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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...

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  • $\begingroup$ 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. $\endgroup$ – bryn Jul 28 '10 at 1:55
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    $\begingroup$ In category theory, isomorphy is effectively identity. $\endgroup$ – Seamus Jul 31 '10 at 19:09
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    $\begingroup$ 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.) $\endgroup$ – Pete L. Clark Aug 29 '10 at 20:49
  • $\begingroup$ Which is a useful heuristic? My original question or my flippant comment "isomorphy is identity"? $\endgroup$ – Seamus Aug 29 '10 at 21:08
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    $\begingroup$ @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." $\endgroup$ – exchange Nov 16 '16 at 7:37
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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).

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  • $\begingroup$ It's true in Set, though. Maybe we are missing some context? $\endgroup$ – Charles Stewart Jul 21 '10 at 13:30
  • $\begingroup$ It's true in categories enriched over sets. But Eric is correct. $\endgroup$ – BBischof Jul 21 '10 at 14:40
  • $\begingroup$ 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/… $\endgroup$ – Seamus Jul 21 '10 at 15:26
  • $\begingroup$ I'm not sure how this is any different from the question you asked on MO, in fact. $\endgroup$ – Akhil Mathew Jul 21 '10 at 17:02
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    $\begingroup$ Correct. But I just looked into the book and the problem I saw was just asking you to prove this in SET. $\endgroup$ – Eric O. Korman Aug 1 '10 at 18:41
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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.

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

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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.

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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 $B$

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    $\begingroup$ I don't understand. A = B. $\endgroup$ – Eric O. Korman Dec 11 '10 at 16:13
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    $\begingroup$ 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, $[1/3, 2/3]$. $\endgroup$ – Qiaochu Yuan Feb 27 '12 at 20:16

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