I have a commutative unital ring $A$, a short exact sequence $L\to M\to N$ of $A$-modules, a faithful exact contravariant functor $D:\text{Mod}_A\to\text{Mod}_A$ satisfying $$DM=0\iff M=0,$$ and monomorphisms $\chi_L:L\to D^2L$, $\chi_M$ and $\chi_N$ making the following diagram commute.
$$\require{AMScd}
\begin{CD}
L @>>> M @>>> N \\
@V{\chi_L}VV @V{\chi_M}VV @V{\chi_N}VV\\
D^2L @>>> D^2M @>>> D^2N;
\end{CD}$$
I have read that from this setup we obtain a short exact sequence $$\text{coker}(\chi_L)\to\text{coker}(\chi_M)\to\text{coker}(\chi_N)$$ via a 'diagram chase', but cannot quite see how to get this.
First of all we need a monomorphism $\text{coker}(\chi_L)\to\text{coker}(\chi_M)$. We have a short exact sequence $D^2L\to D^2M\to D^2N$ since $D$ is exact and contravariant, and therefore a monomorphism $D^2L\to D^2M$ which maps $\text{im}(\chi_L)\to\text{im}(\chi_M)$, but where to go from here?
Best Answer
Given a morphism $h \colon B \to C$ in $\textsf{Mod}_A$, a submodule $B’$ of $B$, and a submodule $C’$ of $C$; if $h[B’] \subseteq C’$, then $h$ induces a morphism $B/B’ \to C/C’$ defined by $b+B’ \mapsto h(b)+C’$. Indeed, if $b_1+B’=b_2+B’$, then $b_1-b_2 \in B’$, and from $h[B’] \subseteq C’$ it follows that $h(b_1)-h(b_2) \in C’$, so $h(b_1)+C’=h(b_2)+C’$.
Now, consider $L \stackrel g \longrightarrow M \stackrel f \longrightarrow N$. Since the diagram $\newcommand{\im}{\operatorname{im}} \newcommand{\coker}{\operatorname{coker}}$ $$ \require{AMScd} \begin{CD} L @>{g}>> M @>{f}>> N \\ @V{\chi_L}VV @V{\chi_M}VV @V{\chi_N}VV\\ D^2L @>>{D^2g}> D^2M @>>{D^2f}> D^2N; \end{CD} $$ commutes, we have that $$ (D^2g)[\im(\chi_L)] = (D^2g)[\chi_L[L]] = \chi_M[g[L]] \subseteq \chi_M[M] = \im(\chi_M) $$ and therefore $D^2g$ induces a map $\coker(\chi_L) \to \coker(\chi_M)$. Similarly, $D^2f$ induces a map $\coker(\chi_M) \to \coker(\chi_N)$.