$\require{AMScd}$Consider the $3×3$ commutative diagram with exact rows and columns:
$$\begin{CD}
@[email protected]@[email protected]@.@.\\
@.@.@VVV@VVV@VVV@.@.\\
@.0@>>>X@>a>>X'@>a'>>X''@>>>0@.\\
@.@.@VfVV@Vf'VV@Vf''VV@.@.\\
@.0@>>>Y@>b>>Y'@>b'>>Y''@>>>0@.\\
@.@.@VgVV@Vg'VV@Vg''VV@.@.\\
@.0@>>>Z@>c>>Z'@>c'>>Z''@>>>0@.\\
@.@.@VVV@VVV@VVV@.@.\\
@[email protected]@[email protected]@.@.
\end{CD}$$
I was wondering if there exists an exact map $0\to X\xrightarrow{\alpha}X'\oplus Y\xrightarrow{\beta} Y'\xrightarrow{\gamma}Y''\oplus Z'\xrightarrow{\delta} Z''\to 0$ where $\alpha(x)=(a(x),f(x))$, $\beta(x',y)=f'(x')-b(y)$, $\gamma(y')=(b'(y'),g'(y'))$ and $\delta(y'',z')=g''(y'')-c'(z')$. $Im(\alpha)\subseteq Ker(\beta)$ and $Im(\gamma)\subseteq Ker(\delta)$ are immediate but I can't see $Im(\beta)\subseteq Ker(\gamma)$. Can it be done through diagram-chasing and is there anywhere I could find a proof?
Best Answer
For a counterexample, take $X’=X’’=Y’=Y’’=\mathbb{Z}$, with identity maps between them, and all other objects zero:
$\require{AMScd}$
$$\begin{CD} @[email protected]@[email protected]@.@.\\ @.@.@VVV@VVV@VVV@.@.\\ @.0@>>>0@>>>\mathbb{Z}@=\mathbb{Z}@>>>0@.\\ @.@.@VVV@|@|@.@.\\ @.0@>>>0@>>>\mathbb{Z}@=\mathbb{Z}@>>>0@.\\ @.@.@VVV@VVV@VVV@.@.\\ @.0@>>>0@>>>0@>>>0@>>>0@.\\ @.@.@VVV@VVV@VVV@.@.\\ @[email protected]@[email protected]@.@. \end{CD}$$
In fact, if $X''\neq0$ then the sequence is never exact, as $\gamma\beta\neq0$. For if $0\neq x''\in X''$, then $x''=a'(x')$ for some $x'\in X'$, and $f''(x'')\neq0$ since $f''$ is a monomorphism. So $\gamma\beta(x',0)=f'(x')$, which is nonzero since $b'f'(x')=f''a'(x')=f''(x'')\neq0$. [I'm assuming that the diagram is a diagram of abelian groups or modules, but the same conclusion will hold in any abelian category.]
Similarly, if $Z\neq0$ then the sequence cannot be exact.
However, there is an exact sequence with $X''\oplus Y'\oplus Z$ as the middle term in place of $Y'$.