Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Notation. Let $U_0 \in U_1 \in U_2$ be Grothendieck universes, each containing $\mathbb N$.

Let $\mathbf{Cat}_{U_0}$ be the ($U_1$-small) 2-category of all $U_0$-small categories, $\mathbf{Cat}_{U_1}$ be the ($U_2$-small) (2-)category of all $U_1$-small categories, $\mathbf{2Cat}$ the ($U_2$-small) 3-category of all $U_1$-small 2-categories and $\mathbb{AB}$ (or, if you prefer, $\mathbf{AbCat}$) the ($U_1$-small) 2-category of all abelian categories of $\mathbf{Cat}_{U_0}$ and additive functors. $\mathbb{AB}$ is clearly an object of $\mathbf{Cat}_{U_1}$

Question 1. Is $\mathbb{AB}$ an additive (or even abelian) category of $\mathbf{Cat}_{U_1}$?

Question 2. Is $\mathbb{AB}$ $U_0$-small, thus an object of $\mathbf{Cat}_{U_0}$?

References are wellcome!

Personal thoughts. Zero-object. Since $\mathbf{0}$ (an initial object in $\mathbf{Cat}_{U_0}$) is not abelian (it has no zero-object) and additive functor preserves zero-objects, then $\mathbf 1$ (the terminal object in $\mathbf{Cat}_{U_0}$) is a candidate for a zero-object if $\mathbb{AB}$.

Biproduct. $\mathbf{Cat}_{U_0}$ is bicomplete. I suppose the product of categories to be the candidate biproduct for $\mathbb{AB}$.

Pre-additivity. Let $F, G\colon\mathcal{A\to B}$ be two 1-morphisms in $\mathbb{AB}$ and $f\colon A_1 \to A_2$ a morphism in $\mathcal A$. Clearly, $F(f)\oplus_{\mathcal B} G(f)$ could not be functorial (it is not well-defined!). A strategy that comes to mind is factor through the skeleton. I think I have proved that the skeleton $\text{sk}(\mathcal B)$ of an abelian category $\mathcal B$ is abelian itself.

Question 3. Let $\mathcal C$ be an abelian category. Is any category equivalent to $\mathcal C$ abelian? Are the functors realizing the equivalence necessary additive?

We could try to define first an additive sum-functor $\text{sk}(F+G)\colon \mathcal{A} \to \text{sk}(\mathcal B)$, then compose with an embedding $\mathcal{E}\colon \text{sk}(\mathcal B) \to \mathcal B$ and finally define $F+G$ as $\mathcal{E}\circ \text{sk}(F+G)$.

Unfortunately, I can see a bug. Suppose $f\colon A_1 \to A_2$ is actually an isomorphism in $\mathcal A$ and $A_1 \neq A_2$. We should have an abelian group $(\text{Hom}_{\mathcal A}(A, B), +_{A, B}, -, N)$, where $N\colon\mathcal{A\to B}$ is the neutral functor with respect to the sum $+_{A, B}$. The procedure above would give $F(A) = (F+_{A,B}N)(A)=(F+_{A,B}N)(B) = F(B)$, absurd. Thus, it does not work; unless one gives a different, particular, definition for the sum with a neutral functor.


Edit (see comments below). Difference between additive functors seems to be a key problem.

Question 4. Do you see a good way to enrich $\mathbb{AB}$ over a monoidal category $(\mathbf{Mon}, ?, \mathbb{N}, \alpha, \lambda, \rho)$, where $\mathbf{Mon}$, object of $\mathbf{Cat}_{U_0}$, is the category of monoids?

share|improve this question
1  
You should define morphisms in $\mathbb{A}\mathbb{B}$ (exact functors?) and say what it means for a $2$-category to be additive/abelian. Q1: No, what is the negative of a functor? Q2: Yes, this is well-known. The point is that the sum of two morphisms is uniquely determined by the coproducts and products, which depend only on the category. See the canonical references (Freyd's book on abelian categories, or nlab). –  Martin Brandenburg Jan 31 '13 at 22:00
    
Thank you, Martin! I've edited the OP and added a question. (Thus, Martin Q2 actually now refers to Q3). Well, if $F\colon \mathcal {A\to B}$ is a 1-morphism in $\mathbb{AB}$, for every $f\colon A_1\to A_2$, $-F(f)$ exists in $\mathcal B$ and this assignment seems to be functorial. Even if it does not match with the sum I've thought... –  Andrea Gagna Jan 31 '13 at 22:19
    
I was evidently wrong. $f\mapsto -F(f)$ is not functorial at all. Indeed, for every composable morphisms $(f, g)$ of $\mathcal A$, $-(F(g)\circ F(f)) = (-F(g))\circ F(f) = F(g)\circ (-F(f)$. Sorry. –  Andrea Gagna Jan 31 '13 at 22:46
1  
And the identity is not preserved. –  Martin Brandenburg Feb 1 '13 at 0:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.