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Oct
10
comment The Verdier Quotient
Dear Prof. Rickard, Thanks for your answers. I was wondering then, what is the reason that they ask for strictness in the definition of triangulated subcategory?
Oct
10
accepted The Verdier Quotient
Oct
10
awarded  Yearling
Oct
10
accepted Free objects in the category of dg modules
Oct
10
asked The Verdier Quotient
Sep
19
comment Free objects in the category of dg modules
@MartinBrandenburg Im not so strong with monoidal category theory. But I think the differentials mess things up no? I can't see how you you would have the universal property. Do you have a reference where I can read about it? I am completely on board in the case of graded modules.
Sep
19
revised Free objects in the category of dg modules
added 14 characters in body
Sep
19
asked Free objects in the category of dg modules
Aug
30
comment Universal property of tensor product of dg algebras
Thanks Martin! Sorry it took me so long to accept.
Aug
30
accepted Universal property of tensor product of dg algebras
Jul
25
asked Universal property of tensor product of dg algebras
Jul
2
awarded  Curious
May
23
comment RHom and Koszul complexes.
Hey Aaron, Thanks for you comments and sorry it took my so long to respond. My problem was that actually all the hom's where supposed to be the graded category anyway, I was confusing myself.
May
15
comment RHom and Koszul complexes.
Yes I want to believe this very badly, but I dont see why $Hom^j_A (P_\bullet, A_0\langle -i \rangle [i])$ for $j \neq 0$ is zero. It should just be $hom_A(P_{j+i}, A_0 \langle -j -i \rangle )$ as you point out, no?
May
15
comment RHom and Koszul complexes.
I am sorry for being so slow Aaron, really! So are you saying that the complex I should end up with should in homological degree zero be $ \prod_{i \in \mathbb{Z}} hom_A ( P_i , \langle -i \rangle ) $ and zeros in all other degrees?
May
14
comment RHom and Koszul complexes.
Yes, so for example its true that $ Hom_A ( P_j, A_0 \langle -i \rangle ) = hom_A ( P_j , A_0 \langle -j \rangle )$ where the lower case is means we take only morphisms of degree zero, right? But how does this help, I feel like its telling me that the differentials are zero or something like that.
May
14
asked RHom and Koszul complexes.
Jan
11
awarded  Commentator
Jan
11
comment Definition for a bar resolution for a module over a dg category
@Adeel, I'd buy that. Do you know of a reference just to be sure?
Jan
10
asked Definition for a bar resolution for a module over a dg category