If we take a category $\mathcal{C}$ of objects with functions as morphisms and restrict the morphisms to injections (monomorphisms?), then this defines a partially ordered set of isomorphism classes.

Edit: What I mean is, for $X$, $Y$ isomorphism classes, let $X \leq Y$ if an object in $X$ has an injection to an object in $Y$ (equivalently, every object in $X$ has an injection to every object in $Y$).

Edit: Qiaochu Yuan points out that these may not be isomorphism classes. I think they are equivalence classes of isomorphism classes (or classes of objects, closed under isomorphism), and equal to isomorphism classes in small, nice cases (Sets, Noetherian rings, finite field extensions?).

What do you call the product in this poset?

Explicitly, the ______ of a set of objects $X_i$ is an object $X$ such that: For every object $Y$ with injections $\phi_i : Y \rightarrow X_i$, there exists an injection $\phi : Y \rightarrow X$ and injections $\pi_i : X \rightarrow X_i$ such that $\phi_i = \pi_i \circ \phi$ for all $i$.

This corresponds to:

  • min in the category sets (isomorphism classes of sets are cardinals)

  • "largest common sub-group/ring/field/etc., up to isomorphism"

  • gcd in the category of sets with functions restricted to "multiplications" (injections from $S$ to $T$ iff $|S|$ divides $|T|$)

  • (I believe) the coproduct of the poset if the morphisms were restricted to surjections.

  • Most intuitively, the largest object isomorphic to a subobject of each object (with an injection into each, a function preserving both = and $\neq$).

It seems like a natural idea, but I can't find a reference to it (probably because I don't know what it's called).

Edit: The dual notion (product, surjections or coproduct, injections) would correspond to max, lcm, or the smallest object containing as a subobject an element of each isomorphism class.

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    $\begingroup$ Jason, please do not keep adding the set-theory tag, please. It really does not fit the question. $\endgroup$ – Andrés E. Caicedo Jan 6 '16 at 16:38
  • $\begingroup$ I'm sorry. I didn't see that you had removed it, and I thought I had forgotten to add it. Thank you for fixing the tags. $\endgroup$ – Jason Eliot Jan 6 '16 at 16:39

This is in some sense the "wrong" construction. Categorically "subobject" is not a relation but a class of morphisms: a map $f : x \to y$ equips an object $x$ with the structure (not the property) of a subobject of $y$ iff it's a monomorphism. (What "structure" means here is, among other things, that $x$ may admit more than one subobject-of-$y$ structure, even up to isomorphism. The example to keep in mind here is embeddings of topological spaces.)

Hence for a fixed object $y$ we can consider the preorder $\text{Sub}(y)$ of monomorphisms into $x$ (this really is a preorder, automatically; this is a good exercise). If you really want you can further quotient this by isomorphism. Then you can consider products / coproducts in this preorder. This produces constructions like intersections of subgroups / subrings / etc., as well as smallest subgroups / subrings containing some collection of subgroups / subrings / etc.

  • $\begingroup$ What set is being ordered in this preorder? $\endgroup$ – Jason Eliot Jan 6 '16 at 17:28
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    $\begingroup$ @Jason: for me a preorder is not necessarily a set, so I mean the possibly proper class of monomorphisms $x \to y$ (with $y$ fixed). If you want the corresponding poset, then up to size issues, it's the set of isomorphism classes of monomorphisms $x \to y$ (with $y$ fixed). $\endgroup$ – Qiaochu Yuan Jan 6 '16 at 17:30

The product of a family of objects in a poset is always the meet of the objects (that is the biggest lower bound for the family of objects considered).

Of course the construction take a category and throw away everything except monomorphisms, doesn't really produce a poset (because there could still remain isomorphic objects and there could be more the one monomorphism between two objects).

Hope this helps. :)

  • $\begingroup$ Thank you. As I said, it produces a poset of the isomorphism classes. Is it common to refer to the "largest common subgroup of a set of groups, up to isomorphism" as the "meet" of the groups? $\endgroup$ – Jason Eliot Jan 6 '16 at 16:19
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    $\begingroup$ Actually, you don't even get a preordered set – there can be more than monomorphism between any two objects... $\endgroup$ – Zhen Lin Jan 6 '16 at 16:31
  • $\begingroup$ I'm sorry. What I mean is, for $X$, $Y$ isomorphism classes, let $X \leq Y$ if an object in $X$ has an injection to an object in $Y$ (equivalently, every object in $X$ has an injection to every object in $Y$). $\endgroup$ – Jason Eliot Jan 6 '16 at 16:42
  • $\begingroup$ @Jason: that is still not a poset. For example, there are topological spaces which are not homeomorphic but which embed into each other. You need to take the monomorphisms as extra data, and then consider the preorder of monomorphisms into a fixed object. Quotienting this by isomorphism gives the poset of subobjects of a fixed object, and then we can talk about meets and joins of these. $\endgroup$ – Qiaochu Yuan Jan 6 '16 at 16:51
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    $\begingroup$ @JasonEliot it stops to work because you don't have Cantor-Berstein in theorem in general categories: that's is it is not necessarily true that if you have to monomorphisms $X \to Y$ and $Y \to X$ then you have an isomorphism between the objects $X$ and $Y$. $\endgroup$ – Giorgio Mossa Jan 6 '16 at 17:16

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