If $X$ and $Y$ are objects of a concrete category $\mathcal{C}$, is there an accepted definition of "free product of $X$ and $Y$," generalizing the in the special case where $\mathcal{C}$ is the concrete category of groups? See also, free product of groups.

Edit. In $\mathbf{Grp},$ the free product is just the coproduct, but I see no reason why this should be the "right" definition of "free product" general. Free products ought to have a special relationship to free objects. If we can find that relationship, then we can either prove that coproducts always enjoy that relationship, or else, find a case where they do not.

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    $\begingroup$ Is there some particular feature of the free product as a concrete object you're keen to capture? I'm guessing you already know it's the coproduct in $\mathbf{Grp}$. $\endgroup$ – Malice Vidrine Jun 27 '14 at 22:25
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    $\begingroup$ "but why should coproducts give the "right" answer in general?" For which question? $\endgroup$ – Martin Brandenburg Jun 27 '14 at 22:40
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    $\begingroup$ "Free products ought to have a special relationship to free objects. " No, except for the trivial statement $F(S \coprod T) = F(S) \coprod F(T)$ which holds for any left adjoint $F$. As long as you don't give a precise meaning of "free product", there is no question to be answered. $\endgroup$ – Martin Brandenburg Jun 27 '14 at 22:50
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    $\begingroup$ This is true, but doesn't change the fact that your question is currently quite meaningless. It is $\approx$ "Is there a categorical definition of the coproduct in the category of groups." ... $\endgroup$ – Martin Brandenburg Jun 27 '14 at 22:59
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    $\begingroup$ Since you haven't defined what "free product" means in other categories, it's not clear why you don't accept the coproduct as the "free product." $\endgroup$ – Thomas Andrews Jun 27 '14 at 23:44

Yes of course, even for any category: it is called the coproduct :)

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    $\begingroup$ And for this reason it should not be called free product. I hope that some day this unfortunate terminology will be replaced by the correct one ... (of course I know that the terminology is motivated by the element structure of the coproduct, but often the element structure doesn't really tell us what an object is.) $\endgroup$ – Martin Brandenburg Jun 27 '14 at 22:41
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    $\begingroup$ Well, the name "free product" comes from pre-category thinking about groups, no doubt. And if you are studying groups before category theory, "free product" is a nice generalization of "free group." Trying to explain the term "co-product" might be difficult without category theory. $\endgroup$ – Thomas Andrews Jun 27 '14 at 23:47

Most notions of "freeness" start with a functor $G:\mathcal C\to \mathcal D$ and finding an adjoint $F:\mathcal D\to \mathcal C$ with $\operatorname{Hom}_{\mathcal C}(FD,C)\equiv \operatorname{Hom}_{\mathcal D}(D,GC)$. For example, with $\mathcal C$ the category of groups, and $\mathcal D$ the category of sets, and $G$ the forgetful functor sending a group to the underlying set.

Then the diagonal functor $G:\mathcal C\to \mathcal C^2$ with $C\mapsto (C,C)$ has as its corresponding $F$ the coproduct.

  • $\begingroup$ I replaced a few instances of the word "function" with "functor." Is that okay? $\endgroup$ – goblin Jun 28 '14 at 0:48
  • $\begingroup$ No problem, @user18921. $\endgroup$ – Thomas Andrews Jun 28 '14 at 1:23
  • $\begingroup$ My only issue with this is that we haven't really linked coproducts to free objects at all. I think we should assume $\mathcal{C}$ is concrete over $\mathbf{Set}$, and I think that the forgetful functor $\mathcal{C} \rightarrow \mathbf{Set}$ should play a role. $\endgroup$ – goblin Jun 30 '14 at 0:21

For the question "what's the link between a free product and a free object?", the answer is given in the wikipedia page, before "contents":

The point is that a disjoint union of groups is not a group but it is a groupoid. A groupoid G has a universal group U(G) and the universal group of a disjoint union of groups is the free (= coproduct) of the groups.


  1. It seems that here $U(-)$ denotes the free functor left adjoint to the forgetful functor $ \mathbf{Grp}\longrightarrow \mathbf{Grpoid}$

  2. The disjoint union is the in $\mathbf{Grpoid}$

To me, the OP's question is obviously: is there a distinct categorical notion of free product (distinct from coproduct in general but such that for groups they coincide, exactly as coproduct and products in other examples)?


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