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In category theory, it seems that a monomorphism from $A$ to $B$ and one from $B$ to $A$ should be enough to guarantee isomorphy, but it doesn't seem to be so. (If I'm right then there's something fishy with the standard definition of "subobject")

So here's the counterexample I thought up, please explain where I went wrong.

Consider a category consisting of 2 objects $A$ and $B$. There is a monomorphism $\phi: A \to B$ and another $\psi : B \to A$. "Close" this under composition in much the same way you do when defining a free group (that is, no non-trivial identities are allowed). I claim that this does not guarantee isomorphism. All morphisms are monic, since no identities hold, so the condition for monomorphism is trivially satisfied.

What am I doing wrong here?

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Nothing is wrong, it is simply not true that monomorphisms in both directions guarantee isomorphism. A very concrete example is given by considering the intervals (0,1) and [0,1] in the category of topological spaces. I don't understand your comment about the standard definition of a subobject. – Dan Petersen Jul 18 '10 at 14:58
Another example is that free group of finite rank $>1$ contains free subgroups of any finite (and countable) rank. Hence, one may take two free groups of different finite ranks $>1$. – Torsten Ekedahl Jul 18 '10 at 15:00
In the category of sets, however, it is true that when two sets inject into each other, then there is a bijection. This is the Cantor-Bernstein-Schroeder theorem.…. But it fails in many other naturally arising categories. MO member John Goodrick has investigated this quite a lot. – Joel David Hamkins Jul 18 '10 at 15:06
Often it's clear what map we mean, and so we can get away with leaving it unmentioned. But when it's not assumed, you may need to make it explicit! I think the intuition behind your weirdness may be something like the fact “if there are monomorphisms ABA, and their composite is the identity $1_A$, then these monos are isos”. (Proving this is a nice exercise!) In terms of subobjects: “if A is a s.o. of B, and B is a subobject of A, and these are compatible with our standard way of thinking of A as a subobject of itself, then A and B are isomorphic”. – Peter LeFanu Lumsdaine Jul 18 '10 at 19:54

Dear Seamus, an example of non-isomorphic objects mutually monomorphing into each other is the following, in the category of groups ( I haven't tried to follow your sketch of construction).

Consider the free group on two generators $F_2$. Its commutator subgroup $C\subset F_2$ is a free group on denumerably many generators: $C=F_\infty$. This can be proved elegantly by using topological covering spaces [you can look it up in Massey's Introduction to Algebraic Topology for example].

So you have monomorphisms $F_2 \hookrightarrow F_\infty$ and $F_\infty \hookrightarrow F_2$, although $F_2$ and $F_\infty$ are not isomorphic, since their abelianizations are free $\mathbb Z$ modules on respectively two and denumerably many generators.

I have used that monomorphisms in the category of groups coincide with injective morphisms, which is a not trivial but true result [ Jacobson, Basic Algebra, vol.II, Prop 1.1]

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Regarding the last paragraph: If $C \to Set$ is a faithful functor which has a left-adjoint, then $f$ is a monomorphism in $C$ if and only if it is injective. – Martin Brandenburg Jul 18 '10 at 17:24
A little more "hard" is the analogous assert about epimorphisms (see Mitchell Theory of Categories$ p. 38). – Buschi Sergio Mar 22 '11 at 22:16

Your counterexample is correct; indeed it is the universal one, every other counterexample comes from a functor defined on your category. For a counterexample in the category of fields, see my answer here Counterexamples in Algebra?.

You seem to be worried about subobjects. If $X$ is an object and $U,V \leq X$ are subobjects such that $U \leq V$ and $V \leq U$, then $U = V$. The reason is that the morphisms $U \to V$ and $V \to U$ over $X$ are uniquely determined (since $V \to X, U \to X$ are monomorphisms). Likewise are the compositions $U \to V \to U, V \to U \to V$ uniquely determined, namely the identity. Thus $U = V$. Thus you don't get into trouble.

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So to get this straight: the definition of subobject requires that the monomorphism be unique? – Seamus Jul 18 '10 at 17:43
I don't know which mono you mean, but nowhere uniqueness of something is required. – Martin Brandenburg Jul 18 '10 at 17:53
I don't know what you mean by "the morphisms ... over X are uniquely determined" – Seamus Jul 18 '10 at 20:15
Here's the worry: in the two object category I defined above, each of A and B is a subobject of the other, but they are not isomorphic. Is this just a case where the categorical notion of "subobject" doesn't make sense, or have I misunderstood? – Seamus Jul 18 '10 at 21:35
you're right, but this is not really a misbehavior of subobjects. since when you fix some object $X$, then the subobjects of $X$ behave well (see my answer). – Martin Brandenburg Jul 19 '10 at 7:32

I should state first that this reply has only to do with the above mentioned ideas in the category of models of a first order theory.

John Goodrick's work is referenced in Joel's post above, and I have heard John Goodrick speak about this at least once. Specifically, John mentioned the following (and a lot more that I didn't write down):

Fix some countable, complete first-order theory, $T$. Suppose $T$ has the following property: Whenever we are given two models $\mathcal M_1$ and $\mathcal M_2$ of $T$ which have elementary embeddings into each other, then $\mathcal M_1 \cong \mathcal M_2.$

Then $T$ is superstable and nonmultidimensional (and I know if John replies to this, he can mention many other things, but I don't remember now). In the case that $T$ is actually $\omega -$stable, nonmultidimensional implies the bi-embedding property stated in the above paragraph.

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One of the simplest example , lightest in structure and very easy in checking is :

In the category of monoids take the canonical injection $i$ from $(N,+,0)$ to $(Z,+,0)$.

This is a monomorphism that is also an epimorphism yet not an iso ($i$ is not a surjection).

($N$ and $Z$ are the positive integers and integers respectively)

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Not the same as what the original question was asking for – Yemon Choi Sep 27 '10 at 1:59
You are perfectly right. I reread the question which now seems new and interesting. The only positive point for my miss-answer would be to try to describe the type of question that include both? – Jérôme JEAN-CHARLES Oct 1 '10 at 1:00

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