# Product of Categories ... Product of Groups as an example ..

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I am reading Steve Awodey's book: Category Theory (Second Edition) and am focused on Section 1.5 Isomorphisms ...

I need some further help in order to fully understand some aspects of the definition of the product of two categories as it applies to the category Groups ... ...

The definition of the product of two categories ... reads as follows:

For the category Groups of groups and group homomorphisms, the product category of two categories ##C## and ##D##, namely ##C \times D##, has objects of the form ##(G,H)## where ##G## and ##H## are groups and where ##G \in C## and ##H \in D## ...

Arrows would be of the form

##(f,g) : (G,H) \to (G',H')##

for ##f: G \to G'## and ##g: H \to H'##

... BUT ...

... now ... you would expect ... indirectly at least! ... that the definition of the category and its rules would specify the product ...

##(g_1, h_1) \star (g_2, h_2) = (g_1 \bullet_1 g_2, h_1 \bullet_2 h_2)## ... ...

... BUT! ...

how does the product category definition imply this in the case of groups ...

Hope someone can help ...

Peter

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What are ##g_i\; , \;h_i\,?##

The direct product of the category group ##\mathcal{G}## with itself has pairs of groups ##(G,H)## as objects and morphisms between those pairs:
##(f,g) \, : \,\mathcal{G} \times \mathcal{G} \longrightarrow \mathcal{G} \times \mathcal{G}## and ##(f,g)\, : \,(G,H) \longmapsto (G',H')##.

This has at prior nothing to do with the direct product of groups within the single category ##\mathcal{G}##. And the morphism ##(f,g)## is simply a transformation from one pair of groups to another pair of groups - no elements needed. The elements of categories are the objects, in this case entire groups or pairs of them.

Math Amateur
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What are ##g_i\; , \;h_i\,?##

The direct product of the category group ##\mathcal{G}## with itself has pairs of groups ##(G,H)## as objects and morphisms between those pairs:
##(f,g) \, : \,\mathcal{G} \times \mathcal{G} \longrightarrow \mathcal{G} \times \mathcal{G}## and ##(f,g)\, : \,(G,H) \longmapsto (G',H')##.

This has at prior nothing to do with the direct product of groups within the single category ##\mathcal{G}##. And the morphism ##(f,g)## is simply a transformation from one pair of groups to another pair of groups - no elements needed. The elements of categories are the objects, in this case entire groups or pairs of them.

Hi fresh_42 ... thanks for the help ...

The ##g_i## and ##h_i## are the elements of ##G## and ##H## ... I expected that somehow the category of the direct product of groups ##G## and ##H## in ##\mathcal{G}## would imply that at the level of elements ##g_1, g_2 \in G## and ##h_1, h_2 \in H## we would get

##(g_1, h_1) \star (g_2, h_2) = (g_1 \bullet_1 g_2, h_1 \bullet_2 h_2)##

Presumably you are saying that we are not dealing at the level of elements in ##\mathcal{G}## ... ... ?

I thought that the construction of the product would have implications at the level of elements ...

Still reflecting on what you have said ...

Peter

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No. On the level of categories, the objects are the elements, entire groups. You can build a direct product of different categories, e.g. ##\mathcal{Sets} \times \mathcal{Rings}##. Whether this serves a purpose is another question. But you see on the example, that it is different from direct products you met before. It is another level. I don't know from heart if there can be constructed a connection between the two, but at prior they are completely different constructions.

Math Amateur
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No. On the level of categories, the objects are the elements, entire groups. You can build a direct product of different categories, e.g. ##\mathcal{Sets} \times \mathcal{Rings}##. Whether this serves a purpose is another question. But you see on the example, that it is different from direct products you met before. It is another level. I don't know from heart if there can be constructed a connection between the two, but at prior they are completely different constructions.

Thanks again for the further clarification ...

What got me started on my chain of thoughts above was Awodey's statement at the end of the above quote, viz.: (see above scanned text ...)

" ... ... The reader familiar with groups will recognize that for groups ##G## and ##H##, the product category ##G \times H## is the usual (direct) product of groups ... ... "

How should we interpret this remark? (I obviously interpreted it wrongly ... ... )

Peter

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I don't know what he meant. Maybe that the construction, its definition is basically the same. Or perhaps if we don't take the entire category ##\mathcal{Groups}## and consider a single group ##G## instead, that is: the set of objects contains only one element ##G##, and the morphisms are all group homomorphisms ##G \longrightarrow G##, then this is also a category, although usually not considered, as categories normally deal with objects one level above, then the direct sum between two of such categories ##(G, G \to G)## and ##(H, H \to H)## is the same as ##G \times H##. In this case we still don't bother the elements, as our new object is ##G\times H## and our new morphisms are ##G\times H \longrightarrow G\times H## composed from to homomorphisms ##f \, : \,G \to G## and ##g\, : \,H \to H##. That the direct product of these two categories and the direct product of groups ##G \times H## are identical, however, has still to be checked. As direct product of groups, all happens within the category ##\mathcal{Groups}## and we have a lot more morphisms than we have in the direct sum category ##(G, G \to G) \times (H,H\to H)##. That is one has to show, that all homomorphisms ##h\, : \,G \times H \longrightarrow G \times H## split into ##h=(f,g)##. I haven't checked this, so that is why I wrote that the two are at prior different. So to summarize, prove the following Lemma. (As said, I haven't checked whether it's true, but maybe you want to try.)

Lemma:
1. A given a group ##G## together with all group homomorphisms ##f\, : \,G \longrightarrow G## builds a category ##\mathcal{G}_G##.

2. The direct sum of two such categories ##\mathcal{G}_G \times \mathcal{G}_H## is a direct sum ##G \times H## in the category of groups ##\mathcal{G}##.

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Math Amateur
1. A given a group ##G## together with all group homomorphisms ##f\, : \,G \longrightarrow G## builds a category ##\mathbb{G}_G##

Does the notation "##f\, : \,G \longrightarrow G##" imply the domain of ##f## is the entire group ##G##?

Math Amateur
steenis
The author says, see post #1:

“The reader familiar with groups will recognize that for groups ##G## and ##H##, the product category ##G \times H## is the usual (direct) product of groups.”

According to the author Awodey, ##G## and ##H## are groups, but ##G \times H## is a product category. This is confusing, especially for people who did not follow the line of the book. It should be mentioned, that ##G## and ##H## are not only groups, but they are groups viewed as categories. In this line, the Cartesian product ##G \times H## of the groups ##G## and ##H##, which is a group, should be viewed as a category. The author asked to show that the group ##G \times H## viewed as category, corresponds with ##G \times H## viewed as a group.

In short, ##G## is group. Viewed as a category, this group consists of one object, say ##\star##, and each element of ##G## becomes an arrow ##g:\star \rightarrow \star## in ##G## viewed as a category. The unit arrow, inverses of elements, composition, etc., are easily defined.

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Does the notation "##f\, : \,G \longrightarrow G##" imply the domain of ##f## is the entire group ##G##?
Good question. I assume so, as subgroups aren't part of the category in this case. I think the comparison as such is unfortunate, I only tried to save it somehow. I haven't checked in detail, so it could well be that it doesn't work. Maybe it's better to see it completely different, namely that the direct products of groups (as built in the category of groups) can also be interpreted as an object of the product category of groups with itself. This way the artificial construct of a category ##(G,f:G\to G)## would be obsolete. I don't know how to sort out those remarks by Awodey. They deserve at least a closer look, if one doesn't agree on simply to ignore them.

steenis
Maybe it's better to see it completely different, ...

I don't know how to sort out those remarks by Awodey. They deserve at least a closer look, if one doesn't agree on simply to ignore them.

See post #8

Math Amateur
In what sense a group is considered a category by the author is defined in Definition 1.4, two pages before the definition of a product of categories.

Math Amateur
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Thanks to fresh_42, steenis, Stephen and Martin ...

... ... for your thoughts and clarifications ...

Peter

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MHB
Maybe initially I should have had the confidence to simply translate Awodey's definition of the product of two categories into the case for two groups ##G## and ##H## ... visually ##G \times H## ... where the two groups and their product are viewed as categories ...

Now the Awodey's definition ( with a couple of trivial amendments to the notation ... ) reads as follows:

"The product of two categories ##C## and ##D##, written as ##C \times D## has objects of the form ##(A, B)## for ##A \in C## and ##B \in D##, and arrows of the form

##(f, g) : (A, B) \to (A', B')##

for ##f : A \to B## and ##g : A' \to B'##

Composition and units are defined componentwise, that is

##(f', g') \circ (f, g) = (f' \circ f, g' \circ g )##

##1_{ (C, D) } = ( 1_C, 1_D )##

... ... ...

Now ... following the above ... the product of two groups ##G, H## viewed as categories ... written as ##G \times H## has objects of the form ##( \bullet, \star )## for ##\bullet \in G## and ##\star \in H## ... and arrows of the form

##(x, y) : ( \bullet, \star ) \to ( \bullet, \star )##

for ##x : \bullet \to \bullet## and ##y : \star \to \star##

Composition and identities/units are defined componentwise

##(x', y') \circ (x, y) = (x' \circ x , y' \circ y )##

where ##(x' \circ x , y' \circ y ) = ( x' \times_1 x , y' \times_2 y )##

and

##1_{ (G, H) } = ( 1_G, 1_H)##

Thus ... for groups ##G## and ##H## the product category ##G \times H## is the usual external direct product for groups ...

Essentially the composition of arrows in the product category is interpreted as 'multiplication' in the direct product ...

Is the above a correct interpretation ... ?

Peter

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Now ... following the above ... the product of two groups ##G, H## viewed as categories ... written as ##G \times H## has objects of the form ##( \bullet, \star )## for ##\bullet \in G## and ##\star \in H## ... and arrows of the form

##(x, y) : ( \bullet, \star ) \to ( \bullet, \star )##
for ##x : \bullet \to \bullet## and ##y : \star \to \star##
I see it differently. I thought of categories ##\mathcal{G}## and ##\mathcal{H}## where the objects of ##\mathcal{G}## is the set ##\{\,G\,\}## and likewise for ##\mathcal{H}##, and the morphisms are thus ##\operatorname{Aut}(G)##. However, this could also be wrong, but it makes more sense to me as your morphisms on elements of ##G##, or are subsets allowed as well? I think it is necessary to look up:
In what sense a group is considered a category by the author is defined in Definition 1.4, two pages before the definition of a product of categories.

Math Amateur
The definition is the one in post #8.

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The definition is the one in post #8.
Thanks. Rather artificial for my taste. I think I stay with the definition as a solution to a universal mapping problem for direct (co-)products.

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@Math Amateur

As mentioned in post #8, a group can be viewed as a category with one object in which all of the morphisms are isomorphisms.

This idea is different from the idea of the category of groups and group homomorphisms. Here, each group is a category not an an object in a category.

If ##G## is a group defined in terms of a law of multiplication, then for each of its elements one has a "morphism" ##m_{a}:G→G## defined by ##m_{a}(x)=a⋅x##. ##m_{id}## is the identity morphism and ##m_{a}## is an isomorphism since its inverse is ##m_{a^{-1}}##. The composition ##m_{a} \circ m_{b}## is the morphism ##m_{a⋅b}##. So left multiplication by the elements of ##G## turns ##G## into a category in which each morphism is an isomorphism.

From this point of view one does not have multiplication of group elements in the usual sense but rather composition of morphisms. A homomorphism between two groups ##G## and ##H## becomes a covariant functor between the category ##G## and the category ##H##. For instance, the functor that sends each morphism in ##G## to the identity morphism in ##H## corresponds to the trivial homomorphism.

Examples:

The group with one element is a category with a single morphism and a single object.

The group with two elements is a category with one object and two morphisms each of which is an isomorphism.

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Math Amateur
steenis
Thanks. Rather artificial for my taste. I think I stay with the definition as a solution to a universal mapping problem for direct (co-)products.

I think you are mixing up:
- a group viewed as a category (defined in post #8)
- the product of objects of a category (which is done with the universal mapping property)
- the product of categories, as defined in post #1

Awodey askes to consider the product category of two groups ##G## and ##H##, viewed as categories, as defined in post #1
“The reader familiar with groups will recognize that for groups ##G## and ##H##, the product category ##G \times H## is the usual (direct) product of groups.”

So you have groups ##G## and ##H##
- define ##G## viewed as a category, my notation ##\overline{G}##
- define ##H## viewed as a category, my notation ##\overline{H}##
- construct the product category, as defined in post #1, of the categories ##\overline{G}## and ##\overline{H}##, that is ##\overline{G} \times \overline{H}##
- define ##G \times H## viewed as a category, my notation ##\overline{G \times H}##
- and, finally, show that

##\overline{G \times H} \cong \overline{G} \times \overline{H}##​

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I think you are mixing up:
- a group viewed as a category (defined in post #8)
- the product of objects of a category (which is done with the universal mapping property)
- the product of categories, as defined in post #1
No, I got it on a review of your previous post #8. I simply don't think that such a view carries any insights, beside the fact that categories can be rather abstract. It simply appears a bit enforced to me, especially if you have to explain the difference between a direct product and a direct sum with such an example in mind!

Math Amateur and lavinia
I simply don't think that such a view carries any insights, beside the fact that categories can be rather abstract.
My guess is that this is the whole point. To show the reader, early on, examples that are not the usual ones, where you have objects and morphisms that you have seen many times before, like groups with group homomorphisms.

Math Amateur
steenis
Viewing groups (or monoids) as categories is quite a common exercise, check Simmons, Lawvere, MacLane, Adamek.

I see no other way to interprete the exercise of Awodey:

“The reader familiar with groups will recognize that for groups ##G## and ##H##, the product category ##G \times H## is the usual (direct) product of groups.”

as I did in post #18

But I am always interested in your interpretion and, above all, your solution.

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My guess is that this is the whole point. To show the reader, early on, examples that are not the usual ones, where you have objects and morphisms that you have seen many times before, like groups with group homomorphisms.
Maybe, but then I would have chosen something absurd like (cars, selling them) or so, or what most authors do, take ##Set##, which can be used to explain two different structures on the same basis of sets, and explain the forget functor. To confuse first readers by choosing groups doesn't seem a good idea to me.

Math Amateur
steenis
I have no idea what this discussion is going about. If you don’t mind I will wait until this thread is on-topic again.

And the topic is Awodey’s exercise, see post #1:

“The reader familiar with groups will recognize that for groups ##G## and ##H##, the product category ##G \times H## is the usual (direct) product of groups.”

Math Amateur