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

Question

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...

Answer 1

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).

Answer 2

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.

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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.

Answer 3

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.

Answer 4

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.

Answer 5

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 $B$