Suppose I have have a short exact sequence of finitely generated Abelian groups $0 \longrightarrow G \overset{f}\longrightarrow H \overset{g}\longrightarrow K \longrightarrow 0$. Suppose I have a projection $p$ from $K$ to $L$, where $L$ is again finitely generated abelian group whose rank is lower than that of $K$. Then is the following one a short exact sequence?

$0 \longrightarrow G \overset{f}\longrightarrow H \overset{p \circ g}\longrightarrow L \longrightarrow 0$

  • $\begingroup$ Well, what is the relation between $\ker(g)$ and $\ker(p\circ g)$? $\endgroup$
    – Berci
    Aug 6, 2014 at 18:16
  • $\begingroup$ It is not - $\ker(p \circ g) = g^{-1}(\ker p) \supsetneq g^{-1}(0) = \ker g = \text{im}(f)$ $\endgroup$
    – zcn
    Aug 6, 2014 at 18:18

1 Answer 1


Any short exact sequence whose first map is $f$ will be isomorphic to your original short exact sequence. Thus, your conjecture can only be true if the projection $K \to L$ is an isomorphism.

For general computation, you can use the ker-coker sequence. e.g. one way to arrange everything is

$$ \begin{matrix}0 &\to& G &\to& H &\to& K &\to 0 \\ & & \downarrow & & \downarrow & & \downarrow \\ & & 0 &\to& L &\to& L &\to 0 \end{matrix} $$

The ker-coker sequence relates the kernels of the vertical maps to the cokernels of the vertical maps, and you get a long exact sequence

$$ 0 \to G \to \ker\{H\to L\} \to \ker\{K \to L\} \to 0 \to \mathop{\mathrm{coker}}\{H \to L \} \to 0 \to 0 $$

While this tells us that $\mathop{\mathrm{coker}}\{H \to L \}$ is indeed a zero group, and thus $H \to L$ will be surjective, its kernel will generally not be $G$ (more precisely, $G$ with the inclusion $G \to H$), but instead be an extension of $\ker\{K \to L\}$ by $G$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .