# 23. Transfers, coefficients and homology of projective spaces

###### Definition 23.1.

Let $n$ be a nonnegative integer. We say that a map $p\colon X\to Y$ is an $n$-sheeted covering if it is a covering map, and $|p^{-1}\{y\}|=n$ for all $y\in Y$.

###### Example 23.2.

For any $n>0$ we have an $n$-sheeted covering $p\colon{\mathbb{C}^{\times}}\to{\mathbb{C}^{\times}}$ given by $p(z)=z^{n}$. This restricts to give an $n$-sheeted covering $S^{1}\to S^{1}$.

###### Example 23.3.

For any space $Y$ and any discrete set $F$ with $|F|=n$, the projection $Y\times F\to Y$ is an $n$-sheeted covering.

###### Example 23.4.

For any $n>0$, the projection $p\colon S^{n}\to{\mathbb{R}}P^{n}$ is a $2$-sheeted covering.

###### Lemma 23.5.

Let $p\colon X\to Y$ be an $n$-sheeted covering, and let $u\colon\Delta_{k}\to Y$ be continuous. Then there are precisely $n$ different continuous maps $\Delta_{k}\to X$ lifting $u$.

###### Proof.

By assumption, the set $F=p^{-1}\{u(e_{0})\}$ has size $n$, say $F=\{x_{1},\dotsc,x_{n}\}$. Proposition 22.13 tells us that for each $i$ there us a unique lift $\widetilde{u}_{i}\colon\Delta_{k}\to X$ with $\widetilde{u}_{i}(e_{0})=x_{i}$. If $\widetilde{u}\colon\Delta_{k}\to X$ is an arbitrary lift of $u$, then $p(\widetilde{u}(e_{0}))=u(e_{0})$ so $\widetilde{u}(e_{0})\in F$ so $\widetilde{u}(e_{0})=x_{i}$ for some $i$, so $\widetilde{u}=\widetilde{u}_{i}$. ∎

###### Definition 23.6.

Let $p\colon X\to Y$ be an $n$-sheeted covering. For any continuous map $u\colon\Delta_{k}\to Y$, we define $\tau(u)$ to be the sum of all the lifts of $u$, considered as an element of $C_{k}(X)$. More generally, given an element $u=m_{1}u_{1}+\dotsb+m_{r}u_{r}\in C_{k}(Y)$, we define $\tau(u)=m_{1}\tau(u_{1})+\dotsb+m_{r}\tau(u_{r})\in C_{k}(X)$. This defines a homomorphism $\tau\colon C_{k}(Y)\to C_{k}(X)$, which is called the transfer.

###### Example 23.7.

Define $p\colon{\mathbb{C}^{\times}}\to{\mathbb{C}^{\times}}$ by $p(z)=z^{3}$, so this is a $3$-sheeted covering. Define $u\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ by $u(1-t,t)=8\exp(2\pi it)$. Then $u(e_{0})=8$ so $p^{-1}\{u(e_{0})\}=\{2,2e^{2\pi i/3},2e^{4\pi i/3}\}$. Define $v_{j}\colon\Delta_{1}\to{\mathbb{C}^{\times}}$ by $v_{j}(1-t,t)=2\exp(2\pi i(t+j)/3)$ for $j=0,1,2$. These are the three lifts of $u$, so $\tau(u)=v_{0}+v_{1}+v_{2}\in C_{1}({\mathbb{C}^{\times}})$.

###### Proposition 23.8.

Let $p\colon X\to Y$ be an $n$-sheeted covering. Then the associated transfer map $\tau\colon C_{*}(Y)\to C_{*}(X)$ is a chain map, and satisfies $p_{\#}(\tau(u))=nu$ for all $u\in C_{k}(Y)$.

###### Proof.

Consider a continuous map $u\colon\Delta_{k}\to Y$, and let $v_{1},\dotsc,v_{n}$ be the continuous lifts of $u$, so $\tau(u)=\sum_{j=1}^{n}v_{j}$. This means that $\partial(\tau(u))=\sum_{i=0}^{k}\sum_{j=1}^{n}(-1)^{i}(v_{j}\circ\delta_{i})$. Now note that $p\circ(v_{j}\circ\delta_{i})=u\circ\delta_{i}$, so $v_{j}\circ\delta_{i}$ is one of the lifts of $u\circ\delta_{i}$. If $v_{j}\circ\delta_{i}=v_{j^{\prime}}\circ\delta_{i}$ then $v_{j}$ and $v_{j^{\prime}}$ agree at $\delta_{i}(e_{0})$ so they must be the same so $j=j^{\prime}$. This proves that the list $v_{1}\circ\delta_{i},\dotsc,v_{n}\circ\delta_{i}$ is the complete list of lifts of $u\circ\delta_{i}$, so $\tau(u\circ\delta_{i})=\sum_{j=1}^{n}(v_{j}\circ\delta_{i})$. From this we get

 $\tau(\partial(u))=\sum_{i=0}^{k}(-1)^{i}\tau(u\circ\delta_{i})=\sum_{i=0}^{k}% \sum_{j=1}^{n}(-1)^{i}(v_{j}\circ\delta_{i})=\partial(\tau(u)).$

This proves that $\tau$ is a chain map. As $p\circ v_{j}=u$ for all $j$ we also have $p_{\#}(\tau(u))=p_{\#}(\sum_{j=1}^{n}v_{j})=\sum_{j=1}^{n}u=n\,u$. ∎

###### Remark 23.9.

It follows that we have an induced map $\tau_{*}\colon H_{*}(Y)\to H_{*}(X)$, which satisfies $p_{*}(\tau_{*}(u))=n\,u$ for all $u\in H_{k}(Y)$.

We would like to use the transfer to obtain homological information about ${\mathbb{R}}P^{n}$. For this, it is convenient to use a slightly different version of homology.

###### Definition 23.10.

We define $C_{k}(X;{\mathbb{Z}}/2)$ to be the set of formal linear combinations $m_{1}u_{1}+\dotsb+m_{r}u_{r}$ where each $u_{i}$ is a continuous map $\Delta_{k}\to X$, but now the coefficients $m_{i}$ lie in ${\mathbb{Z}}/2$ rather than ${\mathbb{Z}}$. We again make this a chain complex by defining $\partial(u)=\sum_{i=0}^{k}(u\circ\delta_{i})$. (We have left out the sign $(-1)^{i}$ because it makes no difference mod $2$.) We define $H_{*}(X;{\mathbb{Z}}/2)$ to be the homology of this chain complex.

###### Remark 23.11.

If all the groups $H_{i}(X)$ are free abelian groups, one can check that $H_{i}(X;{\mathbb{Z}}/2)=H_{i}(X)/2H_{i}(X)$ for all $i$. If some groups $H_{i}(X)$ are not free abelian, then the relationship between $H_{i}(X)$ and $H_{i}(X;{\mathbb{Z}}/2)$ is a little more complicated. In particular, this applies when $X={\mathbb{R}}P^{n}$, because we have already seen that $H_{1}({\mathbb{R}}P^{n})={\mathbb{Z}}/2$ for $n>1$, and this is not a free abelian group.

###### Remark 23.12.

Any element $u\in C_{k}(X;{\mathbb{Z}}/2)$ can be expressed as a formal linear combination $m_{1}u_{1}+\dotsb+m_{r}u_{r}$ with $m_{i}\in{\mathbb{Z}}/2$. If $u_{i}=u_{j}$ for some $i\neq j$ then we can combine the corresponding terms. We can then discard all terms with coefficient zero. As ${\mathbb{Z}}/2=\{0,1\}$, and remaining terms must have coefficient $1$. This means that $u$ can be expressed as $u_{1}+\dotsb+u_{s}$, where the elements $u_{i}$ are distinct maps from $\Delta_{k}$ to $X$.

###### Remark 23.13.

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

 $H_{k}(S^{n};{\mathbb{Z}}/2)=\begin{cases}{\mathbb{Z}}/2&\text{ if }k=0\text{ % or }k=n\\ 0&\text{ otherwise. }\end{cases}$
###### Lemma 23.14.

Let $p\colon S^{n}\to{\mathbb{R}}P^{n}$ be the usual projection, which is a $2$-sheeted covering. Then the sequence

 $C_{*}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\xrightarrow{\tau}C_{*}(S^{n};{\mathbb{% Z}}/2)\xrightarrow{p_{\#}}C_{*}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)$

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

 $H_{i}(S^{n};{\mathbb{Z}}/2)\xrightarrow{p_{*}}H_{i}({\mathbb{R}}P^{n};{\mathbb% {Z}}/2)\xrightarrow{\Delta}H_{i-1}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)% \xrightarrow{\tau_{*}}H_{i-1}(S^{n};{\mathbb{Z}}/2)\xrightarrow{p_{*}}H_{i-1}(% {\mathbb{R}}P^{n};{\mathbb{Z}}/2)$
###### Proof.

First suppose that $u\in C_{k}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)$. As in Remark 23.12, we can write $u=u_{1}+\dotsc+u_{r}$ for some list of distinct maps $u_{i}\colon\Delta_{k}\to{\mathbb{R}}P^{n}$. Let $u^{\prime}_{i}$ and $u^{\prime\prime}_{i}$ be the two lifts of $u_{i}$. Note that $p_{\#}(\sum_{i}u^{\prime}_{i})=u$; this proves that $p_{\#}$ is surjective. Note also that $\tau(u)=\sum_{i}(u^{\prime}_{i}+u^{\prime\prime}_{i})$. If $i\neq j$ then

 $p\circ u^{\prime}_{i}=p\circ u^{\prime\prime}_{i}=u_{i}\neq u_{j}=p\circ u^{% \prime}_{j}=p\circ u^{\prime\prime}_{j},$

so neither $u^{\prime}_{i}$ nor $u^{\prime\prime}_{i}$ can be equal to $u^{\prime}_{j}$ or $u^{\prime\prime}_{j}$. Also, $u^{\prime}_{i}\neq u^{\prime\prime}_{i}$ by construction. Thus, there can be no cancellation in our expression for $\tau(u)$, so $\tau(u)\neq 0$ except in the case where our original expression for $u$ had no terms. This proves that $\tau$ is injective. We also have $p_{\#}(\tau(u))=2u$, which is zero as we are working modulo $2$. This shows that $\operatorname{img}(\tau)\leq\ker(p_{\#})$. Conversely, suppose we have an element $v\in\ker(p_{\#})$. As in Remark 23.12, we can write $v=v_{1}+\dotsc+v_{r}$ for some distinct continuous maps $v_{i}\colon\Delta_{k}\to S^{n}$. Put $u_{i}=p\circ v_{i}\colon\Delta_{k}\to{\mathbb{R}}P^{n}$, so that $p_{\#}(v)=u_{1}+\dotsc+u_{r}$. We are assuming that $v\in\ker(p_{\#})$, so the sum $u_{1}+\dotsc+u_{r}$ must cancel down to zero. After reordering the terms if necessary, we can assume that $r=2r^{\prime}$ for some $r^{\prime}$ and $u_{2i-1}=u_{2i}$ for $i=1,\dotsc,r^{\prime}$. This means that $v_{2i-1}$ and $v_{2i}$ must be the two different lifts of $u_{2i}$, so $v=\tau(u_{2}+u_{4}+\dotsc+u_{2r^{\prime}})$. We conclude that $\operatorname{img}(\tau)=\ker(p_{\#})$. 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

 $H_{k}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)=\begin{cases}{\mathbb{Z}}/2&\text{ if % }0\leq k\leq n\\ 0&\text{ otherwise. }\end{cases}$

Moreover, the map $\tau_{*}\colon H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\to H_{n}(S^{n};{\mathbb% {Z}}/2)$ is an isomorphism, as are the maps $\Delta\colon H_{i}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\to H_{i-1}({\mathbb{R}}P^% {n};{\mathbb{Z}}/2)$ for $1\leq i\leq n$.

###### Proof.

We proved in Proposition 20.9 that

 $H_{i}({\mathbb{R}}P^{n})=\begin{cases}{\mathbb{Z}}&\text{ if }i=0\\ {\mathbb{Z}}/2&\text{ if }i=1\\ 0&\text{ if }i>n.\end{cases}$

Essentially the same argument shows that

 $H_{i}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)=\begin{cases}{\mathbb{Z}}/2&\text{ if % }i=0,1\\ 0&\text{ if }i>n.\end{cases}$

Next, we have a long exact sequence

 $H_{i}(S^{n};{\mathbb{Z}}/2)\xrightarrow{p_{*}}H_{i}({\mathbb{R}}P^{n};{\mathbb% {Z}}/2)\xrightarrow{\Delta}H_{i-1}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)% \xrightarrow{\tau_{*}}H_{i-1}(S^{n};{\mathbb{Z}}/2)$

For $2\leq i\leq n-1$ we have $H_{i}(S^{n};{\mathbb{Z}}/2)=H_{i-1}(S^{n};{\mathbb{Z}}/2)=0$, so the map $\Delta$ is an isomorphism. It follows by induction on $i$ that $H_{i}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)={\mathbb{Z}}/2$ for $0\leq i\leq n-1$. Finally, we have an exact sequence

 $H_{n+1}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\xrightarrow{\Delta}H_{n}({\mathbb{R}% }P^{n};{\mathbb{Z}}/2)\xrightarrow{\tau_{*}}H_{n}(S^{n};{\mathbb{Z}}/2)% \xrightarrow{p_{*}}H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\xrightarrow{\Delta}% H_{n-1}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\xrightarrow{\tau_{*}}H_{n-1}(S^{n};{% \mathbb{Z}}/2).$

After filling in the known groups, this becomes

 $0\xrightarrow{}H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\xrightarrow{\tau_{*}}{% \mathbb{Z}}/2\xrightarrow{p_{*}}H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)% \xrightarrow{\Delta}{\mathbb{Z}}/2\xrightarrow{}0.$

This shows that the first map $\tau_{*}$ is injective, so $H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)$ is isomorphic to a subgroup of ${\mathbb{Z}}/2$, so it is either trivial or of order two. If it was trivial then the sequence could not be exact, so we must have $H_{n}({\mathbb{R}}P^{n};{\mathbb{Z}}/2)\simeq{\mathbb{Z}}/2$ as claimed. Given this, the only way the sequence can be exact is if $\tau_{*}$ and $\Delta$ are isomorphisms, and $p_{*}=0$. ∎