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For the following example taken from Hungerford graduate Algebra text, can someone provide clarification where the example says:

"However, in the third example after Definition 7.1, is the function $\sigma$ assigns to the group $G$ that usual underlying set $G$, then the category in question is not a concrete category(since the morphisms are not functions on the set $G$)." In the context of the third example about multiplicative group $G$, I don't know what is considered of the underlying set, specifically is there a function $\sigma$ for the case of the multiplicative group, and as a result how is not considered as a concrete category. I feel like Hungerford is short on details in this particular example.

Example:appears after Definition 7.6, pg 55 $G$, then the category in question is not a concrete category Definition 7.6 pg 55 The category of groups, equipped with the function that assigns to each group its underlying set in the usual sense, is a concrete category. Similarly the categories of abelian groups and partially ordered sets, with the obvious underlying sets, are concrete categories. However, in the third example third example, multiplicative $G$, pg 53 after Definition 7.1, definition 7.1, pp 52-53. is the function $\sigma$ assigns to the group $G$ that usual underlying set paragraph preceding Definition 7.6, pg 55 $G$, then the category in question is not a concrete category Definition 7.6 pg 55 (since the morphisms are not functions on the set $G$).

Thank you in advance

Seth
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1 Answers1

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Hungerford's definition of a concrete category is nonsense. A concrete category is a category $C$ equipped with a faithful functor $F : C \to \text{Set}$; to specify the data of a concrete category one must specify both what this functor $F$ does to objects and what it does to morphisms, and Hungerford's definition completely fails to specify what $F$ does to morphisms. His condition "every morphism $A \to B$ of $C$ is a function on the underlying sets" doesn't make any sense.

His counterexample is also wrong. Hungerford considers the category $BG$ with one object $\bullet$ and endomorphisms given by a group $G$, and says that if we assign this object $\bullet$ the underlying set of $G$ then we don't get a concrete category. But this is nonsense: he hasn't specified what $F$ does to morphisms! (He is also being unnecessarily confusing by identifying $\bullet$ with $G$. $\bullet$ is just a point.) And in fact there is a meaningful way to define $F$ here: you can send $\bullet$ to the underlying set $G$ and send $g \in G \cong \text{End}(\bullet)$ to the left multiplication map

$$L_g : G \ni x \mapsto gx \in G.$$

This actually does make $BG$ into a concrete category; this is essentially a restatement of Cayley's theorem, and follows abstractly from the Yoneda lemma.

Qiaochu Yuan
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  • There's a whole rant I could get into about the nonsense people spout when they try to assign exercises showing that something is, for example, "not a group." Being a group is not a property! The meaningful question here is to specify the data of a group, namely a set $G$, an element $e \in G$, and a function $m : G \times G \to G$, and ask whether that data satisfies the axioms of a group (associativity etc). But sometimes people are very imprecise about this and e.g. do not specify $m$, which renders the question meaningless. This is similar. – Qiaochu Yuan Aug 25 '22 at 02:18
  • thank you for the detail answer. Apologizes for all the link references. I read over that bit of hungerford dozen times, I was not able to parse out what specifically he was referring to about the example he was pointing to. – Seth Aug 25 '22 at 06:33