Suppose that A is an abelian category with Ab4*, i.e., products exists and are exact. Then a product of quasi-isomorphisms is a quasi-isomorphism and we can define products in D(A) just by taking the product of underlying complexes. If A has just Ab3* (i.e., products exist) then this doesn’t work.<\/p>\n
Let A be a Grothendieck abelian category. Then A has Ab3* (this does not follow directly from the definitions, but rather is an example of what Akhil was referring to here<\/a>). In a nice short paper entitled Resolution of unbounded complexes in Grothendieck categories<\/a>, C. Serp\u00e9 shows that the category of unbounded complexes over A has enough K-injectives. There are other references; I like this one because its proof is a modification of Spaltenstein’s argument in his famous paper Resolutions of unbounded complexes<\/a>. Combining these results we can show products exist in D(A).<\/p>\n In fact, I claim that products exist in D(A) if A has Ab3* and enough K-injective complexes. Namely, suppose that we have a collection of complexes K^*_\u03bb in A parametrized by a set \u039b. Choose quasi-isomorphisms K^*_\u03bb —> I^*_\u03bb into K-injective complexes I^*_\u03bb and consider the termwise product<\/p>\n \u03a0_{\u03bb \u2208 \u039b} I^*_\u03bb<\/p><\/blockquote>\n I claim this is a product of the objects K^*_\u03bb in D(A). Namely, it is a result in the Spaltenstein paper that the product of K-injective complexes is K-injective. Hence to check our assertion we need only check this on the level of maps up to homotopy, where it is clear.<\/p>\n OK, now what I want to know is this: Let A be a Grothendieck abelian category and let B ⊂ A be a subcategory such that D_B(A) makes sense. When does D_B(A) have products? Are there some reasonable assumptions we can make to guarantee this?<\/p>\n","protected":false},"excerpt":{"rendered":" Suppose that A is an abelian category with Ab4*, i.e., products exists and are exact. Then a product of quasi-isomorphisms is a quasi-isomorphism and we can define products in D(A) just by taking the product of underlying complexes. If A … Continue reading