Let $\mathcal{D}=Mod-A$ be a category of module over an Algebra $A$. and $\mathcal{C}$ be a subcategory of $\mathcal{D}$, I have the following questions:-

  1. Is $C$ complete category? Is $\mathcal{C}$ is small?( I know $\mathcal{D}$ is complete).
  2. If we have inclusion functor $F:\mathcal{C} \rightarrow \mathcal{D}$ satisfies the following SSC(Solution Set Condition). For each object $ D \in Ob(D)$ there is a set $I$ and an $I$-indexed family of arrows $g_{i} : F(\mathcal{C}_{i}) \rightarrow {D}$ such that every arrow $h : F(\mathcal{C}) \rightarrow \mathcal{D} $ can be written as a composite $h = g_{i} \circ F(t)$ for some index $i$ and some $t : \mathcal{C} \rightarrow {C_{i}}$.

Update:- I am more interested the the case where $A$ is infinite-dimension algebra over a field $K$.

  • 1
    $\begingroup$ For 1, it's clearly hopeless if you don't impose conditions on $\mathcal{C}$. $\endgroup$ – Captain Lama Apr 23 '16 at 17:13
  • $\begingroup$ I have only condition on the Algebra, which is infinite-dimensional. $\endgroup$ – henry Apr 23 '16 at 17:15
  • $\begingroup$ What is exactly the question in 2? Are you asking if the inclusion functor verifies the solution set condition? $\endgroup$ – Giorgio Mossa Apr 23 '16 at 17:29
  • $\begingroup$ @GiorgioMossa Yes. this is what I want to say. $\endgroup$ – henry Apr 23 '16 at 17:30

Here the answers to your questions:

  1. Clearly not. If you take a generic sub-category of $\mathcal D$ there is no reason why it should be complete: for instance you could take the full sub-category of $\mathcal D=Mod-\mathbb Z$ spanned by the free $\mathbb Z$-modules (that arbitrary direct sums of copies of $\mathbb Z$). This category is clearly not small, for each cardinal $k$ we have a module $\mathbb Z^{\oplus k}$, and it is not complete because no infinite product is a free-module (this is a well known fact of module theory, although I have to admit I have no proof of that at the moment).

  2. The answer is again no. A counter-example is the following: consider $\mathcal D$ a category of modules and let $\mathcal C$ be the discrete category (every morphism is an identity) having all $A$-modules as objects, the inclusion functor $F \colon \mathcal C \to \mathcal D$ does not satisfy the small set condition.

Hope this helps.

  • $\begingroup$ Thanks this will help. But I just now I forget to add a condition on $A$ in which is infinite-dimensional algebra over a field $K$. are still the answer is no? Thanks $\endgroup$ – henry Apr 24 '16 at 7:12
  • $\begingroup$ @henry, if you read Giorgio's answr you will see that nothing will change his conclusions... $\endgroup$ – Mariano Suárez-Álvarez Apr 24 '16 at 7:51
  • $\begingroup$ @henry answer 2 applies to every category of modules. About answer 1 it seems that the same argument could be applied to the case of $\mathbb Q[x]$-modules (direct product of infinite free modules, copies of $\mathbb Q[x]$, need not to be free). Note that $\mathbb Q[x]$ is an infinite dimensional $\mathbb Q$-algebra. $\endgroup$ – Giorgio Mossa Apr 24 '16 at 10:55

There are obvious non-small and non-complete subcategories of $\mathcal D$, e.g. $\mathcal{D}$ itself and the full subcategory spanned by two object $a,b$ and morphisms between them. The solution set condition is also not automatically satisfied. If $\mathcal C$ is a large discrete subcategory, then no map $F(C)\to D$ can factor nontrivially through any $F(C')$, whereas $D$ admits a large class of such maps, namely the zero maps from each $F(C)$.


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