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For an object $A$ in a category $\mathfrak{C}$, consider the following property.

($*$) For every object $B$ in $\mathfrak{C}$, the set of morphisms $\text{Hom}(B,A)$ is either empty or consists solely of isomorphisms.

How does one call objects $A$ satisfying ($*$)?

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    $\begingroup$ Why do you think that such terminology exists? $\endgroup$
    – Oskar
    Mar 7, 2016 at 13:21
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    $\begingroup$ Also, could you give an example of object with such property, which is not a strict initial object and not an object in a groupoid? $\endgroup$
    – Oskar
    Mar 7, 2016 at 13:28
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    $\begingroup$ From my point of view, property ($*$) is to the property "being terminal" just like "maximal" is to "greatest" in partially ordered sets. This analogy leads me to suspecting that such a terminology exists. $\endgroup$
    – William of Baskerville
    Mar 7, 2016 at 13:28
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    $\begingroup$ Seems like this property is actual only for preorders: if $C$ is locally small and complete, then such object $A$ has no non-trivial automorphisms... $\endgroup$
    – Oskar
    Mar 7, 2016 at 13:43
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    $\begingroup$ If $\mathfrak{C}$ is locally small and cocomplete, an object $A$, satisfying (*), is simply a strict initial object of some full subcategory $\mathfrak{B}$ of $\mathfrak{C}$. Indeed, if $B$ is some other object of $\mathfrak{C}$, such that $\text{hom}(A,B)$ is nonempty, then we have: $$ \text{hom}(A,B)\cong\text{hom}((\coprod_{\text{hom}(A,B)}A),B)\cong\prod_{\text{hom}(A,B)}\text{hom}(A,B)\cong\text{hom}(A,B)^{\text{hom}(A,B)}, $$ which implies that $\text{hom}(A,B)\cong\{*\}$. $\endgroup$
    – Oskar
    Mar 7, 2016 at 14:22

1 Answer 1

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If you define the connected component $C(A)$ of an object $A$ in a category $C$ as the full subcategory of all objects $B$ such that $Hom(B,A)$ is not empty, then the property (*) is equivalent to saying that $C(A)$ is a groupoid.

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  • $\begingroup$ This is an interesting condition. And if it holds then the object $A$ is called "..."? $\endgroup$
    – William of Baskerville
    Mar 7, 2016 at 18:30
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    $\begingroup$ If you define the connected component... then I would be using well established mathematical language in a very peculiar way. $\endgroup$
    – Paul Taylor
    Mar 7, 2016 at 18:53
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    $\begingroup$ As @PaulTaylor says, that's not the standard meaning of "connected component". But along the same lines one could say that the co-slice category $A/C$ is a groupoid. $\endgroup$
    – Mike Shulman
    Mar 7, 2016 at 19:39
  • $\begingroup$ @MikeShulman For $\mathfrak C=$ transitive $G$-sets, $A=G$ (the example from my comment) the co-slice is not a groupoid. $\endgroup$
    – მამუკა ჯიბლაძე
    Mar 7, 2016 at 20:26
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    $\begingroup$ @მამუკაჯიბლაძე I think Mike meant slice-category... $\endgroup$
    – Oskar
    Mar 7, 2016 at 21:10

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