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An old splitting theorem for (Hausdorff) locally compact abelian (LCA) groups says that any LCA group $L$ is isomorphic to a direct product of $\mathbb{R}^n$ and $L_1$, where $L_1$ contains a compact-open subgroup (this factorization is unique up to isomorphism of $L_1$).

Taking $L = L_1\times\mathbb{R}^n$ as above, we have the following properties:

(i) There is no embedding of $\mathbb{R}^n$, or any closed subgroup of $\mathbb{R}^n$, into $L_1$.

(ii) If $f:\mathbb{R}^n\to L_1$ and $g:L_1\to\mathbb{R}$ are continuous homomorphisms, then $fg$ is trivial.

Does anyone happen to know if the following generalization has been studied, or if there are some other known examples:

An additive category $\mathcal{C}$ such that every object $A$ can be expressed as a product $A = B\times C$ where:

(i') There is no embedding of any sub-object of $C$ into $B$.

(ii') If $f:B\to C$ and $g:C\to B$ are morphisms, then $fg$ is trivial (maps onto the zero-object).

EDIT: Reworded in terms of additive categories, not abelian.

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This might be stupid, but why can't you just take C to be a product of two pointed categories, say $B\times D$. Then every element $\alpha \in C$ is $(\beta,0)\times (0,\delta)$ where $\beta \in B$ and $\delta \in D$. There is no a priori connection between B and D, so (i') is satisfied. (ii) is trivial because each component goes through the zero object –  David White Sep 7 '11 at 20:02
    
Sorry for the deleted comments. For some reason the software doesn't like the star character in math mode, so I wrote 0 instead for the zero-object (which is both initial and terminal because the categories are pointed) –  David White Sep 7 '11 at 20:05
    
@David - you can use \ast instead of *. –  David Roberts Sep 7 '11 at 22:04
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1 Answer

First, you should make clear whether LCA groups are assumed to be Hausdorff or not.

Next, whichever way you answer the previous question, the category of LCA groups will not be an abelian category. Indeed, it is a standard lemma that in an abelian category, every morphism that is both a monomorphism and an epimorphism has an inverse. In LCA the identity map $$(\mathbf{R},\text{discrete topology}) \to (\mathbf{R},\text{usual metric topology})$$ is both mono and epi, but there is no inverse.

If you want to work with abelian categories, then the theory of torsion theories (see http://ncatlab.org/nlab/show/torsion+theory for example) does more or less what you ask for. However, this theory would need to be extended before it could cover the nonabelian category of LCAs.

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Thanks Neil. I suppose what I meant was an additive category. LCA does form such a category, does it not? If so, I'll go back and re-edit my question appropriately –  Iian Smythe Sep 8 '11 at 14:58
    
Yes, LCA is additive but not abelian. –  Neil Strickland Sep 8 '11 at 15:02
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