# Categories with canonical factorizations into products satisfying two particular properties

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

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.