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I have a several categories some of which are subcategories of others. I want to research properties of products in these categories but don't know where to start.

How direct products in a category and its subcategories are related?

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1 Answer 1

up vote 6 down vote accepted

Let $\mathcal{D}$ be a subcategory of $\mathcal{C}$ and let $A$ and $B$ be objects in $\mathcal{D}$. If $A \times_{\mathcal{D}} B$ is the product in $\mathcal{D}$ and $A \times_{\mathcal{C}} B$ is the product in $\mathcal{C}$, then there is a canonical morphism $A \times_{\mathcal{D}} B \to A \times_{\mathcal{C}} B$, but there is not much more we can say than that in general. Here's a somewhat extreme example: let $\mathcal{C}$ be the category of topological spaces and let $\mathcal{D}$ be the frame of open subsets of a topological space $X$; then, $\mathcal{D}$ is a non-full subcategory of $\mathcal{C}$, but the product in $\mathcal{D}$ is the intersection while the product in $\mathcal{C}$ is the cartesian product!

If $\mathcal{D}$ is a full subcategory of $\mathcal{C}$ and $A \times_{\mathcal{C}} B$ is (isomorphic to) an object of $\mathcal{D}$, then it is isomorphic (in $\mathcal{D}$) to $A \times_{\mathcal{D}} B$. In other words, the embedding of a full subcategory reflects products. (In fact, it reflects all limits and colimits.)

If the embedding of $\mathcal{D}$ into $\mathcal{C}$ preserves products, then $A \times_{\mathcal{D}} B$ is isomorphic to $A \times_{\mathcal{C}} B$ in $\mathcal{C}$, provided both of them exist. It is possible for $A \times_{\mathcal{C}} B$ to exist while $A \times_{\mathcal{D}} B$ does not. For example, we could take $\mathcal{D}$ to be the category of fields and $\mathcal{C}$ to be the category of rings.

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Very comprehensive and complete answer. 1+ –  Martin Brandenburg May 21 '12 at 13:42
    
Another example of products in a non-full subcategory being very different from those in the containing category: consider the opposite of the category of groups as a subcategory of the opposite of the category of sets (if you want, you can make the inclusion functor injective on objects by replacing Set with an equivalent category that has as at least as many copies of each set X as there are group structures on X --perform the replacement if that matches your idea of "subcategory" better); in one products are free products of groups and in the other they are disjoint unions of sets. –  Omar Antolín-Camarena May 21 '12 at 14:49
    
Yet another example: consider the opposite of the category of groups as a subcategory of the category of sets (by sending each group to the powerset of its set of elements); now you are comparing free products of groups with cartesian products of sets. –  Omar Antolín-Camarena May 21 '12 at 14:51
    
In math.stackexchange.com/questions/669508/… I ask how to prove it –  porton Feb 9 at 13:01
    
I do not see how if "$A \times_{\mathcal{C}} B$ is (isomorphic to) an object of $\mathcal{D}$, then it is isomorphic (in $\mathcal{D}$) to $A \times_{\mathcal{D}} B$". I agree that the 2 products are isomorphic in $\mathcal{C}$, but why in $\mathcal{D}$? Consider $\mathcal{C}$ the category with 2 isomorphic objects (1 and 2) and $\mathcal{D}$ the full subcategory with just 1 and its identity. Then the terminal object in $\mathcal{D}$ is isomorphic to 2 in $\mathcal{C}$, but not in $\mathcal{D}$ –  magma Feb 9 at 18:02

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