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(Co)chain complexes, abelian Categories, (pre)sheaves, (co)homology in various (possibly highly generalized) settings, spectra, derived functors, resolutions, spectral sequences, homotopy categories. Chain complexes in an abelian category form the heart of homological algebra.

6 votes

Heuristic behind $A_{\infty}$ - algebras

There is one aspect of $A_\infty$-algebras I haven't seen mentioned yet: Massey products and related structures. Suppose you have a differential graded algebra $(A,d)$: a chain complex with a product …
John Palmieri's user avatar
1 vote
Accepted

Boundary operator in the colimit of complexes.

For each integer $n$, the $n$th term $C[n]$ in the colimit complex $C$ is the colimit of the $n$th terms $C_i[n]$. For each $i \in \mathcal{I}$ and each integer $n$, the $n$th boundary map in $C_i$ i …
John Palmieri's user avatar
2 votes
Accepted

Why do we use the diagonal for diagonal approximations ?

The diagonal map $\Delta$ is "coassociative": the two maps $(\Delta \otimes 1) \circ \Delta$ and $(1 \otimes \Delta) \circ \Delta$ from $\mathbb{Z}G$ to $(\mathbb{Z}G)^{\otimes 3}$ are equal. Therefor …
John Palmieri's user avatar
2 votes

$\inf\{i\in \mathbb N \cup \{0\}\cup\infty\mid Ext^i_R(R/I,R)\neq 0\}=0 ?$

I think the answers to both questions are yes. Put a grading on $R$ so that it is connected (zero in negative degrees, $k$ in degree 0): for example, put $x_n$ in degree $n$. (It also seems safest to …
John Palmieri's user avatar
3 votes
0 answers
39 views

Filtering a pre-Koszul algebra to get a homogeneous associated graded algebra

In Priddy's paper "Koszul resolutions", on p. 42 he defines an algebra $A$ to be pre-Koszul if it can be written as a quotient of a free algebra $F = F\langle x_i \rangle$ with generators $\{x_i\}$ by …
John Palmieri's user avatar
4 votes
Accepted

Adams spectral sequence and short exact sequences. Some clarifications

The red dot in (3,0) comes from a map $\Sigma^3 D \to \mathbb{F}_2$, and this map is the image of a map $\mathbb{R}P^\infty \to \mathbb{F}_2$, so it goes to zero under the coboundary map. This agrees …
John Palmieri's user avatar
15 votes
Accepted

Is homology finitely generated as an algebra?

Another counterexample: let $A$ be the algebra $\mathbb{Q}[y,z]/(y^2) \otimes \bigwedge(x)$ with $x$ in degree 1, $y$ and $z$ in degree 2. Put a differential on this by $z \mapsto xy$. This is a commu …
John Palmieri's user avatar