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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.

3 votes
2 answers
373 views

Is there an elementary reason that this colocalisation map of complexes is a quasi-isomorphism?

A fact about triangulated categories is that (exact) localisation functors and so-called colocalisation functors come in pairs, making an exact localisation triangle. I've tried to come up with less t …
Justin Bloom's user avatar
10 votes
Accepted

Factoring through projective modules is an equivalence relation

This equivalence relation is just a quotient of abelian groups. It will be cleaner to show that the $\mathrm{PHom}_R(M, N)$ is a subgroup of $\mathrm{Hom}_R(M,N).$ Indeed, if $f:M\to P_1 \to N$, $g: …
Justin Bloom's user avatar
4 votes

Extensions of $G$-modules parametrized by $H^1$

No. If $G$ is the trivial group and $q =p$ is a prime, $\operatorname{Ext}^1_G(V, W) = \operatorname{Ext}^1_{\mathbb{Z}}(\mathbb{Z}/ p, \mathbb{Z} / p) \cong \mathbb{Z}/p$ but $H^1(G, V^\vee\otimes_\m …
Justin Bloom's user avatar
4 votes

The Krull dimension of the tensor product of rings

You are looking at infinite tensor products so we really shouldn't expect the infinite product to have finite dimension in most cases. Your second example was simply an infinite tensor product of $\ma …
Justin Bloom's user avatar