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A branch of algebraic topology concerning the study of cocycles and coboundaries. It is in some sense a dual theory to homology theory. This tag can be further specialized by using it in conjunction with the tags group-cohomology, etale-cohomology, sheaf-cohomology, galois-cohomology, lie-algebra-cohomology, motivic-cohomology, equivariant-cohomology, ...

13 votes

Should cohomology of $\mathbb{C} P^\infty$ be a polynomial ring or a power series ring?

This has the advantage that the formula works for all complex-oriented cohomology theories, e.g. for complex K-theory: $$KU^*(\mathbb{CP}^\infty)=\lim_n KU^*[x]/x^n$$ in which case it does not coincide … quick comment about one of your addenda: you can always think of homotopy types as the ind-category of finite homotopy types (precisely, this is true at the level of the ∞-category of spaces), so every cohomology
Denis Nardin's user avatar
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11 votes
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Roadmap for L-Theory

I apologize for the self promotion -- I hope the content of this answer can be useful anyway... My favourite introduction to L-theory is Lurie's notes on Algebraic L-theory and surgery (warning: aggr …
Denis Nardin's user avatar
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11 votes
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Understanding Homology Operations and how to compute them

Basically, when you restrict to finite spectra homology and cohomology are pretty much the same thing: you just need a Spanier-Whitehead dual to pass from one to the other. … They are also the "dual" of the cohomology operations. …
Denis Nardin's user avatar
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8 votes
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A "non-abelian excision" statement for mapping out of a space

It depends on what exactly you mean by "subspace" and "fiber". Let me put some theorems down for you: Theorem: Let $U,V\subseteq X$ open subspaces. Then the following is a homotopy pushout square: $ …
Denis Nardin's user avatar
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5 votes
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The contravariant mapping space represented by a homotopical classifying space (e.g. BG)

Let $G$ be a topological group and $X$ be a paracompact Hausdorff topological space. For simplicity let us assume that $G$ has the homotopy type of a CW complex, although a lot of this answer does not …
Denis Nardin's user avatar
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