Let be a nonnegative integer. We say that a map is an -sheeted covering if it is a covering map, and for all .
For any we have an -sheeted covering given by . This restricts to give an -sheeted covering .
For any space and any discrete set with , the projection is an -sheeted covering.
For any , the projection is a -sheeted covering.
Let be an -sheeted covering, and let be continuous. Then there are precisely different continuous maps lifting .
By assumption, the set has size , say . Proposition 22.13 tells us that for each there us a unique lift with . If is an arbitrary lift of , then so so for some , so . ∎
Let be an -sheeted covering. For any continuous map , we define to be the sum of all the lifts of , considered as an element of . More generally, given an element , we define . This defines a homomorphism , which is called the transfer.
Define by , so this is a -sheeted covering. Define by . Then so . Define by for . These are the three lifts of , so .
Let be an -sheeted covering. Then the associated transfer map is a chain map, and satisfies for all .
Consider a continuous map , and let be the continuous lifts of , so . This means that . Now note that , so is one of the lifts of . If then and agree at so they must be the same so . This proves that the list is the complete list of lifts of , so . From this we get
This proves that is a chain map. As for all we also have . ∎
It follows that we have an induced map , which satisfies for all .
We would like to use the transfer to obtain homological information about . For this, it is convenient to use a slightly different version of homology.
We define to be the set of formal linear combinations where each is a continuous map , but now the coefficients lie in rather than . We again make this a chain complex by defining . (We have left out the sign because it makes no difference mod .) We define to be the homology of this chain complex.
If all the groups are free abelian groups, one can check that for all . If some groups are not free abelian, then the relationship between and is a little more complicated. In particular, this applies when , because we have already seen that for , and this is not a free abelian group.
Any element can be expressed as a formal linear combination with . If for some then we can combine the corresponding terms. We can then discard all terms with coefficient zero. As , and remaining terms must have coefficient . This means that can be expressed as , where the elements are distinct maps from to .
Essentially everything that we have done previously works in the same way with coefficients . In particular, the groups are functorial and homotopy invariant, and we have Mayer-Vietoris sequences and transfers. For we have
Let be the usual projection, which is a -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 . As in Remark 23.12, we can write for some list of distinct maps . Let and be the two lifts of . Note that ; this proves that is surjective. Note also that . If then
so neither nor can be equal to or . Also, by construction. Thus, there can be no cancellation in our expression for , so except in the case where our original expression for had no terms. This proves that is injective. We also have , which is zero as we are working modulo . This shows that . Conversely, suppose we have an element . As in Remark 23.12, we can write for some distinct continuous maps . Put , so that . We are assuming that , so the sum must cancel down to zero. After reordering the terms if necessary, we can assume that for some and for . This means that and must be the two different lifts of , so . We conclude that . 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. ∎
For any we have
Moreover, the map is an isomorphism, as are the maps for .
Next, we have a long exact sequence
For we have , so the map is an isomorphism. It follows by induction on that for . Finally, we have an exact sequence
After filling in the known groups, this becomes
This shows that the first map is injective, so is isomorphic to a subgroup of , so it is either trivial or of order two. If it was trivial then the sequence could not be exact, so we must have as claimed. Given this, the only way the sequence can be exact is if and are isomorphisms, and . ∎