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Apologies in advance for what is likely a very simple, pedantic question. My question is if there are concrete categories whose objects are not sets.

Here's my thinking: the definition I am following says that a category $\mathcal{C}$ is concrete provided there is a faithful functor ${\mathcal{C}}\xrightarrow{\sigma}\textsf{Set}$. I understand that one can realize any concrete category as a category whose objects are sets, but - on the face of it - I see no reason that the objects of ${\mathcal{C}}$ need to be sets themselves.

For example, let $\textsf{Pt}$ be the category with a single object and morphism. It can be realized by letting its object be a singleton and morphism be the identity map on that singleton. However, I see no reason why you could not let the only object of $\textsf{Pt}$ be a proper class, the identity morphism be a singleton, and let composition be the obvious one. Does this work? If it does, are there any less trivial examples?

Thanks for any insight/help!

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  • $\begingroup$ As I have seen the phrase "concrete category" used, it means a category where the objects are sets (possibly with additional structure) and the arrows are functions (maps of set elements). So the only wiggle room I can see in that definition would be whether extra structure qualifies the something-more-than-a-set as being not-a-set. $\endgroup$ – hardmath May 13 '18 at 2:59
  • $\begingroup$ @hardmath Yes, this is where my confusion stems from. In popular usage this is the interpretation of a concrete category (from what I have seen), but I could not deduce this from the definition. However, Eric's answer below makes more sense to me. $\endgroup$ – Dan Normand May 13 '18 at 3:10
  • $\begingroup$ It would help to cite the source where you found the alternative definition of concrete category. If forced to guess, I would speculate the alternative is motivated by a setup where set theory is not taken as the foundation but is to be derived from a categorical framework. $\endgroup$ – hardmath May 13 '18 at 5:50
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    $\begingroup$ @hardmath: The definition in terms of a faithful functor is completely standard and is pretty much the only precise definition of a concrete category I've ever seen. Your definition is fine as an informal notion but is not precise without saying what "possibly with additional structure" actually means. (To be clear, though, the usual definition is not just that there exists a faithful functor to Set but that one such faithful functor has been chosen, so a concrete category is a category together with the extra structure of such a functor.) $\endgroup$ – Eric Wofsey May 13 '18 at 17:25
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By the usual definitions (working in ZFC, say), this is not allowed. A category is defined to have a class of objects, and by definition a class is a definable collection of sets. So, every object of a category is a set.

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  • $\begingroup$ I believe I see my technical difficulty. Following this, to ask if a category is concrete is not to ask whether its objects are sets, rather if there exists a faithful functor into $\textsf{Set}$, correct? $\endgroup$ – Dan Normand May 13 '18 at 3:05
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Eric Wofsey May 13 '18 at 3:06
  • $\begingroup$ Thank you Eric, this was very clear. For some reason I've always believed that objects need not be sets, hence the need for concrete categories (though I've long wondered how that would work ... the more you know). $\endgroup$ – Dan Normand May 13 '18 at 3:24
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    $\begingroup$ The point of concrete categories is not that the objects are sets, but that the morphisms are functions between those sets. In a general category, the morphisms have no set-theoretic connection to the objects at all. $\endgroup$ – Eric Wofsey May 13 '18 at 3:26
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    $\begingroup$ Note in addition that $C$ can have a class of objects where each object is given by some set, and a faithful functor $F : C \to \text{Set}$ making $C$ a concrete category can give each object $c \in C$ an interpretation as an unrelated set $F(c)$. When people say that the objects of a concrete category "are" sets, they're referring to $F(c)$, not $c$ itself. $\endgroup$ – Qiaochu Yuan May 13 '18 at 3:28

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