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

 A: This was an error in the original book, and I added a correction to the 
errata in 2007. Homology does not commute with direct sums unless (AB4)
holds, as Sam points out.
-Chuck Weibel
A: 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!
