MAS61015 Algebraic Topology

23. Transfers, coefficients and homology of projective spaces

Definition 23.1.

Let n be a nonnegative integer. We say that a map p:XY is an n-sheeted covering if it is a covering map, and |p-1{y}|=n for all yY.

Example 23.2.

For any n>0 we have an n-sheeted covering p:×× given by p(z)=zn. This restricts to give an n-sheeted covering S1S1.

Example 23.3.

For any space Y and any discrete set F with |F|=n, the projection Y×FY is an n-sheeted covering.

Example 23.4.

For any n>0, the projection p:SnPn is a 2-sheeted covering.

Lemma 23.5.

Let p:XY be an n-sheeted covering, and let u:ΔkY be continuous. Then there are precisely n different continuous maps ΔkX lifting u.


By assumption, the set F=p-1{u(e0)} has size n, say F={x1,,xn}. Proposition 22.13 tells us that for each i there us a unique lift u~i:ΔkX with u~i(e0)=xi. If u~:ΔkX is an arbitrary lift of u, then p(u~(e0))=u(e0) so u~(e0)F so u~(e0)=xi for some i, so u~=u~i. ∎

Definition 23.6.

Let p:XY be an n-sheeted covering. For any continuous map u:ΔkY, we define τ(u) to be the sum of all the lifts of u, considered as an element of Ck(X). More generally, given an element u=m1u1++mrurCk(Y), we define τ(u)=m1τ(u1)++mrτ(ur)Ck(X). This defines a homomorphism τ:Ck(Y)Ck(X), which is called the transfer.

Example 23.7.

Define p:×× by p(z)=z3, so this is a 3-sheeted covering. Define u:Δ1× by u(1-t,t)=8exp(2πit). Then u(e0)=8 so p-1{u(e0)}={2,2e2πi/3,2e4πi/3}. Define vj:Δ1× by vj(1-t,t)=2exp(2πi(t+j)/3) for j=0,1,2. These are the three lifts of u, so τ(u)=v0+v1+v2C1(×).

Proposition 23.8.

Let p:XY be an n-sheeted covering. Then the associated transfer map τ:C*(Y)C*(X) is a chain map, and satisfies p#(τ(u))=nu for all uCk(Y).


Consider a continuous map u:ΔkY, and let v1,,vn be the continuous lifts of u, so τ(u)=j=1nvj. This means that (τ(u))=i=0kj=1n(-1)i(vjδi). Now note that p(vjδi)=uδi, so vjδi is one of the lifts of uδi. If vjδi=vjδi then vj and vj agree at δi(e0) so they must be the same so j=j. This proves that the list v1δi,,vnδi is the complete list of lifts of uδi, so τ(uδi)=j=1n(vjδi). From this we get


This proves that τ is a chain map. As pvj=u for all j we also have p#(τ(u))=p#(j=1nvj)=j=1nu=nu. ∎

Remark 23.9.

It follows that we have an induced map τ*:H*(Y)H*(X), which satisfies p*(τ*(u))=nu for all uHk(Y).

We would like to use the transfer to obtain homological information about Pn. For this, it is convenient to use a slightly different version of homology.

Definition 23.10.

We define Ck(X;/2) to be the set of formal linear combinations m1u1++mrur where each ui is a continuous map ΔkX, but now the coefficients mi lie in /2 rather than . We again make this a chain complex by defining (u)=i=0k(uδi). (We have left out the sign (-1)i because it makes no difference mod 2.) We define H*(X;/2) to be the homology of this chain complex.

Remark 23.11.

If all the groups Hi(X) are free abelian groups, one can check that Hi(X;/2)=Hi(X)/2Hi(X) for all i. If some groups Hi(X) are not free abelian, then the relationship between Hi(X) and Hi(X;/2) is a little more complicated. In particular, this applies when X=Pn, because we have already seen that H1(Pn)=/2 for n>1, and this is not a free abelian group.

Remark 23.12.

Any element uCk(X;/2) can be expressed as a formal linear combination m1u1++mrur with mi/2. If ui=uj for some ij then we can combine the corresponding terms. We can then discard all terms with coefficient zero. As /2={0,1}, and remaining terms must have coefficient 1. This means that u can be expressed as u1++us, where the elements ui are distinct maps from Δk to X.

Remark 23.13.

Essentially everything that we have done previously works in the same way with coefficients /2. In particular, the groups H*(X;/2) are functorial and homotopy invariant, and we have Mayer-Vietoris sequences and transfers. For n>0 we have

Hk(Sn;/2)={/2 if k=0 or k=n0 otherwise. 
Lemma 23.14.

Let p:SnPn be the usual projection, which is a 2-sheeted covering. Then the sequence


is a short exact sequence of chain complexes and chain maps. It therefore gives a long exact sequence of homology groups


First suppose that uCk(Pn;/2). As in Remark 23.12, we can write u=u1++ur for some list of distinct maps ui:ΔkPn. Let ui and ui′′ be the two lifts of ui. Note that p#(iui)=u; this proves that p# is surjective. Note also that τ(u)=i(ui+ui′′). If ij then


so neither ui nor ui′′ can be equal to uj or uj′′. Also, uiui′′ by construction. Thus, there can be no cancellation in our expression for τ(u), so τ(u)0 except in the case where our original expression for u had no terms. This proves that τ is injective. We also have p#(τ(u))=2u, which is zero as we are working modulo 2. This shows that img(τ)ker(p#). Conversely, suppose we have an element vker(p#). As in Remark 23.12, we can write v=v1++vr for some distinct continuous maps vi:ΔkSn. Put ui=pvi:ΔkPn, so that p#(v)=u1++ur. We are assuming that vker(p#), so the sum u1++ur must cancel down to zero. After reordering the terms if necessary, we can assume that r=2r for some r and u2i-1=u2i for i=1,,r. This means that v2i-1 and v2i must be the two different lifts of u2i, so v=τ(u2+u4++u2r). We conclude that img(τ)=ker(p#). This completes the proof that we have a short exact sequence of chain complexes and chain maps, and the Snake Lemma gives the claimed long exact sequence of homology groups. ∎

Theorem 23.15.

For any n>0 we have

Hk(Pn;/2)={/2 if 0kn0 otherwise. 

Moreover, the map τ*:Hn(Pn;/2)Hn(Sn;/2) is an isomorphism, as are the maps Δ:Hi(Pn;/2)Hi-1(Pn;/2) for 1in.


We proved in Proposition 20.9 that

Hi(Pn)={ if i=0/2 if i=10 if i>n.

Essentially the same argument shows that

Hi(Pn;/2)={/2 if i=0,10 if i>n.

Next, we have a long exact sequence


For 2in-1 we have Hi(Sn;/2)=Hi-1(Sn;/2)=0, so the map Δ is an isomorphism. It follows by induction on i that Hi(Pn;/2)=/2 for 0in-1. Finally, we have an exact sequence


After filling in the known groups, this becomes


This shows that the first map τ* is injective, so Hn(Pn;/2) is isomorphic to a subgroup of /2, so it is either trivial or of order two. If it was trivial then the sequence could not be exact, so we must have Hn(Pn;/2)/2 as claimed. Given this, the only way the sequence can be exact is if τ* and Δ are isomorphisms, and p*=0. ∎