How can the product of two categories help us understand groups?

[Math Amateur]

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:View attachment 8356
View attachment 8357
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 ...$$(g_1, h_1) \star (g_2, h_2) = (g_1 \bullet_1 g_2, h_1 \bullet_2 h_2)$$ ... ...

... where we are dealing with a product of two groups $$G$$ and $$H$$ ... namely $$G \times H$$ ...

... where $$g_1, g_2 \in G$$ and $$h_1, h_2 \in H$$ are elements of the two groups ...... BUT! ...

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

Note that 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?Hope someone can help ...

Peter

 

[steenis]

Let $G$ be a group, let us use the notation $\overline{G}$, that is, G is a group viewed as a group and $\overline{G}$ is the group G viewed as category.

Given a group $G=(G, 1, \cdot )$
$\overline{G}$ is defined as follows:
$\overline{G}$ has one object, say $\star$
Each element $g$ of $G$ corresponds with an arrow $g: \star \rightarrow \star$ in $\overline{G}$
The composition $g \circ h$ in $\overline{G}$ corresponds with the product $g \cdot h$ in $G$
The identity $1_\star$ in $\overline{G}$ corresponds with the unit $1$ in $G$
Idem for inverses and associativity

Show that $\overline{G \times H}$ corresponds with the group $G \times H$
$\overline{G \times H}$ has one object, say $\bullet$
Each element $(g, h)$ of $G \times H$ corresponds with an arrow $\alpha = (g, h): \bullet \rightarrow \bullet$ in $\overline{G \times H}$
The composition $(g, h) \circ (p, q)$ in $\overline{G \times H}$ corresponds with the product $(g, h) \cdot (p, q) = (gp, hq)$ in $G \times H$
The identity $1_\bullet$ in $\overline{G \times H}$ corresponds with the unit $(1, 1)$ in $G \times H$
Idem for inverses and associativity

I think this is enough to show the correspondence

(One can also consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
In that case $(g, h) \circ (p, q) = (g \circ p, h \circ q)$ makes sense
too much work for me)

 

[steenis]

Elementwise products like $(g, h) \star (p, q)=(g \bullet p, h \bullet q)$ make sense in, for instance groups. They make no sense in general categories, because an object $C$ of a category $\mathscr{C}$ in general does not have elements

In fact, looking closely at a group $G$ viewed as a category, you see that this category does not possesses elements: besides one object, it only contains arrows. And products like $(g, h) \star (p, q)=(g \bullet p, h \bullet q)$ make only sense if $g, h, p, q$ are arrows.

 

[Math Amateur]

Let $G$ be a group, let us use the notation $\overline{G}$, that is, G is a group viewed as a group and $\overline{G}$ is the group G viewed as category.

Given a group $G=(G, 1, \cdot )$
$\overline{G}$ is defined as follows:
$\overline{G}$ has one object, say $\star$
Each element $g$ of $G$ corresponds with an arrow $g: \star \rightarrow \star$ in $\overline{G}$
The composition $g \circ h$ in $\overline{G}$ corresponds with the product $g \cdot h$ in $G$
The identity $1_\star$ in $\overline{G}$ corresponds with the unit $1$ in $G$
Idem for inverses and associativity

Show that $\overline{G \times H}$ corresponds with the group $G \times H$
$\overline{G \times H}$ has one object, say $\bullet$
Each element $(g, h)$ of $G \times H$ corresponds with an arrow $\alpha = (g, h): \bullet \rightarrow \bullet$ in $\overline{G \times H}$
The composition $(g, h) \circ (p, q)$ in $\overline{G \times H}$ corresponds with the product $(g, h) \cdot (p, q) = (gp, hq)$ in $G \times H$
The identity $1_\bullet$ in $\overline{G \times H}$ corresponds with the unit $(1, 1)$ in $G \times H$
Idem for inverses and associativity

I think this is enough to show the correspondence

(One can also consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
In that case $(g, h) \circ (p, q) = (g \circ p, h \circ q)$ makes sense
too much work for me)

Thanks for this post Steenis ...

It is most helpful ...

Peter

- - - Updated - - -

Elementwise products like $(g, h) \star (p, q)=(g \bullet p, h \bullet q)$ make sense in, for instance groups. They make no sense in general categories, because an object $C$ of a category $\mathscr{C}$ in general does not have elements

In fact, looking closely at a group $G$ viewed as a category, you see that this category does not possesses elements: besides one object, it only contains arrows. And products like $(g, h) \star (p, q)=(g \bullet p, h \bullet q)$ make only sense if $g, h, p, q$ are arrows.

Thanks steenis ...

Peter

 

[Math Amateur]

Thanks for this post Steenis ...

It is most helpful ...

Peter

- - - Updated - - -Thanks steenis ...

Peter

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

 

[steenis]

Where did I go wrong ?

(One can also consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
In that case $(g, h) \circ (p, q) = (g \circ p, h \circ q)$ makes sense
too much work for me)

I should not only consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
but I must prove this. Only then the compositon rule

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

of the product category is fulfilled

Where did you go wrong ?

You defined, with my notation, $\overline{G}$ and $\overline{H}$ and constructed $\overline{G} \times \overline{H}$,
but you did not show that this is equal to the product $G \times H$ viewed as a category.

So, you have to define $G \times H$ viewed as a category, i.e., $\overline{G \times H}$ and show that this corresponds to
$\overline{G} \times \overline{H}$.

So, you also have to prove that $\overline{G \times H} \cong \overline{G} \times \overline{H}$

 

[Math Amateur]

Where did I go wrong ?
I should not only consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
but I must prove this. Only then the compositon rule

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

of the product category is fulfilled

Where did you go wrong ?

You defined, with my notation, $\overline{G}$ and $\overline{H}$ and constructed $\overline{G} \times \overline{H}$,
but you did not show that this is equal to the product $G \times H$ viewed as a category.

So, you have to define $G \times H$ viewed as a category, i.e., $\overline{G \times H}$ and show that this corresponds to
$\overline{G} \times \overline{H}$.

So, you also have to prove that $\overline{G \times H} \cong \overline{G} \times \overline{H}$[/QUO

Thanks for the above post Steenis ...Thinking through your post and reflecting on what you have said ...

I will, indeed must, think about this some more in the morning ...

Thanks again...

Peter

 

[Math Amateur]

Where did I go wrong ?
I should not only consider to prove $\overline{G \times H} \cong \overline{G} \times \overline{H}$,
but I must prove this. Only then the compositon rule

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

of the product category is fulfilled

Where did you go wrong ?

You defined, with my notation, $\overline{G}$ and $\overline{H}$ and constructed $\overline{G} \times \overline{H}$,
but you did not show that this is equal to the product $G \times H$ viewed as a category.

So, you have to define $G \times H$ viewed as a category, i.e., $\overline{G \times H}$ and show that this corresponds to
$\overline{G} \times \overline{H}$.

So, you also have to prove that $\overline{G \times H} \cong \overline{G} \times \overline{H}$[/QUO

Thanks for the above post Steenis ...Thinking through your post and reflecting on what you have said ...

I will, indeed must, think about this some more in the morning ...

Thanks again...

Peter

Hi steenis ...

I set out merely to translate Awodey's definition of the product of two categories into the case for two groups G, H ... and I think I did that correctly ...

But I did go too far when I stated ...

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

I did not show this! For this I need to follow your advice and as you said ...

" ... ... define $G \times H$ viewed as a category, i.e., $\overline{G \times H}$ and show that this corresponds to
$\overline{G} \times \overline{H}$."

That is ... " ... prove that $\overline{G \times H} \cong \overline{G} \times \overline{H}$ ... "
Thanks for the advice and help ...

Peter

 

[steenis]

No, you did not go to far, You defined the product category of the groups $G$ and $H$, viewed as categories, and you did that correctly.

After that it has to be shown that $\overline{G} \times \overline{H}$ corresponds with $\overline{G \times H}$ because Awodey askes: “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.” (viewed as category - steenis)

So, in my view, the correct things to do are:

- 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

$$\overline{G \times H} \cong \overline{G} \times \overline{H}$$

 

[Math Amateur]

No, you did not go to far, You defined the product category of the groups $G$ and $H$, viewed as categories, and you did that correctly.

After that it has to be shown that $\overline{G} \times \overline{H}$ corresponds with $\overline{G \times H}$ because Awodey askes: “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.” (viewed as category - steenis)

So, in my view, the correct things to do are:

- 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

$$\overline{G \times H} \cong \overline{G} \times \overline{H}$$

Thanks for all your help in this matter Steenis ...

I now understand this example thanks to you

Peter