Proof check, existence of free product

Question

I used to fear the existence of free products and respect it, after seeing the pretty (but nontrivial) treatment in introduction to manifolds. Afterwards I was wondering why doesn't the following argument to show the existence of free products of the collection $H_i$. I will use the word clearly just to shorten notation but I don't mean any disrespect.

We start by just looking at words (including the empty words) with letters coming from the $H_i$, call this set $R$. On those words we define concation, denoted by

$\cdot:RxR\to R$, which is clearly assosicative.

We define the equivalence relation formed by allowing to remove the unit element a group from a word. We also allow combining 2 elements from the same group that are adjacent. Then we note that if for 2 words $V,W$, $V \cong V'$, and $W\cong W'$, then $V\cdot W\cong V'\cdot W'$. Thus this is well defined on the equivalence classes. Since it was assosicative before, it is now. Also now there are clearly inverses, and the identity still acts as such. So we have this group.

There are also the obvious things we want to show are embeddings of the $H_i$. For now we just claim they are clearly homomorphisms.

Now I claim that given a group $G$ and a set of homormophisms from each $H_i$ to $G$ , we can extend it uniquely to our group (in particular the embeddings are injective, and their images disjoint, since we can take $G=Z_2$, and set the appropriate maps).

First we define the obvious map from $R$ to $G$. The resulting map from $R$ to $G$ is clearly a homomorphism (in the sense that it commutes with $\cdot$ Now note that it remains the same under each step allowed in our relation, and so it is constant under equivalence classes, and thus defines a well defined map from our group, and we know it sends the empty word to $1$, and is a homomorphism.

So this just misses that there is a unique reduced word, but I don't exactly see where this is critical (although it's certainly good to know). Thanks

Answer 1

Yes, this works. More generally, essentially the same argument shows that "free products" (a.k.a. coproducts) exist for any sort of structure with algebraic operations which must satisfy some equational axioms (a "variety", in the language of universal algebra). So in some sense, the existence of free products is just "easy abstract nonsense". In contrast, the explicit description of the equivalence classes in the free product using reduced words is a stronger result that takes more work to prove.

Answer 2

$R$ is clearly the free product in the category of monoids. The induced monoid homomorphism $R\to G$ clearly factors over the equivalence relation $\sim$. Hence $R/\sim$ (which is a group) plays the role of free product in the category of groups.