# 15. Homology of spheres

Consider a topological space $X$ with open subsets $U$ and $V$ such that $X=U\cup V$. There are many cases like this where we already understand three of the groups $H_{*}(U)$, $H_{*}(V)$, $H_{*}(U\cap V)$ and $H_{*}(U\cup V)=H_{*}(X)$, and we want to determine the fourth one.

Video (Theorem 15.1 and Proposition 15.2)

We name the corresponding inclusion maps as shown on the left below; they induce homomorphisms of homology groups as shown on the right.

As the left hand diagram commutes and homology is functorial, we have $k_{*}i_{*}=l_{*}j_{*}\colon H_{*}(U\cap V)\to H_{*}(U\cup V)$.

We can now combine the above maps to get maps as follows:

 $H_{n}(U\cap V)\xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{n}(U)\oplus H_{n}(V)\xrightarrow{\left[% \begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}H_{n}(U\cup V)$

Here the notation $\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]$ refers to the map sending $a$ to $(i_{*}(a),-j_{*}(a))$, and the notation $\left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]$ refers to the map sending $(b,c)$ to $k_{*}(b)+l_{*}(c)$. The composite of these two maps therefore sends $a$ to $k_{*}i_{*}(a)-l_{*}j_{*}(a)$, but this is zero because $k_{*}i_{*}=l_{*}j_{*}$. Thus, the above sequence has a chance to be exact. In fact, we have the following result:

###### Theorem 15.1(Mayer-Vietoris).

In the above context, there are natural maps $\delta\colon H_{n}(U\cup V)\to H_{n-1}(U\cap V)$ such that the resulting sequence

 $H_{n+1}(U\cup V)\xrightarrow{\delta}H_{n}(U\cap V)\xrightarrow{\left[\begin{% smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{n}(U)\oplus H_{n}(V)\xrightarrow{\left[% \begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}H_{n}(U\cup V)% \xrightarrow{\delta}H_{n-1}(U\cap V)$

is exact for all $n$.

In this section, we will explain some homology calculations based on the above theorem. The proof will be given in Section 19, using preliminary results from Sections 17 and 18.

We first note that the exact sequences in the theorem can be chained together to make an infinite sequence as follows:

The following slightly modified form is often convenient.

###### Proposition 15.2.

Suppose that $U\cap V$ is path connected (and therefore nonempty). Then we can modify the Mayer-Vietoris sequence by replacing all the $H_{0}$ terms by zero, and the resulting sequence is still exact. Moreover, if $U$ and $V$ are also path connected then the same is true of $U\cup V$, so all the $H_{0}$ groups are isomorphic to ${\mathbb{Z}}$.

###### Proof.

Pick a point $a\in U\cap V$, so $H_{0}(U\cap V)={\mathbb{Z}}.[a]$. Now $H_{0}(U)$ has a basis corresponding to the elements of $\pi_{0}(U)$, and $i_{*}[a]$ is one of these basis elements, so the map $i_{*}\colon H_{0}(U\cap V)\to H_{0}(U)$ is injective. It follows that the map $\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]\colon H_{0}(U\cap V)\to H_{0}(U)\oplus H_{0}(V)$ is also injective. Now consider the tail end of the Mayer-Vietoris sequence:

 $H_{1}(U)\oplus H_{1}(V)\xrightarrow{\left[\begin{smallmatrix}k_{*}&l_{*}\end{% smallmatrix}\right]}H_{1}(U\cup V)\xrightarrow{\delta}H_{0}(U\cap V)% \xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{0}(U)\oplus H_{0}(V).$

The map $\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]$ is injective, and so the kernel is zero. As the sequence is exact, we see that the image of $\delta$ is also zero, so $\delta$ is the zero homomorphism. Using exactness again, we deduce that the map $\left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]$ must be surjective. Using this together with the exactness of the original MV sequence, we see that the sequence

 $H_{1}(U\cap V)\xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{1}(U)\oplus H_{1}(V)\xrightarrow{\left[% \begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}H_{1}(U\cup V)% \xrightarrow{}0\xrightarrow{}0\xrightarrow{}0$

is exact. The rest of the modified MV sequences is the same as the original MV sequence, and so is also exact, as claimed.

Now suppose that $U$ and $V$ are also path connected. Any point in $U$ can be joined to $a$ by a path in $U$, and any point in $V$ can be joined to $a$ by a path in $V$, so any point in the space $X=U\cup V$ can be joined to $a$ by a path in $X$. This shows that $X$ is also path connected. We therefore see that the groups $H_{0}(X)$, $H_{0}(U)$, $H_{0}(V)$ and $H_{0}(U\cap V)$ are all the same: a copy of ${\mathbb{Z}}$, generated by $[a]$. ∎

We now want to calculate the homology groups of spheres $S^{n}$ for $n\geq 0$. We will do this by induction, starting with $n=0$. Note that the point $e_{0}=(1,0,\dotsb,0)$ always lies in $S^{n}$ and so gives an element $a_{n}=[e_{0}]\in H_{0}(S^{n})$. Note also that

 $S^{0}=\{x\in{\mathbb{R}}\;|\;|x|=1\}=\{e_{0},-e_{0}\}.$

As in Remark 10.24, it follows that the elements $a_{0}=[e_{0}]$ and $[-e_{0}]$ give a basis for $H_{0}(S^{0})$, and that $H_{k}(S^{0})=0$ for $k\neq 0$. However, it turns out to be more convenient to put $b_{0}=[-e_{0}]-[e_{0}]\in H_{0}(S^{0})$; the elements $a_{0}$ and $b_{0}$ also form a basis.

###### Theorem 15.3.

For all $n\geq 1$ there is an element $b_{n}\in H_{n}(S^{n})$ such that

 $H_{k}(S^{n})=\begin{cases}{\mathbb{Z}}a_{n}&\text{ if }k=0\\ {\mathbb{Z}}b_{n}&\text{ if }k=n\\ 0&\text{ otherwise. }\end{cases}$
###### Proof.

Recall that

 $S^{n}=\{(x_{0},\dotsc,x_{n})\in{\mathbb{R}}^{n+1}\;|\;\sum_{i}x_{i}^{2}=1\}.$

We will think of a point $x\in{\mathbb{R}}^{n+1}$ as a pair $(y,z)$ with $y\in{\mathbb{R}}^{n}$ and $z\in{\mathbb{R}}$, so $\|x\|^{2}=\|y\|^{2}+z^{2}$, and $x$ lies in $S^{n}$ iff $\|y\|^{2}+z^{2}=1$. We put $N=(0,1)$ (the “North pole”) and $U=S^{n}\setminus\{N\}=\{(y,z)\in S^{n}\;|\;z\neq 1\}$ and $V=S^{n}\setminus\{-N\}=\{(y,z)\in S^{n}\;|\;z\neq-1\}$. These are open sets with $U\cup V=S^{n}$ and

 $U\cap V=S^{n}\setminus\{N,-N\}=\{(y,z)\in S^{n}\;|\;-1

Just as in Example 4.13 we have a homeomorphism $f\colon U\to{\mathbb{R}}^{n}$ given by $f(y,z)=y/(1-z)$ with $f^{-1}(y)=(2y,\|y\|^{2}-1)/(\|y\|^{2}+1)$. This proves that $U$ is contractible. The map $(y,z)\mapsto(y,-z)$ gives a homeomorphism between $U$ and $V$, so $V$ is also contractible. It follows that $H_{0}(U)=H_{0}(V)={\mathbb{Z}}$ and $H_{d}(U)=H_{d}(V)=0$ for $d>0$.

We also have a homeomorphism $g\colon U\cap V\to S^{n-1}\times(-1,1)$ given by $g(y,z)=y/\|y\|$ and $g^{-1}(y,z)=(\sqrt{1-z_{2}}y,z)$. (It is valid to divide by $\|y\|$ here because $y\neq 0$ whenever $(y,z)\in U\cap V$.) As $(-1,1)$ is contractible, it follows that $U\cap V$ is homotopy equivalent to $S^{n-1}$. More precisely, we can define $p\colon S^{n-1}\to U\cap V$ by $p(y)=(y,0)$, and we find that $p$ is a homotopy equivalence, so the map $p_{*}\colon H_{*}(S^{n-1})\to H_{*}(U\cap V)$ is an isomorphism.

Now consider the Mayer-Vietoris sequence

 $H_{d}(U)\oplus H_{d}(V)\to H_{d}(S^{n})\xrightarrow{\delta}H_{d-1}(U\cap V)\to H% _{d-1}(U)\oplus H_{d-1}(V).$
• (a)

For $d\geq 2$ we have $H_{d}(U)=H_{d}(V)=H_{d-1}(U)=H_{d-1}(V)=0$. Thus, Lemma 12.19 tells us that in these cases the map $\delta\colon H_{d}(S^{n})\to H_{d-1}(U\cap V)=H_{d-1}(S^{n-1})$ is an isomorphism. This almost proves the induction step, apart from tweaks needed when $d=0,1$. We are assuming that $n\geq 1$ so $S^{n}$ is path connected so $H_{0}(S^{n})={\mathbb{Z}}.a_{n}$. Thus, we just need to deal with the case $d=1$.

• (b)

Using the known values of $H_{d}(U)$ and $H_{d}(V)$ for $d=0,1$, the bottom end of the Mayer-Vietoris sequence is as follows:

 $0\to H_{1}(S^{n})\xrightarrow{\delta}H_{0}(S^{n-1})\xrightarrow{\left[\begin{% smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}{\mathbb{Z}}\oplus{\mathbb{Z}}% \xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{0}(S^{n})\to 0.$
• (c)

Consider the case where $n=1$. Point (a) tells us that for $d\geq 2$ we have $H_{d}(S^{1})=H_{d-1}(S^{0})=0$. Point (b) gives an exact sequence

 $0\to H_{1}(S^{1})\xrightarrow{\delta}{\mathbb{Z}}\{a_{0},b_{0}\}\xrightarrow{% \left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]}{\mathbb{Z}}% \oplus{\mathbb{Z}}\xrightarrow{\left[\begin{smallmatrix}i_{*}\\ -j_{*}\end{smallmatrix}\right]}H_{0}(S^{1})\to 0.$

This means that $\delta$ gives an isomorphism from $H_{1}(S^{1})$ to the kernel of the map $\left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]$.

Here we have identified $H_{0}(U)$ with ${\mathbb{Z}}$. More precisely, as $U$ is connected and contains both $e_{0}$ and $-e_{0}$, we see that $H_{0}(U)={\mathbb{Z}}.[e_{0}]$ and that $[e_{0}]=[-e_{0}]$ in $H_{0}(U)$. It follows that $k_{*}(a_{0})=[e_{0}]$ and $k_{*}(b_{0})=[-e_{0}]-[e_{0}]=0$ in $H_{0}(U)$. For essentially the same reason we have $l_{*}(a_{0})=[e_{0}]$ and $l_{*}(b_{0})=0$ in $H_{0}(V)$. Thus, the kernel of $\left[\begin{smallmatrix}k_{*}&l_{*}\end{smallmatrix}\right]$ is ${\mathbb{Z}}.b_{0}$. The exact sequence therefore means that there is a unique element $b_{1}\in H_{1}(S^{1})$ such that $\delta(b_{1})=b_{0}$, and that $H_{1}(S^{1})={\mathbb{Z}}.b_{1}$. As $S^{1}$ is path connected, we also know that $H_{0}(S^{1})={\mathbb{Z}}.a_{1}$. This proves that $H_{*}(S^{1})$ is as claimed.

• (d)

Now suppose instead that $n\geq 2$, and that we have already proved the claim for $H_{*}(S^{n-1})$. Point (a) gives an isomorphism $\delta\colon H_{n}(S^{n})\to H_{n-1}(S^{n-1})={\mathbb{Z}}.b_{n-1}$, so there is a unique element $b_{n}\in H_{n}(S^{n})$ with $\delta(b_{n})=b_{n-1}$, and we have $H_{n}(S^{n})={\mathbb{Z}}.b_{n}$. Point (a) also shows that $H_{d}(S^{n})=0$ for $d\geq 2$ with $d\neq n$. As $S^{n}$ is path connected, we have $H_{0}(S^{n})={\mathbb{Z}}.a_{n}$. This just leaves $H_{1}(S^{n})$. The bottom end of the modified Mayer-Vietoris sequence has the form $0=H_{1}(U)\oplus H_{1}(V)\to H_{1}(S^{n})\to 0$, and exactness forces $H_{1}(S^{n})$ to be zero as expected.

###### Remark 15.4.

If $u\in H_{d}(X)$ then we sometimes say that $u\in H_{*}(X)$ with $|u|=d$, and we call $d$ the degree of $u$. We could then say that the elements $a_{n}$ and $b_{n}$ form a basis for $H_{*}(S^{n})$ with $|a_{n}|=0$ and $|b_{n}|=n$. This formulation is valid for $n=0$ as well as for $n>0$.