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I'm trying to understand the definition of group objects in categories, this is an extract of Paolo Aluffi's book:

QUESTIONS

  1. Can I say that $e(1)$ is the identity in our group $G$ we have just built?
  2. Why $1$ should be a final object? in another words, why the morphism $G\to 1$ have to be unique?
  3. Why this category has to have finite products? it's not suffice just products with two components? (for the multiplication)

Thanks in advance

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The fact that $1$ is a final object implies that $G\times 1 \cong G$ functorially. But I don't know what would happen if we ask for some other object $1'$ not final fitting the diagrams. –  Aldo Guzmán Sáenz Apr 7 at 6:33
    
Regarding 3, existence of two-factor products is the same as existence of finite products. $G_1\times \ldots \times G_n\cong (\ldots(G_1\times G_2)\times \ldots )\times G_n$ canonically. (The diagrams already use three-factor product $(G\times G)\times G$, in the form of repeated two-factor products, and the canonical isomorphism with $G\times(G\times G)$). –  Hagen von Eitzen Apr 7 at 6:39
    
Regarding 1, whhat do you mean with $e(1)$? This need not be a concrete category. –  Hagen von Eitzen Apr 7 at 6:39
    
@HagenvonEitzen I'm sorry, I meant this in the category of groups, is that true that $e(1)$ is the identity in this category? –  user42912 Apr 7 at 8:45

3 Answers 3

  1. What do you mean by "$e(1)$"? The identity in your group is the morphism $e$ itself.

  2. In any category $C$ with a terminal object $1$, a morphism $1 \to c$ for an object $c \in C$ is called a global point of $c$. In some, but not all, categories, this often gives a reasonable "underlying set" functor $\text{Hom}(1, -) : C \to \text{Set}$, although it may not be the obvious one: for example, if $C$ is the category $G\text{-Set}$ of $G$-sets for a group $G$, then a global point is a fixed point.

    So this is one reason you might not think it's totally crazy to ask that picking out a particular element should abstractly correspond to picking out a morphism from $1$. But there are at least two better reasons.

    First, $e$ should describe the outcome of taking the product of zero elements of $G$, and the domain of this operation should be the product of zero copies of $G$, which is the terminal object.

    Second, using the Yoneda lemma, you can show that a group object in a category $C$ is precisely the same thing as a "representable presheaf of groups on $C$," or more precisely a functor $C^{op} \to \text{Grp}$ which is representable in the sense that it is representable after composing with the forgetful functor $\text{Grp} \to \text{Set}$. Meditate in particular on the example of the group object $\text{GL}_n(-)$ in affine schemes, which is the functor which, given a commutative ring $R$, returns the group $\text{GL}_n(R)$.

  3. As Najib pointed out in the comments to Marco's answer, having a terminal object and binary products is equivalent to having all finite products. But finite products is morally the correct condition anyway, because the morally correct definition involves all finite products: see this blog post and this blog post for details.

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regarding #1, I didn't understand, $e$ is a morphism not an element of $G$. –  user42912 Apr 7 at 7:10
    
@user: one should think of morphisms into an object as "generalized elements" of that object. For example, if $X$ is a set, then a morphism $1 \to X$ is precisely an element of $X$. This intuition is particularly useful, among other places, in algebraic geometry. –  Qiaochu Yuan Apr 7 at 7:11
    
@user: for example, if $G$ is a group object I can "multiply generalized elements" of $G$ as follows: if $a : A \to G, b : B \to G$ are two elements, then I can form $a \times b : A \times B \to G \times G$, then compose with the multiplication to get $m \circ (a \times b) : A \times B \to G$, another generalized element of $G$. –  Qiaochu Yuan Apr 7 at 7:13
    
If $G$ is a group in the category $grp$, $e(1)$ is the identity element of $G$? –  user42912 Apr 7 at 7:16
    
@user: if I'm interpreting your question correctly, yes (but the correct terminology is "a group object in the category of sets." A group object in the category of groups turns out to be an abelian group; this is the Eckmann-Hilton argument). –  Qiaochu Yuan Apr 7 at 7:22

It may be useful to see an example on a more interesting category.

Let $X$ be a topological space, and consider the category of all bundles over $X$. Recall that a bundle over $X$ is the same thing as a continuous map $E \to X$; $E$ is called the "total space" of the bundle. A morphism of between two bundles $E \to X$ and $E' \to X$ is a continuous map $E \to E'$ that fills in a commutative triangle.

In this category, $\mathbf{1}$ is the bundle $1_X : X \to X$. The product of two bundles $E \to X$ and $E' \to X$ is the one whose total space is the pullback $E \times_X E'$ (and whose structure map is the obvious map to $X$).

A simple example of this is the bundle $A$ defined by the map $\pi_1 : \mathbf{R}^3 \times \mathbf{R}^3 \to \mathbf{R}^3$ given by the projection map onto the first coordinate.

The maps $\mathbf{1} \to A$ are global sections of this bundle: continuous functions $s : \mathbf{R}^3 \to \mathbf{R}^3 \times \mathbf{R}^3$ that are the identity on the first coordinate. These, of course, can be identified with the composite $\pi_2 \circ s$, which you might know these under the more familiar name of a "vector field", and you may know the total space under the more familiar name of the "tangent bundle to $\mathbf{R}^3$".

If you grind out that the usual algebraic structure on the tangent bundle (where we only consider $+$ and $-$ and not scalar multiplication, since we're talking about groups), you'll see that it is an example of this abstract group in the category of bundles over $\mathbf{R}^3$.

And the group identity is not an "element" of the total space: it is a section of the bundle. (although, you'll eventually learn that sections provide a very good notion of the concept of "element" in categories like this)

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1.- You can not talk about the "identity" of $G$, because $G$ may not be a set. But thats the intuition.

2.- $1$ is a final object because thats the definition, not a consequence. The morphism $G \to 1$ is unique because $1$ is a final object.

3.- The two assertions are equivalent. A category has finite products if and only if has the product of every two objects and a terminal object.

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2  
Regarding #3: products of two objects and empty product (aka terminal object). –  Najib Idrissi Apr 7 at 6:48
    
Yes, thank you. –  Marco Armenta Apr 7 at 6:50
    
regarding #2, I know that's the definition, but why the reason beyond it? –  user42912 Apr 7 at 8:58
    
It's like when you said that if $e(1)$ was the indentity, deep in your heart you now that if $1$ is the identity group, then the map $e : 1 \to G$ gives you the identity in $G$. And if you look carefully you can notice that every diagram gives you an axiom of the definition of group. –  Marco Armenta Apr 7 at 17:50
    
Sigh. I miss when things were simple and a function was a set of ordered pairs............... –  Mathemagician1234 Apr 7 at 17:55

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