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Let $ \text{Vect} $ be the category of finite dimensional vector spaces over an algebraically closed field. The idempotent completion of a $\text{Vect}$-category $ \mathcal{C} $ may be though of in two different ways:

  • As a category with objects $ (X,e_X) $ where $ X $ is in $ \mathcal{C} $ and $ e_X $ is an idempotent in $ \text{End}(X) $.
  • As the the additive subcategory of $ \text{Fun}(\mathcal{C}^{\text{op}}, \text{Vect}) $ (linear functors from $\mathcal{C}^{\text{op}} $ to $ \text{Vect}$) generated by direct summands of $ \mathbb{Y}(X) $ for $ X $ in $ \mathcal{C} $ (where $ \mathbb{Y} $ is the Yoneda embedding).

To an object-idempotent pair $ (X,e_X) $ one can associate the (contravariant) functor

$$ \begin{align} (X,e_X)^\sharp \colon \mathcal{C} &\to \text{Vect} \\ Y &\mapsto \{ f \in \text{Hom}(Y,X) \mid e_X \circ f = f \} \end{align} $$

which provides an equivalence between these two categories.

Question

From the above discussion we see that the following are equivalent

  • $ \mathbb{Y}\colon \mathcal{C} \to \text{Fun}(\mathcal{C}^{\text{op}}, \text{Vect}) $ is an idempotent completion.
  • Every contravariant functor from $\mathcal{C}$ to $\text{Vect}$ is naturally isomorphic to $ (X,e_X)^\sharp $ for some $X$ in $\mathcal{C}$ and some idempotent $e_X$ in $\text{End}(\mathcal{C}) $.

Under what assumption on $ \mathcal{C} $ do these statements hold? For example, they hold if $ \mathcal{C} $ is semisimple with finitely many simple objects. I would like to know if they still hold without the assumption that $ \mathcal{C} $ has finitely many simple objects. I would also like to know if they hold when $\text{Fun}(\mathcal{C}^{\text{op}}, \text{Vect})$ is semisimple (even if $ \mathcal{C} $ isn't).

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    $\begingroup$ This should almost never hold, and it certainly does not suffice to assume semisimplicity. Are you sure this is the question you meant to ask? $\endgroup$ Commented Mar 27, 2019 at 19:08
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    $\begingroup$ Explicitly, if $C = \text{Vect}$, consider the presheaf $V \mapsto V^{\ast} \otimes W$ for infinite-dimensional $W$. This corresponds to linear maps $V \to W$ of finite rank, and it is not a direct summand of a representable presheaf, e.g. because it does not send colimits to limits. So semisimplicity does not suffice. It's also easy to construct counterexamples if you don't require your presheaves to be linear, which you don't specify in your question. $\endgroup$ Commented Mar 27, 2019 at 20:24
  • $\begingroup$ The question was badly formulated, I have edited it appropriately. $\endgroup$
    – Arthur
    Commented Mar 27, 2019 at 23:30

1 Answer 1

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Edit: This answer currently addresses a previous version of the question; here $\text{Vect}$ denotes the category of all vector spaces.

We can replace $C$ with its idempotent completion WLOG, so the question becomes: for $C$ an idempotent complete $k$-linear category, when is every ($k$-linear) presheaf $C^{op} \to \text{Vect}$ representable?

The answer is: iff $C$ is the zero category, by which I mean the $k$-linear category with only the zero object, or the empty category. (I'm a little confused as to whether the empty category should be regarded as being idempotent complete.)

Suppose $C$ is neither of these, so it has at least one nonzero object $c$. Then the representable presheaf $\text{Hom}(-, c) : C^{op} \to \text{Vect}$ takes at least one nonzero value. Now consider the presheaf $F(-) = \text{Hom}(-, c) \otimes W$ for an infinite-dimensional vector space $W$. We want to show that $F$ is not representable. It will suffice to show that $\text{Hom}(F, -)$ does not preserve colimits.

$F$ itself is the filtered colimit of the presheaves $\text{Hom}(-, c) \otimes V$ as $V$ runs over all finite-dimensional subspaces of $W$, and if $\text{Hom}(F, -)$ preserved this filtered colimit then every natural transformation $F \to F$ would factor through $\text{Hom}(-, c) \otimes V$ for some finite-dimensional $V$. But the identity natural transformation does not factor in this way, as we can see by plugging in $(-) = c$ (since $c$ is nonzero, $\text{End}(c)$ is also nonzero).

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    $\begingroup$ Sorry, I should have specified that $ \text{Fun}(\mathcal{C}^{\text{op}}, \text{Vect}) $ is the linear functors and that $ \text{Vect} $ is the category of finite dimensional vector spaces and that $ \mathcal{C} $ is only enriched over $ \text{Vect} $. I have edited the question appropriately. $\endgroup$
    – Arthur
    Commented Mar 27, 2019 at 22:52
  • $\begingroup$ To see that in this case semisimplicity implies the desired property consider a functor $ F $ in $ \text{Fun}(\mathcal{C}^{\text{op}}, \text{Vect}) $. The map $$ F(X) = \bigoplus\limits_S \text{Hom}(S,X)^* \otimes F(S) = \bigoplus\limits_S \text{Hom}(X,S)\otimes \text{Hom}(\mathbb{Y}(S),F)\\ = \bigoplus\limits_S \text{Hom}(\mathbb{Y}(S),F) \otimes \mathbb{Y}(S)(X) $$ identifies $ F $ with an object in the additive subcategory generated by direct summands of $ \mathbb{Y}(X) $. $\endgroup$
    – Arthur
    Commented Mar 27, 2019 at 23:05
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    $\begingroup$ @Arthur Doesn’t your argument require that $\mathcal{C}$ has only finitely many isomorphism classes of simple objects? $\endgroup$ Commented Mar 28, 2019 at 11:29

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