In an abelian category, can there be a commutative diagram of (vertical/horizontal) exact sequences $$ \require{AMScd} \begin{CD} 0 @>>> N @>>> M\\ @. @VVV @VVV \\ 0 @>>> X @>>> Y\\ @. @VVV \\ @. 0 \end{CD} $$

such that the following conditions are true?

  • All named objects are nonzero.
  • No morphisms between named objects are zero morphisms, including $N \to Y$ (implied by the previous and the diagram).
  • There are no morphisms $X \to M$ or $M \to X$ that commute with the other maps.

(I am trying to "replace" $N$ by $\ker(N \to Y)$ and $Y$ by $\text{coker}(N \to Y)$ in the diagram to get back a commutative diagram of exact sequences, and my attempts so far require the existence of a compatible map between $X$ and $M$ in at least one direction, so I would like to figure out what happens if I have no such maps.)

Original question

"Modules with morphisms 0 -> N -> M and N -> X -> 0 but no compatible maps between M and X?"

I wish for three modules N, M, and X with exact sequences $0 \to N \to M$ and $N \to X \to 0$, but there are no such morphisms $M \to X$ or $X \to M$ which commute with the maps above.

(Not homework. I'm trying to replace X, N, and M in a bigger diagram so that X becomes 0 and N and M become something maximal (universal). If the above kind of module triples exists, I might be in trouble.)


1 Answer 1


Consider $0 \to \mathbb{Z} \to \mathbb{Q}$, and $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$. Then the only map in either direction between $\mathbb{Q}$ and $\mathbb{Z}/2\mathbb{Z}$ is the zero map, since $\mathbb{Q}$ is torsionfree, and every quotient of $\mathbb{Q}$ is divisible.

  • $\begingroup$ It works. You showed me that my conditions are not enough for what I want. I will add another condition to the question. $\endgroup$
    – leewz
    Commented Sep 25, 2014 at 4:45

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