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)?

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    $\begingroup$ Your question would gain in clarity if you explained the meaning of "wide". $\endgroup$ Aug 4, 2012 at 15:37
  • $\begingroup$ 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). $\endgroup$ Aug 4, 2012 at 17:35

1 Answer 1


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 $\mathcal C$.


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