# Find a coproduct in a cauchy complete category

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 $$Z=\ker g$$ ?

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

$fg$ 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.