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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.

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    Monoids, groups, and sets can all be viewed as special types of categories. The e$x$ercise asks you to show that the product category of two monoids (for e$x$ample) is the same as the regular product.2011-09-22

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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.

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    @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.2011-09-23