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$\newcommand{\Mod}{\operatorname{Mod}}$

Let $A, B$ be two rings, and let $F:\Mod A\times \Mod A \to \Mod B$ be a biadditive bifunctor.

I want to extend $F$ naturally to a bifunctor from complexes over $A$ to complexes over $B$.

There are two ways I know how do to it, one is direct sum totalization, setting

$$F^\oplus (M,N)^n = \bigoplus_{p+q=n} F(M^p,N^q)$$

The other one is direct product totalization,

$$F^\prod (M,N)^n = \prod_{p+q=n} F(M^p,N^q).$$

For example, if $A$ is commutative, and $F$ is tensor product over $A$, one uses direct sum totalization to get the tensor product of complexes.

If $A$ is commutative and $F$ is the $\operatorname{Hom}_A(-,-)$ bifunctor, then one uses direct product totalization to get the Hom between two complexes.

My question - how does one knows which of these to use, in order to get the "correct" extension? why do we use direct sum for tensor and direct product for hom? and what to do in general for other biadditive bifunctors?

Michael Hardy
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the L
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  • Note the difference between $\displaystyle\oplus_{p+q=n}$ and $\displaystyle\bigoplus_{p+q=n}$. I changed it. – Michael Hardy May 01 '14 at 19:07
  • «How does one knows which of these to use?» Well: usually only one of the two options works in the context one wants to use the construction! – Mariano Suárez-Álvarez May 01 '14 at 19:09
  • So you are saying that there is no general rule to decide this? just try and see what produces reasonable results for the specific functor? – the L May 01 '14 at 19:17
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    You can view the tensor product as a colimit and the internal hom as a limit. This explains the differences. – Ma Ming May 01 '14 at 19:21
  • @MaMing, thanks, that makes sense! I am teaching this material and trying to figure out how to present this to my students. – the L May 01 '14 at 19:34
  • @Math Man, sorry this is not about your question but, where are you teaching this? – Marco Armenta May 04 '14 at 06:36

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