17. The Snake Lemma

First, as the sequence $U_{*}\stackrel{𝑖}{\to }{V}_{*}\stackrel{𝑝}{\to }{W}_{*}$ is exact we have $p\circ i=0$. It follows that $p_{*}\circ i_{*}={(p\circ i)}_{*}=0_{*}=0$, so $\mathrm{img}(i_{*})\le \ker (p_{*})$.

Conversely, suppose we are given an element $b\in \ker (p_{*})$; 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 $U_{*}\stackrel{𝑖}{\to }{V}_{*}\stackrel{𝑝}{\to }{W}_{*}$ 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

so $v-d(v^{\prime })\in \ker (p)$. We also have $\ker (p)=\mathrm{img}(i)$ 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 $d(v^{\prime })+i(u)=v$. ∎

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 $U_{*}\stackrel{𝑖}{\to }{V}_{*}\stackrel{𝑝}{\to }{W}_{*}$ 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 $d(v)\in \ker (p)$, but $\ker (p)=\mathrm{img}(i)$ 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 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 $v-d(v^{\prime })\in \ker (p)=\mathrm{img}(i)$. 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 $a=[u]=0$. ∎

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 $\mathrm{img}(p_{*})\le \ker (\delta )$.

Conversely, consider an element $c\in \ker (\delta )\le H_n(W)$. 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 $c\in \mathrm{img}(p_{*})$. ∎

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 $\mathrm{img}(\delta )\le \ker (i_{*})$.

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 $a\in \mathrm{img}(\delta )$. ∎