Bounds on homological dimension of functor categories

Question

Let $A$ be a Grothendieck abelian category. I will say that $A$ is of global dimension less or equal to $n$ if $Ext^{k}_{A}(a, b) = 0$ for $k > n$ and all $a, b \in A$. This is equivalent to saying that any object of $A$ admits an injective resolution of length at most $n$.

Let $I$ be a small diagram category, so that the category of functors $Fun(I, A)$ is again Grothendieck. Are there any general bounds on the dimension of this functor category in terms of the dimension of $A$ and some invariant of the diagram category?

Note that I do not expect this dimension would be finite for arbitrary $I$, I am rather looking for examples of "nice" $I$ for which we have such a bound for arbitrary Grothendieck category. In the simplest possible case, I would like to know the answer even if $A$ is a category of modules over a ring.

Answer 1

Claim: If the simplicial nerve of $I$ has dimension $m$ and $A$ has global dimension $n$, then $\operatorname{Fun}(I, A)$ has global dimension (at most) $m+n$.

To prove it, you can use the standard simplicial resolution of a functor $F\colon I \to A$ by representable functors.

$$F(-)\leftarrow \bigoplus_{c_0\in Ob(I)} F(c_0)\times \operatorname{mor}_I(c_0,-) \Leftarrow \bigoplus_{c_0\to c_1} F(c_0)\times \operatorname{mor}_I(c_1,-)\cdots$$

Taking $\operatorname{nat}(-, G)$ we obtain a cosimplicial resolution of $\operatorname{nat}(F, G)$. The cosimplicial object has dimension $m$, and each term has homological dimension $n$. It follows that $\operatorname{Ext}^k(F, G)=0$ for $k>m+n$.