If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic?

In category theory, a subobject of $X$ is defined as an object $Y$ with a monomorphism, from $Y$ to $X$. If $A$ is a subobject of $B$, and $B$ a subobject of $A$, are they isomorphic? It is not true in general that having monomorphisms going both ways between two objects is sufficient for isomorphy, so it would seem the answer is no.

I ask because I'm working through the exercises in Geroch's Mathematical Physics, and one of them asks you to prove that the relation "is a subobject of" is reflexive, transitive and antisymmetric. But it can't be antisymmetric if I'm right...

• I don't know the answer to your question but being antisymmetric is stronger than what you have here. If 'being a subject of' is antisymmetric then A and B should not just be isomorphic but identical. – bryn Jul 28 '10 at 1:55
• In category theory, isomorphy is effectively identity. – Seamus Jul 31 '10 at 19:09
• in your first few weeks of study of the subject, that's a useful heuristic. After that it's distinctly misleading. I highly recommend that you read math.harvard.edu/~mazur/preprints/when_is_one.pdf. (There is, among other things, philosophical content here.) – Pete L. Clark Aug 29 '10 at 20:49
• Which is a useful heuristic? My original question or my flippant comment "isomorphy is identity"? – Seamus Aug 29 '10 at 21:08
• @Seamus Thanks so much for asking this question, I worked through the book as well and encountered exactly the same problem. The 1985 version really asks to proof it for arbitrary categories: "Define a similar "<" for subobjects of a fixed object in an arbitrary category and prove that these three properties again hold." – exchange Nov 16 '16 at 7:37

I don't think this can be true in general. What if we just take the category consisting of two objects $A$, $B$ and take morphisms $f:A\to B$, $g:B\to A$ with no relations between the morphisms, but forcing associativity. Then certainly $f$ and $g$ are monomorphisms but $A$ and $B$ are not isomorphic (since there are no relations between the morphisms).

• It's true in Set, though. Maybe we are missing some context? – Charles Stewart Jul 21 '10 at 13:30
• It's true in categories enriched over sets. But Eric is correct. – BBischof Jul 21 '10 at 14:40
• It's true in SET. That's what motivates Geroch's exercise, but as Eric's counterexample shows, it's not true more generally. I was just wondering whether I was missing something about subobjecthood over and above monomorphism. I asked a similar question on MathOverflow: mathoverflow.net/questions/32368/… – Seamus Jul 21 '10 at 15:26
• I'm not sure how this is any different from the question you asked on MO, in fact. – Akhil Mathew Jul 21 '10 at 17:02
• Correct. But I just looked into the book and the problem I saw was just asking you to prove this in SET. – Eric O. Korman Aug 1 '10 at 18:41

Here is an excellent paper on that question for a bunch of different categories. It's true for any set-based category of "finite" things.

The paper (The Cantor–Schroeder–Bernstein property in categories by Don Laackman) defines a category $\mathcal{C}$ to have the CSB property to be if whenever $f : A \to B$ and $g : B \to A$ are monomorphisms in $\mathcal{C}$, then $A$ and $B$ are isomorphic. The categories of sets and well-ordered sets have this property, while the categories of topological spaces, groups, posets, or abelian groups don't. The first theorem of this paper is:

Theorem. If a category $\mathcal C$ has a faithful functor $F : \mathcal{C} \to \mathsf{FinSet}$ to the category of finite sets, then $\mathcal C$ has the CSB property.

• I edited your answer to make it more self-contained (in case the link you give ever becomes unavailable). – Najib Idrissi Sep 21 '15 at 13:04

The free group on two letters contains as subgroups groups that are isomorphic free group on any finite number of letters. The free group on $n \geq 2$ letters contains the free group on two letters as a subgroup. So if we consider the category of groups, with $A = F_2$ and $B = F_n$ ($n > 2$) we get a counterexample.

I suspect you did not read carefully that exercise. A subobjects is not an object with some special property. A subobject is an object and a monomorphism, like $(A_0,m_0)$. So if you define $A_0\le A_1$ iff $\exists f:A_0\to A_1. mono(f)$, this does not hire $m_0, m_1$, and this is intuitively wrong. And there is a standard solution: define $(A_0,m_0)\le (A_1,m_1)$ iff $m_0$ factors through $m_1$ ($\exists f:A_0\to A_1. m_0 = m_1\circ f$). $(A_0,m_0)\le (A_1,m_1) \land (A_1,m_1)\le (A_0,m_0)$ and factoring gives you morphisms to construct an isomorphism between $A_0,A_1$.

For the context where subobjects are used, see “Robert Goldblatt. Topoi. The Categorial Analysis of Logic.” or any other textbook on categorical logic or Wikipedia.

In Top (topological speces) take the subspaces of the real line: $B=[0,1]$ and $A=[0,1]\cup \{4/3\}$ considering the inclusion $B \subset A$ and imbedding map $f: A\to B: x \mapsto x/2$. But $A$ isn't isomorphic (i.e. homeomorphic) to $B$

• I don't understand. A = B. – Eric O. Korman Dec 11 '10 at 16:13
• A correct counterexample in topological spaces is $B = [0, 1], A = (0, 1)$. There is an inclusion $A \to B$ and also an inclusion $B \to A$ sending $B$ to, for example, $[1/3, 2/3]$. – Qiaochu Yuan Feb 27 '12 at 20:16