# Does there exist a wide but not full abelian subcategory of an abelian category?

Does there exist an example of an abelian category A and abelian subcategory B, where B is wide but not full (as a subcategory of A)?

• Your question would gain in clarity if you explained the meaning of "wide". Aug 4, 2012 at 15:37
• By wide subcategory one usually means a subcategory which contains all the objects of the original category (but not necessarily all the morphisms, as in this question). A better notion is that of essentially wide which just means that the inclusion functor is essentially surjective on objects (the subcategory contains at least one object for each isomorphism class). Aug 4, 2012 at 17:35

Take $\textrm{Mod-}A$, where $A$ is a finite dimensional local $k$-algebra ($k$ a field). Let $\mathcal C$ be the category whose objects are the same as those of $\textrm{Mod-}A$, but whose homomorphisms are replaced by all $k$-vector space homomorphsims, i. e. $\textrm{Hom}_{\mathcal C}(X,Y) := \textrm{Hom}_k(X,Y)$. By construction, $\textrm{Mod-}A$ is a wide subcategory of $A$, and clearly it isn't full unless $A$ is equal to $k$. Moreover, $\mathcal C$ is equivalent to the category of vector spaces over $k$ (since every $k$-vector space can be given the structure of a $k$-module by letting $A$ act via $A/\textrm{Rad}(A)\cong k$). Hence $\mathcal C$ is abelian. Even kernels and cokernels in $\textrm{Mod-}A$ coincide with those in $\mathcal C$.