I am a bit confused regarding the definition of a split exact sequence, whose definition is for example available here (http://ncatlab.org/nlab/show/split+exact+sequence). Let's work in an abelian category, and suppose to have a s.e.s. $0 \to B \xrightarrow{f} A \oplus{ B} \xrightarrow{g} A \to 0$ where the maps $f$ and $g$ are not supposed to be the inclusion and the projection. Does the sequence split?

I have tried using the universal property of direct sum...but...mmm...I feel as I am missing something. Please, could you help me?

Thank you in advance! Cheers

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    $\begingroup$ I'm fairly confident the answer is no, but I don't have a counterexample off the top of my head. Is that the answer you were expecting? $\endgroup$ – Qiaochu Yuan May 21 '15 at 20:58
  • $\begingroup$ Yes, and actually I was having an idea using the Yoneda isomorphism - but "Crostul" managed to find a simpler counterexample faster :) $\endgroup$ – user233650 May 21 '15 at 21:51

A counterexample follows in the category of abelian groups. Let $A=\bigoplus_{n=0}^{\infty} \Bbb{Z}/2 \Bbb{Z}$, $B = \Bbb{Z}$.

Define $g: B \oplus A \longrightarrow A$ to be the map $$(z, x_0, x_1,x_2, \dots) \mapsto (z \bmod{2}, x_0, x_1, x_2, \dots)$$ and $f: B \longrightarrow B \oplus A$ to be the map $$z \mapsto (2z, 0, 0, 0, \dots) $$

Then $0 \to B \to B \oplus A \to A \to 0$ is exact and does not split.

  • $\begingroup$ Yes, you are right! Thank you a lot. Cheers $\endgroup$ – user233650 May 21 '15 at 21:44

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