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I.Bucur and A.Deleanu in their "Introduction to the theory of categories and functors" define integral object in a category ${\mathcal C}$ as an arbitrary object $I$ that satisfies the following two conditions:

1) for any object $X$ there is a morphism $i:I\to X$, and

2) for any pair of parallel morphisms $u,v:X\to Y$ the condition $u\ne v$ implies the existense of a morphism $i:I\to X$ such that $$ u\circ i\ne v\circ i. $$

Let now ${\mathcal C}$ be an arbitrary closed monoidal category. My question:

Is the unit object $I$ in ${\mathcal C}$ always integral?

In the examples that I know this is true, but apparently I don't see some tricks, I can't prove this for arbitrary category ${\mathcal C}$.

I would appreciate very much if somebody could clarify this to me.

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  • $\begingroup$ I haven't heard the term "integral object" before. Your second condition corresponds to (the ugly, ugly contrapositive of) the notion of a separating object. I'd probably describe $I$ as a weakly initial, separating object. $\endgroup$ – Derek Elkins May 31 at 20:23
  • $\begingroup$ @DerekElkins for me this word sounds strange as well. But Bucur and Deleanu use it (at page 31). I did not know that this is called a separating object, I'll have in mind, thank you. $\endgroup$ – Sergei Akbarov May 31 at 21:08
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No. For instance, taking the category of sets with the Cartesian product, the unit object (a singleton) is not integral since it has no morphism to the empty set. More generally, if you take the category of sheaves (of sets, or of abelian groups if you want an abelian category say) on some site, both properties (1) and (2) will typically fail since a sheaf can have no global sections or not be generated by global sections. For instance, if you take the category of sheaves of sets on $\mathbb{R}$ and let $X$ be the sheaf given by $X(U)=\{*\}$ if $U\subseteq(0,1)$ and $X(U)=\emptyset$ otherwise, the two inclusions $X\to X\coprod X$ are distinct but $X$ has no morphisms from the unit object (which is the constant singleton sheaf).

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  • $\begingroup$ Ah, yes. Eric, if we remove the first condition will this be true nevertheless? $\endgroup$ – Sergei Akbarov May 31 at 14:52
  • $\begingroup$ Eric, I am not sure that I understand why the condition (2) fails in the case of sheaves. Could you, please, give a concrete example? $\endgroup$ – Sergei Akbarov May 31 at 15:16

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