In an abelian category, does every family of subobjects have an intersection?

Question

I'm studying abelian categories in Borceux, so I'm using the same definitions as him. I already proved that the intersection of two subobjects always exists in an abelian category. My question is : does the intersection of an infinite family of subobjects always exists in abelian categories ? My intuition is "No" but I can't find any counterexample. Any help?

Answer 1

Here is a counterexample. Let $k$ be your favorite field, and let $C$ be the category of "eventually constant" sequences of vector spaces over $k$. That is, an object of $C$ is a sequence $(V_n)$ of vector spaces over $k$ such that there exists $N$ for which $V_n=V_m$ for all $n,m\geq N$. A morphism $(V_n)\to (W_n)$ of $C$ is a sequence of morphisms $T_n:V_n\to W_n$ such that there exists $N$ for which $T_n=T_m$ for all $n,m\geq N$.

This category is abelian, as kernels and cokernels can just be computed coordinatewise. Now consider the object $A$ which is the constant sequence $(k)$. For each $m$, we have a subobject $A_m$ of $A$ whose $m$th term is $0$ but all other terms are $k$. The collection of subobjects $A_{2m}$ then has no intersection in $C$. Indeed, if $(V_n)$ were such an intersection, then $V_n$ would have to be $0$-dimensional for all even $n$ and $1$-dimensional for all odd $n$, which is impossible. (It must be $1$-dimensional for odd $n$ since we have a subobject of $A$ whose $n$th term is $k$ and all other terms are $0$, and this subobject is contained in every $A_{2m}$.)

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Here's another way to get a counterexample. Again, let $k$ be your favorite field, and let $\mathtt{vect}$ be the category of finite-dimensional vector spaces over $k$. Now let $C$ be an ultrapower of the category $\mathtt{vect}$ by a nonprincipal ultrafilter on $\mathbb{N}$. Since the axioms of abelian categories are first-order, $C$ is abelian. The idea is now that $C$ consists of "hyperfinite-dimensional vector spaces". But the hypernatural numbers are not Dedekind-complete, so you can find a nested family of such "vector spaces" whose dimensions do not have an infimum, and then they cannot have an intersection in $C$.

In detail, let ${}^*\mathbb{N}$ denote the ultrapower of the natural numbers by the same ultrafilter. For any $n\in{}^*\mathbb{N}$, we have an object $k^n$ in $C$ (explicitly, if $n$ is represented by a sequence $(n_i)$, then $k^n$ is the object of $C$ represented by the sequence $(k^{n_i})$). We also have natural monic "inclusion" maps $k^m\to k^n$ whenever $m\leq n$. There is a function $\dim:Ob(\mathtt{vect})\to\mathbb{N}$ sending a vector space to its dimension, and this extends to a "dimension" function on the ultrapowers $\dim:Ob(C)\to{}^*\mathbb{N}$. For any $n\in{}^*\mathbb{N}$ we have $\dim(k^n)=n$, and if $W$ is a proper subobject of $V$ in $C$ then $\dim W<\dim V$.

Now fix a nonstandard $n\in{}^*\mathbb{N}$ and consider the subobjects $k^m$ of $k^n$ for each nonstandard $m\leq n$. Suppose these subobjects had an intersection $V$. Then $V$ would contain $k^\ell$ for every standard $\ell$, but would be contained in $k^m$ for every nonstandard $m\leq n$. This means that $\dim V$ would have to be greater than or equal to every standard natural number but less than or equal to every nonstandard natural number. This is impossible (there is no greatest standard number or least nonstandard number), so no such $V$ can exist.