j, and this completes the proof. ∎