We now return to the task of constructing the Mayer-Vietoris sequence. There are two key ingredients: the Snake Lemma (in this section) and subdivision (in the next section). The videos cover this material in a slightly different order than the notes: the first video is attached to Definition 17.3 below.
The basic input for the Snake Lemma is as follows: we have chain complexes ${U}_{*}$, ${V}_{*}$ and ${W}_{*}$ and chain maps
$${U}_{*}\stackrel{\mathit{i}}{\to}{V}_{*}\stackrel{\mathit{p}}{\to}{W}_{*}$$ |
which form a short exact sequence. One might hope that the resulting sequence
$${H}_{*}(U)\stackrel{{i}_{*}}{\to}{H}_{*}(V)\stackrel{{p}_{*}}{\to}{H}_{*}(W)$$ |
would also be a short exact sequence, but that is not quite right. We will show that the above sequence is exact (in the sense that $\mathrm{img}({i}_{*})=\mathrm{ker}({p}_{*})$), but ${i}_{*}$ need not be injective, and ${p}_{*}$ need not be surjective. In other words, $\mathrm{ker}({i}_{*})$ need not be zero, and $\mathrm{img}({p}_{*})$ need not be all of ${H}_{*}(W)$. We can still obtain a great deal of information about $\mathrm{ker}({i}_{*})$ and $\mathrm{img}({p}_{*})$, but that will require some preparation. For the moment we will just prove the easier statement mentioned above.
Let ${U}_{\mathrm{*}}\stackrel{\mathit{i}}{\mathrm{\to}}{V}_{\mathrm{*}}\stackrel{\mathit{p}}{\mathrm{\to}}{W}_{\mathrm{*}}$ be a short exact sequence of chain maps between chain complexes. Then in the resulting sequence ${H}_{\mathrm{*}}\mathit{}\mathrm{(}U\mathrm{)}\stackrel{{i}_{\mathrm{*}}}{\mathrm{\to}}{H}_{\mathrm{*}}\mathit{}\mathrm{(}V\mathrm{)}\stackrel{{p}_{\mathrm{*}}}{\mathrm{\to}}{H}_{\mathrm{*}}\mathit{}\mathrm{(}W\mathrm{)}$ we have $\mathrm{img}\mathit{}\mathrm{(}{i}_{\mathrm{*}}\mathrm{)}\mathrm{=}\mathrm{ker}\mathit{}\mathrm{(}{p}_{\mathrm{*}}\mathrm{)}$.
First, as the sequence ${U}_{*}\stackrel{\mathit{i}}{\to}{V}_{*}\stackrel{\mathit{p}}{\to}{W}_{*}$ is exact we have $p\circ i=0$. It follows that ${p}_{*}\circ {i}_{*}={(p\circ i)}_{*}={0}_{*}=0$, so $\mathrm{img}({i}_{*})\le \mathrm{ker}({p}_{*})$.
Conversely, suppose we are given an element $b\in \mathrm{ker}({p}_{*})$; must show that it lies in $\mathrm{img}({i}_{*})$. If $b\in {H}_{r}(V)$ then we have $b=[v]$ for some $v\in {V}_{r}$ with $d(v)=0$. We are assuming that ${p}_{*}b=0$, which means that $[p(v)]$ is zero in the quotient group ${Z}_{r}(W)/{B}_{r}(W)$, which means that $p(v)\in {B}_{r}(W)$, which means that $p(v)=d({w}^{\prime})$ for some ${w}^{\prime}\in {W}_{r+1}$. Also, we are assuming that the sequence ${U}_{*}\stackrel{\mathit{i}}{\to}{V}_{*}\stackrel{\mathit{p}}{\to}{W}_{*}$ is short exact, which means in particular that $p$ is surjective. We can therefore choose ${v}^{\prime}\in {V}_{r+1}$ with $p({v}^{\prime})={w}^{\prime}$. We now have
$$p(v-d({v}^{\prime}))=p(v)-p(d({v}^{\prime}))=p(v)-d(p({v}^{\prime}))=p(v)-d({w}^{\prime})=0,$$ |
so $v-d({v}^{\prime})\in \mathrm{ker}(p)$. We also have $\mathrm{ker}(p)=\mathrm{img}(i)$ by our exactness assumption, so we can find $u\in {U}_{r}$ with $i(u)=v-d({v}^{\prime})$. From our initial assumptions we have $d(v)=0$, and also ${d}^{2}=0$ so $d(d({v}^{\prime}))=0$, so $d(i(u))=0$. As $i$ is a chain map this gives $i(d(u))=0$, and $i$ is injective so $d(u)=0$. This means we have an element $a=[u]\in {H}_{r}(U)$. This satisfies ${i}_{*}(a)=[i(u)]=[v-d({v}^{\prime})]$ but $d({v}^{\prime})\in {B}_{r}(V)$ so $[v-d({v}^{\prime})]$ is the same as $[v]$, which is $b$. We conclude that ${i}_{*}(a)=b$, so $b\in \mathrm{img}({i}_{*})$ as claimed.
We can display the relevant groups and elements as follows:
The two dotted arrows are supposed to indicate the relation $d({v}^{\prime})+i(u)=v$. ∎
For ${U}_{\mathrm{*}}\stackrel{\mathit{i}}{\mathrm{\to}}{V}_{\mathrm{*}}\stackrel{\mathit{p}}{\mathrm{\to}}{W}_{\mathrm{*}}$ as above, there is a natural map $\delta \mathrm{:}{H}_{n}\mathit{}\mathrm{(}W\mathrm{)}\mathrm{\to}{H}_{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}U\mathrm{)}$ such that the sequence
is exact for all $n$.
The proof will be broken into a number of steps. The map $\delta $ will be defined in Definition 17.6, and Propositions 17.1, 17.10 and 17.11 will show that the resulting long sequence is exact.
A snake for the above sequence is a system $(c,w,v,u,a)$ such that
$c\in {H}_{n}(W)$;
$w\in {Z}_{n}(W)$ is a cycle such that $c=[w]$;
$v\in {V}_{n}$ is an element with $p(v)=w$;
$u\in {Z}_{n-1}(U)$ is a cycle with $i(u)=d(v)\in {V}_{n-1}$;
$a=[u]\in {H}_{n-1}(U)$.
More specifically, we say that a system $(c,w,v,u,a)$ as above is a snake from $c$ to $a$.
For any $c\mathrm{\in}{H}_{n}\mathit{}\mathrm{(}W\mathrm{)}$, there is a snake starting with $c$.
Consider an element $c\in {H}_{n}(W)$. As ${H}_{n}(W)={Z}_{n}(W)/{B}_{n}(W)$ by definition, we can certainly choose $w\in {Z}_{n}(W)$ such that $c=[w]$. As the sequence ${U}_{*}\stackrel{\mathit{i}}{\to}{V}_{*}\stackrel{\mathit{p}}{\to}{W}_{*}$ is short exact, we know that $p:{V}_{n}\to {W}_{n}$ is surjective, so we can choose $v\in {V}_{n}$ with $p(v)=w$. As $p$ is a chain map we have $p(d(v))=d(p(v))=d(w)=0$ (the last equation because $w\in {Z}_{n}(W)$). This means that $d(v)\in \mathrm{ker}(p)$, but $\mathrm{ker}(p)=\mathrm{img}(i)$ because the sequence is exact, so we have $u\in {U}_{n-1}$ with $i(u)=d(v)$. Note also that $i(d(u))=d(i(u))=d(d(v))=0$ (because $i$ is a chain map and ${d}^{2}=0$). On the other hand, exactness means that $i$ is injective, so the relation $i(d(u))=0$ implies that $d(u)=0$. This shows that $u\in {Z}_{n-1}(U)$, so we can put $a=[u]\in {H}_{n-1}(U)$. We now have a snake $(c,w,v,u,a)$ starting with $c$ as required. ∎
Suppose we have two snakes that have the same starting point; then they also have the same endpoint.
Suppose we have two snakes that start with $c$. We can then subtract them to get a snake $(0,w,v,u,a)$ starting with $0$. It will be enough to show that this ends with $0$ as well, or equivalently that $a=0$. The first snake condition says that $[w]=0$, which means that $w=d({w}^{\prime})$ for some ${w}^{\prime}\in {W}_{n+1}$. Because $p$ is surjective we can also choose ${v}^{\prime}\in {V}_{n+1}$ with ${w}^{\prime}=p({v}^{\prime})$ , and this gives $w=d({w}^{\prime})=d(p({v}^{\prime}))=p(d({v}^{\prime}))$. The next snake condition says that $p(v)=w$. We can combine these facts to see that $p(v-d({v}^{\prime}))=0$, so $v-d({v}^{\prime})\in \mathrm{ker}(p)=\mathrm{img}(i)$. We can therefore find ${u}^{\prime}\in {U}_{n}$ with $v-d({v}^{\prime})=i({u}^{\prime})$. We can apply $d$ to this using ${d}^{2}=0$ and $di=id$ to get $d(v)=i(d({u}^{\prime}))$. On the other hand, the third snake condition tells us that $d(v)=i(u)$. Subtracting these gives $i(u-d({u}^{\prime}))=0$, but $i$ is injective, so $u=d({u}^{\prime})$, so $u\in {B}_{n-1}(U)$. The final snake condition now says that $a=[u]=u+{B}_{n-1}(U)$, but $u\in {B}_{n-1}(U)$ so $a=[u]=0$. ∎
For any $c\in {H}_{n}(W)$, we define $\delta (c)\in {H}_{n-1}(U)$ to be the endpoint of any snake that starts with $c$. (This is well-defined by the last two lemmas.)
It is easy to see that the sum of two snakes is a snake, and from that we can deduce that $\delta $ is a homomorphism.
The slogan behind the definition is that $\delta ={i}^{-1}d{p}^{-1}$. In more detail, suppose we have $c\in {H}_{n}(W)$. To calculate $\delta (c)$, we must find a snake of the form $(c,w,v,u,a)$, then $\delta (c)=a$. The slogan glosses over the distinction between $w$ and $c=[w]$, and the distinction between $u$ and $a=[u]$. The condition $p(v)=w$ means that $v$ is a choice of ${p}^{-1}w$, and the condition $i(u)=d(v)$ means that $u$ is essentially ${i}^{-1}(d(v))={i}^{-1}(d({p}^{-1}(w)))$. The point of the above definitions and lemmas is to make this slogan precise.
The Snake Lemma (in a slightly different incarnation) is probably the most advanced piece of mathematics ever to appear in a mainstream movie:
The sequence ${H}_{n}\mathit{}\mathrm{(}V\mathrm{)}\stackrel{{p}_{\mathrm{*}}}{\mathrm{\to}}{H}_{n}\mathit{}\mathrm{(}W\mathrm{)}\stackrel{\mathit{\delta}}{\mathrm{\to}}{H}_{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}U\mathrm{)}$ is exact (or equivalently, $\mathrm{img}\mathit{}\mathrm{(}{p}_{\mathrm{*}}\mathrm{)}\mathrm{=}\mathrm{ker}\mathit{}\mathrm{(}\delta \mathrm{)}$).
First, suppose that $b\in {H}_{n}(V)$, so $b=[v]$ for some $v\in {V}_{n}$ with $d(v)=0$. We find that $({p}_{*}(b),p(v),v,0,0)$ is a snake starting with ${p}_{*}(b)$, so $\delta ({p}_{*}(b))=0$. From this we get $\delta \circ {p}_{*}=0$ and $\mathrm{img}({p}_{*})\le \mathrm{ker}(\delta )$.
Conversely, consider an element $c\in \mathrm{ker}(\delta )\le {H}_{n}(W)$. As $c\in \mathrm{ker}(\delta )$, there must exists a snake of the form $(c,w,v,u,0)$. The last snake condition says that $[u]=0$, so we must have $u=d({u}^{\prime})$ for some ${u}^{\prime}\in {U}_{n}$. Another snake condition says that $d(v)=i(u)=i(d({u}^{\prime}))=d(i({u}^{\prime}))$, so we have $d(v-i({u}^{\prime}))=0$. This means that $v-i({u}^{\prime})$ is a cycle, so we have a homology class $b=[v-i({u}^{\prime})]\in {H}_{n}(V)$. This satisfies ${p}_{*}(b)=[p(v-i({u}^{\prime}))]$, but $pi=0$ and $p(v)=w$ so this simplifies to ${p}_{*}(b)=[w]=c$, so $c\in \mathrm{img}({p}_{*})$. ∎
The sequence ${H}_{n}\mathit{}\mathrm{(}W\mathrm{)}\stackrel{\mathit{\delta}}{\mathrm{\to}}{H}_{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}U\mathrm{)}\stackrel{{i}_{\mathrm{*}}}{\mathrm{\to}}{H}_{n\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}V\mathrm{)}$ is exact (or equivalently, $\mathrm{img}\mathit{}\mathrm{(}\delta \mathrm{)}\mathrm{=}\mathrm{ker}\mathit{}\mathrm{(}{i}_{\mathrm{*}}\mathrm{)}$).
First suppose we have an element $c\in {H}_{n}(W)$. Choose a snake $(c,w,v,u,a)$ starting with $c$, so $\delta (c)=a=[u]$. We then have ${i}_{*}\delta (c)={i}_{*}[u]=[i(u)]$, but one of the snake conditions says that $i(u)=d(v)\in {B}_{n-1}(V)$, so $[i(u)]=0$, so ${i}_{*}\delta (c)=0$. This proves that ${i}_{*}\circ \delta =0$ and so $\mathrm{img}(\delta )\le \mathrm{ker}({i}_{*})$.
Conversely, suppose that $a\in \mathrm{ker}({i}_{*})$. We can choose $u\in {Z}_{n-1}(U)$ such that $a=[u]$. Now $[i(u)]={i}_{*}(a)=0$, so $i(u)\in {B}_{n-1}(V)$, so there exists $v\in {V}_{n}$ with $d(v)=i(u)$. Put $w=p(v)\in {W}_{n}$. We then have $d(w)=d(p(v))=p(d(v))=p(i(u))$, and this is zero because $p\circ i=0$. This means that $w\in {Z}_{n}(W)$, so we can define $c=[w]\in {H}_{n}(W)$. We now see that $(c,w,v,u,a)$ is a snake, so $a=\delta (c)$, so $a\in \mathrm{img}(\delta )$. ∎