MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C$ be a category with filtered projective limits and coproducts (denoted by $+$). Then for all filtered projective systems $\{X_i\}$, $\{Y_i\}$ in $C$ there is a canonical morphism

$$\alpha ~:~ \lim_i X_i + \lim_i Y_i \longrightarrow \lim_i (X_i + Y_i).$$

Question. For which categories $C$ is this an isomorphism? In other words, when does the coproduct functor $+ : C \times C \to C$ preserve filtered projective limits? More specifically, let's say $C$ is algebraic, i.e. the category of algebras of some type $(\Sigma,E)$. For which $(\Sigma,E)$ is $\alpha$ an isomorphism?

Examples. $\bullet$ For the categories $(\mathrm{Set})$, $(\mathrm{Set}_*)$, $(\mathrm{Mod}(R))$, $(\mathrm{CMon})$ it is an isomorphism (direct computation). $\bullet$ For $(\mathrm{CRing})$ it is no isomorphism (take $X_n = \mathbb{Z}/p^n$ and $Y_n = \mathbb{Z}/q^n$ for primes $p \neq q$, then $\alpha$ is $\mathbb{Z}_p \otimes_{\mathbb{Z}} \mathbb{Z}_q \to 0$). $\bullet$ It also fails for $(\mathrm{Grp})$ (take $X_n = Y_n$ the free groups with generators $x_1,\dotsc,x_n$ resp. $y_1,\dotsc,y_n$ with $X_{n+1} \twoheadrightarrow X_n$ mapping $x_{n+1} \to 1$. Then $(x_1 \cdot y_1 \cdot \dotsc x_n \cdot y_n)_n \in \lim_n (X_n * Y_n)$ does not lie in the image of $\alpha$). A similar counterexample works in $(\mathrm{Mon})$.

The examples mentioned above indicate that the property is related to disjoint coproducts. If there is no general necessary/sufficient criterion available, can you give some more examples?

Added. The dual question is also interesting: Let $C$ be a category with filtered colimits and products. When is the canonical morphism $\beta : \mathrm{colim}_i (X_i \times Y_i) \to (\mathrm{colim}_i X_i) \times (\mathrm{colim}_i Y_i)$ an isomorphism? This is true for $C=(\mathrm{Set})$, and then for every algebraic category since filtered colimits and products are reflected. This property guarantees that for every topological space $X$ the associated-sheaf functor $\mathrm{PSh}(X;C) \to \mathrm{Sh}(X;C)$ preserves products and that the stalk maps $\mathrm{Sh}(X;C) \to C$, $F \mapsto F_x$ preserve products.

share|cite|improve this question
It seems like any Grothendieck abelian category satisfies this, almost by definition... So (1) Ind(C) for any small abelian category, (2) Qcoh(X), (3) Ch(A) for one of those A's. – Dylan Wilson Jan 31 '12 at 10:23
Thanks! Can you add this as an answer? I don't see why it should be true for Grothendieck categories (the axiom don't tell us anything about infinite projective limits), so it would be nice to include a proof :). – Martin Brandenburg Jan 31 '12 at 10:27
Any additive category surely satisfies this, as in an additive category finite coproducts are the same thing as finite products (and products, being a particular case of projective limits, always commute with projective limits). – Leonid Positselski Jan 31 '12 at 11:24
@Leonid: Of course! Thanks. – Martin Brandenburg Jan 31 '12 at 12:36

Your Answer


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

Browse other questions tagged or ask your own question.