In an arbitrary abelian category, does chain complex homology commute with coproduct?

2011-09-18T22:53:22Z

On page 55 of Weibel's Introduction to homological algebra the following passage appears:

Here are two consequences that use the fact that homology commutes with arbitrary direct sums of chain complexes

I understand why homology commutes with arbitrary direct sums when the direct sum of a collection of monics is a monic (i.e the direct sum functor is exact) but I was under the impression that there were abelian categories where the direct sum functor is not exact. After a bit of thought, I realised that I don't know an example of an abelian category in which the coproduct functor is not exact.

Sheaves of abelian groups on a fixed topological space give an example of an abelian category in which the product functor is not exact.

Question 1: Is the passage from Weibel's book correct? If so, then why?

Question 2: Is there an example of an abelian category where the direct sum functor is not exact?

Answer by Sam Gunningham for In an arbitrary abelian category, does chain complex homology commute with coproduct?

2011-09-19T01:41:44Z

I couldn't think of a natural example of an abelian category in which direct sums are not exact (I think this is called axiom AB4). For example, sheaves of abelian groups and R-modules both have this property. However there are natural examples of abelian categories where direct products are not exact (i.e. not satisfying AB4*), for example, the category of abelian sheaves on a space.

Taking the opposite category of such a category will then give an example of a category not satisfying AB4 (albeit, not a very nice one).

Once you have such an example, homology of chain complexes in this category will not commute with direct sum:

if $A_i \to B_i$ is a sequence of monos such that $\bigoplus (f_i :A_i \to B_i)$ is not a mono, then consider the sequence of two-term complexes

$A_i \to B_i$.

$H^0$ of each of these complexes is zero, but $H^0$ of the direct sum is the kernel of $\bigoplus f_i$.

Here is one way to see that Sh(X) does not satisfy AB4* (probably not the easiest). Assume for simplicity X = [0,1]. Take a finite open cover, $\mathcal U_i$ of X by balls of radius $1/i$. Let $A_i$ be the sheaf

$\prod _{U \in \mathcal U_i} j_{U!} \mathbb Z_U$.

This has an epimorphism to $\mathbb Z_X$, but the direct product of all of them together is not epimorphic: taking sections over any open set $V$ will kill off any $A_i$ when no $1/i$-ball contains $V$.

I hope this is correct!

In an arbitrary abelian category, does chain complex homology commute with coproduct? - MathOverflow

In an arbitrary abelian category, does chain complex homology commute with coproduct?

Answer by Sam Gunningham for In an arbitrary abelian category, does chain complex homology commute with coproduct?