Consider the commutative diagram of finite abelian groups $\require{AMScd}$ \begin{CD} 0@>>> A @>i>> B@>\pi>> C@>>> 0\\ \ @VV 0 V@VVfV@VV 0 V\\ 0@>>>A @>>i> B@>>\pi> C@>>> 0 \end{CD} where all maps are homomorphisms, the rows are exact, and the leftmost and the rightmost vertical map are zero? Is the middle map $f$ also zero?

It is clear that $f\circ f=0$ but this is all that seems to follow from lazy diagram chasing, hence I suspect $f$ need not be zero, in general. If so, what is a counterexample?


1 Answer 1


Take the sequence $0\to Z/2Z\to Z/4Z\to Z/2Z\to 0$ and the vertical map multiplication by 2.

  • $\begingroup$ Oh, thanks, I should have seen this myself. $\endgroup$ Sep 8, 2020 at 18:38
  • 5
    $\begingroup$ Or take $0\to A\to A\oplus C\to C\to0$, then $f(a,c)=(h(c),0)$ for any homomorphism $h:C\to A$ $\endgroup$ Sep 8, 2020 at 18:44

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