Let $C$ be a category with a zero object $0$, small products, and small coproducts. Let $(A_i)_{i \in I}$ be a (possibly infinite) list of objects. There is a canonical map $\amalg_{i \in I} A_i \to \prod_{i \in I} A_i$. If this map is an isomorphism, we denote the product / coproduct by $\oplus_{i \in I} A_i$ and call it the biproduct of the $A_i$. If the $A_i$'s are all the same, call this the $I$-fold bipower $A^{(I)}$ of $A$. When $I$ is finite, I think I have a pretty good handle on $I$-indexed biproducts. But when $I$ is infinite, I get more confused. Here are some things I know or think I know about this situation:

  1. Some categories have all small biproducts. The main examples I'm aware of are the category of suplattices and the category of locally presentable categories (and left adjoint functors). These can be modified to get examples of categories which have $\kappa$-small biproducts but no larger.

  2. Certain infinite biproducts may exist in an additive category. For example, in $Ab^{\mathbb N}$ (where $\mathbb N$ is regarded as a set), the constant functor at $\mathbb Z$ is the biproduct of the representables.

  3. On the other hand, I thought I knew an argument at some point that in an additive category, any infinite bipower of an object must be zero. I can't seem to reproduce the argument, but I think it was some sort of Eilenberg swindle sort of thing...

  4. More generally, I want to say that if an infinite biproduct $\oplus_{i \in I} A_i$ exists in an additive category, then we must be "close" to the situation of (2) above, where $Hom(A_i,A_j) = 0$ for $i \neq j$. I don't know how to formulate this, though, if it's correct in any sense at all.

So I suppose my questions are the following:

Question 1: Is it true that in an additive category, any infinite bipower which exists must be zero?

Question 2: In an additive category, if an infinite biproduct $\oplus_{i \in I} A_i$ exists, is there sense in which it must be "close" to the setting of (2) above? For example, must it be the case that for each $i \in I$, $Hom(A_i,A_j) = 0$ for all but finitely many $j \in I$?

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    $\begingroup$ An example where 4 doesn’t hold: the wedge of all negative suspensions of the Eilenberg Mac Lane spectrum of the integers mod 2. However if you just require i<j then it does hold. $\endgroup$ Aug 15 at 19:34
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    $\begingroup$ As for 3, you can define the infinite sum of the identity map, so you can apply an Eilenberg swindle argument to show that it is zero. $\endgroup$ Aug 15 at 19:54
  • $\begingroup$ @FernandoMuro Thanks, I definitely should have thought of the Eilenberg-MacLane example! $\endgroup$
    – Tim Campion
    Aug 15 at 23:47

3 Answers 3


I think your memory is right on Question 1: an Eilenberg swindle implies that if an infinite bipower $\bigoplus_X A$ of an object $A$ exists in an additive category, then $A=0$. The point is that just as addition of morphisms can be defined using finite bipowers, making any category with finite bipowers semiadditive, using infinite bipowers you can define a "sum" of infinitely many morphisms. But if you can also subtract morphisms, then for any $f:A\to A$ we have

$$f = f + \left(\sum_X f - \sum_X f\right) = \left(f + \sum_X f\right) - \sum_X f = \sum_X f - \sum_X f = 0$$

The crucial step $f + \sum_X f = \sum_X f$ follows from the fact that $1+X \cong X$, since $X$ is infinite, and the "associativity" of these sums induced by bipowers. Applying this to $f = 1_A$ we find $A=0$.


For Question 2, here is an example where $Hom(A_i,A_j) \neq 0$ for all $i,j$. So at least the kind of constraints we want have to be in terms of something more subtle.

Take $C$ to be $Ab^\mathbb{N}$, and define $A_i$ by $A_i(j) = \mathbb{Z}$ if the binary expansion of $j$ has a $1$ at the $i$-th place, and $A_i(j)=0$ otherwise.

Then for each $i$,$i'$ there is are many $j$ that have a $1$ both at the $i$ and $i'$ position, which will give you a lot of non-zero maps between $A_i$ and $A_{i'}$, but for each $j$ there is only a finite number of $i$ such that $A_i(j) \neq 0$ and hence the infinite products and coproducts coincide.

  • $\begingroup$ Thanks, this is a great example! It suggests to me that maybe some of the subtleties are set-theoretical in nature... $\endgroup$
    – Tim Campion
    Aug 15 at 23:47

Here's another example for Question 2 that I encountered in nature.

In the derived category of modules for a ring, pick one module $M_i$ for each $i\in\mathbb{Z}$, and let $A_i=M_i[i]$. Then the natural map $$\coprod_{i\in\mathbb{Z}} A_i\to\prod_{i\in\mathbb{Z}} A_i$$ is an isomorphism, since in each degree only finitely many (in fact, only one) of the $A_i$ have nonzero homology, so the map is an isomorphism on homology. So there is a biproduct $\bigoplus_{i\in\mathbb{Z}} A_i$.

But $\text{Hom}(A_i,A_j)=\text{Ext}^{j-i}(M_i,M_j)$, which may be nonzero for all $j\geq i$.


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