Atiyah-Macdonald Chapter 2 (In Progress)

Proposition 2.10 (Snake Lemma). Let

$$\begin{array}{ccccccccc}0& ⟶& M^{\prime }& \stackrel{u}{\to }& M& \stackrel{v}{\to }& M^{\prime \prime }& ⟶& 0\\ & & f^{\prime }\downarrow & & f\downarrow & & f^{\prime \prime }\downarrow & & \\ 0& ⟶& N^{\prime }& \stackrel{u^{\prime }}{\to }& N& \stackrel{v^{\prime }}{\to }& N^{\prime \prime }& ⟶& 0\end{array}$$

be a commutative diagram of $A$-modules and homomorphisms, with the rows exact. Then there exists an exact sequence

$$\begin{array}{rl}0⟶\ker (f^{\prime })& \stackrel{\bar{u}}{\to }\ker (f)\stackrel{\bar{v}}{\to }\ker (f^{\prime \prime })\stackrel{d}{\to }\\ & \mathrm{coker}(f^{\prime })\stackrel{{\bar{u}}^{\prime }}{\to }\mathrm{coker}(f)\stackrel{{\bar{v}}^{\prime }}{\to }\mathrm{coker}(f^{\prime \prime })⟶0,\end{array}$$

in which $\bar{u},\bar{v}$ are restrictions of $u,v$, and ${\bar{u}}^{\prime },{\bar{v}}^{\prime }$ are induced by $u^{\prime },v^{\prime }$.

Proof) The boundary homomorphism $d$ is defined as follows: for any $m^{\prime \prime }\in \ker f^{\prime \prime }$, there must be some $m\in M$ such that $v(m)=m^{\prime \prime }$ (by the surjectivity of $v$). Furthermore, $f(m)$ must then be in the kernel of $v^{\prime }$ by the commutative diagram, which implies there is some $n^{\prime }$ such that $u^{\prime }(n^{\prime })=f(m)$. We define $d$ such that $d(m^{\prime \prime })$ is the residue class of $n^{\prime }$ in $\mathrm{coker}(f^{\prime })$. It is not obvious that $d$ is well-defined; there is an ambiguity in the definition as there may be multiple elements in $M$ that map to $m^{\prime \prime }$. Let $m_1,m_2$ be such that $v(m_1)=v(m_2)$; it suffices to show that $[u^{\prime -1}(f(m_1))]=[u^{\prime -1}(f(m_2))]$. Note that $m_1-m_2$ is in the kernel of $v$, and thus the image of $u$. If $m^{\prime }\in M^{\prime }$ satisfies $u(m^{\prime })=m_1-m_2$, then $u^{\prime }(f^{\prime }(m^{\prime }))=f(m_1)-f(m_2)\Rightarrow f^{\prime }(m^{\prime })=u^{\prime -1}(f(m_1))-u^{\prime -1}(f(m_2))$. As $f^{\prime }(m^{\prime })\in \mathrm{Im}(f^{\prime })$, we get out desired result. The fact that $d$ is a $A$-module homomorphism is trivial given well-definedness.

It is clear from the commutative diagram that $u$ maps $\ker f^{\prime }$ into $\ker f$ into $\ker f^{\prime \prime }$, and as the image $u^{\prime }(f^{\prime }(M^{\prime }))$ must be contained in $f(M)$, the maps between cokernels are well-defined as well. Now that we have established well-definedness of all the relevant maps, we start chasing the exact sequence. $\bar{u}$ must be injective as $u$ is injective. Next, one can see that kernel $\ker \bar{v}=\ker v\cap \ker f$ in $M$, so we want to show $\ker v\cap \ker f=\bar{u}(\ker f^{\prime })=u(\ker f^{\prime })$; an examination of the commutative diagram shows this must be true.

Next, the kernel of $d$ is precisely $\ker (f^{\prime \prime })\cap v(\ker f+\mathrm{Im}(u))=\ker (f^{\prime \prime })\cap v(\ker f)$, which is the image of $\bar{v}$. The image of $d$ is residue classes with representatives in $u^{\prime -1}(f(M))$, which is precisely the kernel of ${\bar{u}}^{\prime }$. The image of ${\hat{u}}^{\prime }$ is the residue classes with representatives in $u^{\prime }(N^{\prime })=\ker v^{\prime }$. This is the kernel of ${\bar{v}}^{\prime }$ as if $[n]$ maps into the image of $f^{\prime \prime }$, by the surjectivity of $v$ there is some $n_0\in f(M)$ such that $v^{\prime }(n_0)=v^{\prime }(n)$. Thus, $n=n_0+k$ where $k\in \ker v^{\prime }$ and $n_0\in f(M)$, as desired. Finally, ${\bar{v}}^{\prime }$ must be surjective as $v^{\prime }$ is surjective.