# amalgam of structures

Trying to refine my question here. This is a response to the questions here: Homomorphisms between structures

My objective is to take a set of $S-$structures and form an amalgam object out of that set with two properties: (a) the amalgam object is a structure and (b) every structure in the set can be embedded within the amalgam object via an injective homomorphism (which has been called a partial isomorphism).

There are a few candidates for what kind of amalgam object I'm looking for and I'm not picky about the nature of that amalgam object; I just want to prove there is such an object.

Currently, I'm looking at direct products. In this case, given a set of $S-$structures $\left\{ \mathcal{A}_{i}:i\in I\right\}$, I want to prove that their direct product exists in the categorical sense of product.

The definition of product in the instance of $S-$structures I am using comes from modifying the one given in Hilton and Stammbach's "A course in homological algebra." Given a set of $S-$structures, the product $(P;\pi_{i})$ is an $S-$structure, together with the maps $\pi_{i}:P\rightarrow A_{i}$ called projections, with the universal property: Given any $S-$structure $\mathcal{Y}$ and homomorphisms $f_{i}:Y\rightarrow A_{i}$, there exists a unique homomorphism $f:Y\rightarrow P$ such that $\pi_{i}\circ f=f_{i}$ for all $i\in I$.

Ebbinghaus, et al., gives a definition for a product it calls the direct product of a set of $S-$structures $\left\{ \mathcal{A}_{i}:i\in I\right\}$. $\prod_{i\in I}\mathcal{A}_{i}$ is defined so that the universe of $\prod_{i\in I}\mathcal{A}_{i}$ is the Cartesian product of the universes, $\prod_{i\in I}A_{i}$, and the following rules tell us how to interpret constant symbols, relation symbols, and function symbols:

(Notation: for $g\in\prod_{i\in I}A_{i}$, we also write $\left\langle g\left(i\right):i\in I\right\rangle$.)

For a constant symbol $c$, $c^{\mathcal{A}}:=\left\langle c^{\mathcal{A}_{i}}:i\in I\right\rangle .$

For an $n-$ary relation symbol $R$ and for $g_{1},...,g_{n}\in\prod_{i\in I}A_{i}$, say that $R^{\mathcal{A}}g_{1}...g_{n}$ iff for all $i\in I$, $R^{\mathcal{A_{\mathit{i}}}}g_{1}\left(i\right)...g_{n}\left(i\right)$.

For an $n-$ary function symbol $f$ and for $g_{1},...,g_{n}\in\prod_{i\in I}A_{i}$, say that $f^{\mathcal{A}}\left(g_{1},...,g_{n}\right):=\left\langle f^{\mathcal{A_{\mathit{i}}}}\left(g_{1}\left(i\right),...,g_{n}\left(i\right)\right):i\in I\right\rangle .$

At this point, we have two notions of product, one from category theory and another from math logic. What I would like to know is if the math logic notion given here is an example of a product in the category theory sense. I can prove that given any $S-$structure $\mathcal{Y}$ and homomorphisms $f_{i}:Y\rightarrow A_{i}$, there exists a unique homomorphism $f:Y\rightarrow P$ such that $\pi_{i}\circ f=f_{i}$ for all $i\in I$.

Fix an $i_{0}\in I$ and apply the universal property to the structure $\mathcal{A}_{i_{0}}=\mathcal{Y}$ to prove the existence of a unique homomorphism $f:A_{i_{0}}\rightarrow P$ such that $\pi_{i_{0}}\circ f=f_{i_{0}}$ where the $f_{i}:A_{i_{0}}\rightarrow A_{i}$ are defined so that if $i=i_{0}$, $f_{i}=1_{A_{i_{0}}}$ and if $i\neq i_{0}$ then I need to define a homomorphism from $A_{i_{0}}\rightarrow A_{i}$. My main problem is defining such a homomorphism. Does the trivial homomorphism, the empty function say, provide me with what I need? I don't care what the homomorphisms $f_{i}$ are when $i\neq i_{0}$, just so long as $f_{i_{0}}=1_{A_{i_{0}}}$ so that I will get that $\pi_{i_{0}}\circ f=1_{A_{0}}$ which will imply $f$ is injective since each $\pi_{i}$ is surjective and the identity is injective. Thus $f$ is the desired injective homomorphism, if I can find homomorphisms $f_{i}:A_{i_{0}}\rightarrow A_{i}$.

I am also quite willing to entertain notions of amalgamation besides direct products, just so long as goals (a) and (b) above are met.

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The concept you want is almost surely the coproduct, not the product. If I'm not mistaken any algebraic theory (i.e. a one-sorted first order theory with no relations, a fixed set of finitary operations, and a fixed set of equational axioms) has coproducts; unfortunately there is no easy concrete description of the coproduct in most cases. – Zhen Lin Feb 23 '12 at 20:14
The "usual" concept of amalgam refers to several structures with common substructures (formally, a family $\{A_i\}$, and for each pair $(i,j)$, an object $B_{ij}$ together with embeddings of $B_{ij}$ into $A_i$ and into $A_j$; with $B_{ii}=A_i$ and the embeddings being the identity, and $B_{ij}=B_{ji}$). An "amalgam" is then an object $C$ and embeddings $A_i\hookrightarrow C$ with the property that composing the embedding of $B_{ij}$ into $A_i$ and $A_i$ into $C$ yields the same as the embedding of $B_{ij}$ into $A_j$ followed by the embedding of $A_j$ into $C$. – Arturo Magidin Feb 23 '12 at 20:22
And products don't always contain isomorphic copies of the factors as substructures (for example, the direct product of unital rings does not contain copies of factors as unital subrings). As Zhen Lin points out, the object you would normally want is the coproduct, which, if for any two objects $A$ and $B$ there is always at least one morphism $A\to B$, does contain copies of each element of the family. – Arturo Magidin Feb 23 '12 at 20:24
Thank you. How would the coproduct of structures be defined? Between two structures the empty function serves as a trivial homomorphism, does that satisfy the criteria "there is always at least one morphism from A to B"? – atat Feb 23 '12 at 22:12
The coproduct is defined by a universal property which can be found in any textbook on category theory; you will find that it is almost exactly your requirements (a) and (b). The empty function does not count as a homomorphism because it is not a total function (unless the domain is empty). – Zhen Lin Feb 24 '12 at 7:26