Where can I find explicit formulas for the higher homotopies, which exhibit the cup product (in singular simplicial cohomology, say) as homotopy commutative on the cochain level? Same question in Cech cohomology.
[Math] Homotopy commutativity of the cup product
at.algebraic-topology
Related Solutions
The cup product in Hochschild cohomology$H^\bullet(A,A)$ is graded commutative for all unitary algebras. If $M$ is an $A$-bimodule, then the cohomology $H^\bullet(A,M)$ with values in $M$ is a symmetric graded bimodule over $H^\bullet(A,A)$.
(If $M$ itself is also an algebra such that its multiplication map $M\otimes M\to M$ is a map of $A$-bimodules, then in general $H^\bullet(A,M)$ is not commutative (for example, take $A=k$ to be the ground field, and $M$ to be an arbitrary non-commutative algebra! I do not know of a criterion for commutatitivity in this case)
These results originally appeared in [M. Gerstenhaber, The cohomology structure of an associative ring, Ann. of Math. (2) 78 (1963), 267–288.], and they are discussed at length in [Gerstenhaber, Murray; Schack, Samuel D. Algebraic cohomology and deformation theory. Deformation theory of algebras and structures and applications (Il Ciocco, 1986), 11--264, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 247, Kluwer Acad. Publ., Dordrecht, 1988.] Both references give proofs of a rather computational nature.
You can find an element-free proof of the graded commutativity in this paper, which moreover applies to the cup-products of many other cohomologies.
As for what happens with non-unital algebras, I do not know. But they are very much used today as they were before. One particular context in which they show up constantly is in the intersection of K-theory and functional analysis, where people study `algebras' which are really just ideals in rings of operators of functional spaces---one egregius example is the algebra of compact operators in a Hilbert space.
By the way, you say that group cohomology is a special case of Hochschild cohomology: it is only in a sense... There is a close relationship between the group cohomology $H^\bullet(G,\mathord-)$ of a group $G$, and the Hochschild chomology $H^\bullet(kG,\mathord-)$ of the group algebra, but they are not the same. You can use the second to compute the first (because more generally if $M$ and $N$ are $G$-modules, then $\mathord{Ext}_{kG}^\bullet(M,N)=H^\bullet(kG, hom(M,N))$, where on the right we have Hochschild cohomology of the group algebra $kG$ with values in the $kG$-bimodule $\hom(M,N)$), but the «principal» group cohomology $H^\bullet(G,k)$ is only a little part of the «principal» Hochschild cohomology $H^\bullet(kG,kG)$.
Finally, in the paper by Sletsjøe the definition for the boundary is given as you say because he only considers commutative algebras and only principal coefficients, that is $H^\bullet(A,A)$.
Cup products $\cup: H^1(X,\mathbb{Z})\times H^1(X,\mathbb{Z}) \to H^2(X,\mathbb{Z})$ are easy to visualize. Given a cellular 2-complex $X$, the 1-skeleton of $X$ is a graph $\Gamma$, and the 2-skeleton is a collection of disks with boundaries attached to the graph $\Gamma$, which (up to homotopy) are encoded by closed paths in $\Gamma$. The 1-cycles $\alpha_1,\alpha_2 \in H^1(X,\mathbb{Z})$ are encoded by maps $\alpha_i: X\to S^1=K(\mathbb{Z},1)$, and the 2-cycle $\alpha_1\cup\alpha_2$ is represented by the map $(\alpha_1\times\alpha_2)^{\ast}: H^2(S^1\times S^1)\to H^2(X)$. Choose the standard cell structure on $S^1 =c_0\cup c_1$. Subdivide $\Gamma$ by putting a vertex in the middle of each edge $e\subset \Gamma$ into intervals $e_1\cup e_2$. Then up to homotopy, we may assume that $\alpha_i$ sends $e_i$ to $c_0$ (and is extended over the 2-cells in any fashion). Then $$\alpha_1\times \alpha_2: \Gamma \to (c_0\times c_1\cup c_1\times c_0)= S^1 \vee S^1\subset S^1\times S^1,$$ in such a way that each edge $e_1$ is sent to a horizontal edge, and each edge $e_2$ is sent to a vertical edge. To figure out the degree of this map, for each 2-cell $f\subset X$, we lift $\alpha_1\times \alpha_2:\partial f\to \mathbb{R}^2$, the universal cover of $S^1\times S^1$. Then we compute the winding number of the path represented by $f$ with respect to the midpoints of the lattice $(\widetilde{S^1\vee S^1} \subset \mathbb{R}^2$ giving $\alpha_1\cup\alpha_2(f)$. For example, the closed path in this picture has winding number 7.
Best Answer
Such homotopies are given by the $\smile_i$-products. Steenrod gives explicit formulas, IIRC, in [Steenrod, N. E. Products of cocycles and extensions of mappings. Ann. of Math. (2) 48, (1947). 290--320. MR0022071], but the easiest is to prove they exist using acyclic models.
(Maybe Steenrod only deals with $\mathbb Z_2$ coefficients? I don't have access to the paper now :( )