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I am stuck on a question because I do not understand what the question is asking.

I am reading Mac Lane's CTFWM (2nd ed.), on p.39, exercise 1 the question reads:

Show that the product of categories includes the following known special cases: the product of monoids, of groups, of sets.

I do not understand what this means, or what I need to show. I understand exercise 2: show that the product of two preorders is a preorder.

What is the difference between these two questions? Thanks.

Arturo Magidin
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Edison
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    There are already existing notions of products of monoids, groups, and sets (the Cartesian product with the natural structure) and the question is to show that these notions are compatible with the categorical product. – Qiaochu Yuan Sep 22 '11 at 04:30
  • @Qiaochu, okay, that makes sense, thank you – Edison Sep 22 '11 at 04:33
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    Monoids, groups, and sets can all be viewed as special types of categories. The exercise asks you to show that the product category of two monoids (for example) is the same as the regular product. – Yuri Sulyma Sep 22 '11 at 06:35

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A group can be viewed as a one-object category in which every arrow is invertible as follows: given a group $G$, the object of the category $\mathbf{G}$ is a single element $\bullet$. There is one arrow of $\mathbf{G}$ for each element of $G$, with composition of arrows corresponding to multiplication of elements. So the composing the arrow corresponding to $g$ with the arrow corresponding to $h$ is $hg$ (do $g$ first, then $h$). The identity arrow corresponds to the identity element of $G$. Conversely, any one-object category $\mathbf{C}$ in which every arrow is invertible corresponds to a group, with underlying set $\mathbf{C}(\bullet,\bullet)$, and multiplication of elements corresponding to composition of arrows.

Likewise, a monoid $M$ can be viewed as a one-object category in the same manner, and any one-object category can be viewed as a monoid (the underlying set being the set of arrows, and operation being composition).

Finally, a set $S$ can be viewed as a category where you have one object for every element of $S$, and the only arrows are the identity morphisms. And any category in which the only arrows are the identity morphisms corresponds to a set, namely the set of objects.

Mac Lane has just finished defining the product of categories. The exercise asks you to show that if you have two groups $G$ and $H$, and view them as categories $\mathbf{G}$ and $\mathbf{H}$, then the product of the categories $\mathbf{G}$ and $\mathbf{H}$ will be a category that corresponds to the group $G\times H$. Likewise for monoids, and for sets.

Arturo Magidin
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  • I have shown that the product of the categories $G$, and $H$, correspond to the group $G\times H$. But I think this exact proof covers the cases for monoids and sets, am I miss understanding something subtle? – Edison Sep 23 '11 at 01:37
  • @ElG: Certainly a proof for monoids should work, mutatis mutando, for groups, since the only difference lies in showing every arrow is invertible, and that should be straightforward. I would expect the proof for sets to be even simpler. No, I don't think there is anything subtle in this, just a straightforward verification that things work out. – Arturo Magidin Sep 23 '11 at 03:10