In an additive Cauchy complete category $\mathcal C$, all idempotents split. Let $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $g \circ f = id_X$. Is there an object $Z$ such that $Y = X \sqcup Z$ (the coproduct of $X$ and $Z$)?

Is $Z=\ker g$ ?

  • $\begingroup$ Is your category additive? $\endgroup$
    – Zhen Lin
    Apr 7, 2015 at 14:39
  • $\begingroup$ Yes, this is an additive category. $\endgroup$
    – milanelo
    Apr 7, 2015 at 14:40
  • 2
    $\begingroup$ Yes, then there is a direct sum decomposition as you say. $\endgroup$
    – Zhen Lin
    Apr 7, 2015 at 14:41
  • $\begingroup$ Where does the condition "all idempotents splits" is used? In the process of showing $Kerg$ is a direct summand of $Y$? $\endgroup$
    – milanelo
    Apr 7, 2015 at 14:45
  • 1
    $\begingroup$ First, why does $\ker(g)$ exist? $\endgroup$ Apr 7, 2015 at 14:47

1 Answer 1


$fg$ is idempotent, so $(1_Y-fg)$ is idempotent (check!), so it has a splitting by idempotent completeness, i.e. $h: Z \to Y$, $k: Y \to Z$ such that $kh = 1_Z$ and $hk = 1-fg$. Then you can check that $h = \mathrm{ker} g$ and that we have a biproduct here.

To see that it's a product, consider $X \overset{a}{\leftarrow} W \overset{b}{\rightarrow} Z$. From this we can construct a map $W \to Y$ as $fa+hb$; in the other direction from a map $W \overset{c}{\to} Y$ we get maps $gc,kc$. And you can check that these mappings are inverse to one another and natural. Similarly, one can show it's a coproduct, and that the product/coproduct fit together into a biproduct.

Also, just to make sure we're clear, let me harp on a terminological point: "Cauchy complete (enriched) category" has a very precise meaning. In a $\mathsf{Set}$-enriched setting, it just means idempotent-complete, but in an $\mathsf{Ab}$-enriched, setting, it means idempotent complete and having biproducts (i.e. additive). So "Cauchy-complete additive category" is slightly redundant, although I still think it's the best way to refer to what you're talking about.

Although I suppose we did show that this was a biproduct without assuming the existence of biproducts. So an idempotent in an idempotent-complete $\mathsf{Ab}$-enriched category gives rise to a biproduct decomposition, even if the category is not additive.

  • $\begingroup$ Glad to be of help! $\endgroup$
    – tcamps
    Apr 10, 2015 at 4:39

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