# 17. The Snake Lemma

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

 $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$

which form a short exact sequence. One might hope that the resulting sequence

 $H_{*}(U)\xrightarrow{i_{*}}H_{*}(V)\xrightarrow{p_{*}}H_{*}(W)$

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 $\operatorname{img}(i_{*})=\ker(p_{*})$), but $i_{*}$ need not be injective, and $p_{*}$ need not be surjective. In other words, $\ker(i_{*})$ need not be zero, and $\operatorname{img}(p_{*})$ need not be all of $H_{*}(W)$. We can still obtain a great deal of information about $\ker(i_{*})$ and $\operatorname{img}(p_{*})$, but that will require some preparation. For the moment we will just prove the easier statement mentioned above.

###### Proposition 17.1.

Let $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$ be a short exact sequence of chain maps between chain complexes. Then in the resulting sequence $H_{*}(U)\xrightarrow{i_{*}}H_{*}(V)\xrightarrow{p_{*}}H_{*}(W)$ we have $\operatorname{img}(i_{*})=\ker(p_{*})$.

###### Proof.

First, as the sequence $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$ is exact we have $p\circ i=0$. It follows that $p_{*}\circ i_{*}=(p\circ i)_{*}=0_{*}=0$, so $\operatorname{img}(i_{*})\leq\ker(p_{*})$.

Conversely, suppose we are given an element $b\in\ker(p_{*})$; must show that it lies in $\operatorname{img}(i_{*})$. 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 $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$ 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 $v-d(v^{\prime})\in\ker(p)$. We also have $\ker(p)=\operatorname{img}(i)$ 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 $b\in\operatorname{img}(i_{*})$ as claimed.

We can display the relevant groups and elements as follows:

The two dotted arrows are supposed to indicate the relation $d(v^{\prime})+i(u)=v$. ∎

###### Theorem 17.2.

For $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$ as above, there is a natural map $\delta\colon H_{n}(W)\to H_{n-1}(U)$ 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.

###### Definition 17.3.

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 $a$.

Video (Lemma 17.4 to Remark 17.8)

###### Lemma 17.4.

For any $c\in H_{n}(W)$, there is a snake starting with $c$.

###### Proof.

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 $U_{*}\xrightarrow{i}V_{*}\xrightarrow{p}W_{*}$ is short exact, we know that $p\colon V_{n}\to W_{n}$ 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 $d(v)\in\ker(p)$, but $\ker(p)=\operatorname{img}(i)$ 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. ∎

###### Lemma 17.5.

Suppose we have two snakes that have the same starting point; then they also have the same endpoint.

###### Proof.

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 $v-d(v^{\prime})\in\ker(p)=\operatorname{img}(i)$. 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 $a=[u]=0$. ∎

###### Definition 17.6.

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.)

###### Remark 17.7.

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.

###### Remark 17.8.

The slogan behind the definition is that $\delta=i^{-1}dp^{-1}$. 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.

###### Remark 17.9.

The Snake Lemma (in a slightly different incarnation) is probably the most advanced piece of mathematics ever to appear in a mainstream movie:

Video (Proposition 17.1, 17.10 and 17.11)

###### Proposition 17.10.

The sequence $H_{n}(V)\xrightarrow{p_{*}}H_{n}(W)\xrightarrow{\delta}H_{n-1}(U)$ is exact (or equivalently, $\operatorname{img}(p_{*})=\ker(\delta)$).

###### Proof.

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 $\operatorname{img}(p_{*})\leq\ker(\delta)$.

Conversely, consider an element $c\in\ker(\delta)\leq H_{n}(W)$. As $c\in\ker(\delta)$, 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 $c\in\operatorname{img}(p_{*})$. ∎

###### Proposition 17.11.

The sequence $H_{n}(W)\xrightarrow{\delta}H_{n-1}(U)\xrightarrow{i_{*}}H_{n-1}(V)$ is exact (or equivalently, $\operatorname{img}(\delta)=\ker(i_{*})$).

###### Proof.

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 $\operatorname{img}(\delta)\leq\ker(i_{*})$.

Conversely, suppose that $a\in\ker(i_{*})$. 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 $a\in\operatorname{img}(\delta)$. ∎