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By structure, I mean that which is defined here:

What I'm looking for is a way of gluing together structures so that each structure used is embedded within the whole glued-together object. (Each--meaning not just "most")

I don't need this embedding to be elementary; just something that "preserves" function, relation, and constant symbols.

I would think that some type of product would work.

If we have a set of structures $S$, then I want to show there exists a structure $U$ with the property that there is an injective homomorphism from every structure $A$ in $S$ into $U$.

The following link will help solidify what I mean:

Are there two cases, depending on the size of $S$, finite and infinite? What I'm hoping for is some type of product of the structures in $S$ where $S$ is going to be a "large" collection of structures.

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I'm not sure what you're asking here. It seems that either a disjoint union or a product would work just fine. – Asaf Karagila Nov 28 '11 at 20:27
Are you trying to glue together several objects that have the same structure, or several objects, each with its own "structure"? (E.g., 'gluing together' a bunch of rings, or 'gluing together' a ring with a group with a lattice?) – Arturo Magidin Nov 28 '11 at 20:28
@Asaf: It is not always the case that you have natural embeddings into the product. For example, in the category of rings with unity, the coordinate embeddings don't give homomorphic embeddings. In categories of algebras you can try to use the coproduct, but under certain circumstances the coproduct may not contains copies (e.g., again in the category of rings with unity, there is no ring with unity that contains isomorphic copies of $F_2$ and $F_3$, since we would need the unity to have both additive order 2 and 3. – Arturo Magidin Nov 28 '11 at 20:31
@Arturo: Well, this is a lot about how you interpret this question. I do agree that you have a valid point there though. When I first read it, it seemed to me to be asking in the most naive way. You can always add a dummy constant to the language with absolutely no axioms whatsoever and no relation to anyone else, and changing the axioms a bit in case needed (suppose $F(x)\neq x$ will turn into $x\neq c\rightarrow F(x)\neq x$ or so). In such way a product would work just fine, when the language is the disjoint union of the languages + the constant. – Asaf Karagila Nov 28 '11 at 20:35
@atat: My question is, are all the elements of $S$ structures of the same type, or not? – Arturo Magidin Nov 28 '11 at 20:46
up vote 4 down vote accepted

Assuming you have a fixed signature (all structures in $S$ are algebras of the same type), there isn't always such an object.

For example, if you are working with rings with unity, it is not the case that given any collection of rings with unity $\{R_i\}$ there is always a ring with unity that contains each $R_i$ as a subring (for this situation, we require that subrings of $R$ have the same unity as $R$). For instance, there is no ring with unity that contains the field with 2 elements and the field with 3 elements as subrings-with-unity, because that would require the multiplicative identity to have both order $2$ and order $3$. On the other hand, in the category of rings (with or without unity, and homomorphisms are not required to preserve the unity), then the direct product and the "coordinate embeddings" work.

More generally, if you are working in a category of algebras where:

  1. For any two objects $A$ and $B$, there is always at least one morphism from $A$ to $B$; and
  2. For any family $\{A_i\}_{i\in I}$ of objects there is a (categorical) product,

then the product of the family of structures will always work (assuming the Axiom of Choice): for given a family $\{A_i\}_{i\in I}$, if $(P,\pi_i)$ is the product, then for pair $(i,j)$ let $f_{ij}\colon A_i\to A_j$ be an arbitrary morphism if $i\neq j$, and let $f_{ij}=\mathrm{id}_{A_i}$ if $i=j$. Then the family $f_{i_0,j}\colon A_{i_0}\to A_j$ induces a homomoprhism into the product $\mathcal{F}_{i_0}\colon A_{i_0}\to P$ such that $f_{i_0,j}=\pi_j\circ \mathcal{F}_{i_0}$ for each $j$; in particular, $\mathrm{id}_{A_{i_0}} = \pi_{i_0}\circ \mathcal{F}_{i_0}$, so $\mathcal{F}_{i_0}$ must be one-to-one, giving the desired immersion.

There are other circumstances where such an object may exist (that is, the conditions above are probably not necessary). But it seems hard to characterize when they do and when they don't in the abstract.

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Thanks. What happens if not all structures in S have the same signature? Moreover, if the structure U is allowed to have a different signature than the signatures of structures in S? How would the various signatures of structures in S be "glued" together? I would think that relaxing the assumption that all structures in S have the same signature would make it easier. – atat Nov 29 '11 at 11:27

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