Let $X$ be a topological space. In elementary algebraic topology, the cup product $\phi \cup \psi$ of cochains $\phi \in H^p(X), \psi \in H^q(X)$ is defined on a chain $\sigma \in C_{p+q}(X)$ by $(\phi \circ \psi)(\sigma) = \phi(_p\sigma)\psi(\sigma_q)$ where $p_\sigma$ and $\sigma_q$ denote the restriction of $\sigma$ to the front $p$-face and the back $q$-face, respectively. (More generally, any diagonal approximation $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ could be used; this is the Alexander-Whitney one.) The cup product defined by the Alexander-Whitney diagonal approximation as above is associative for cochains but skew-commutative only up to homotopy (this results from the fact that the two diagonal approximations $C_{\ast}(X) \to C_{\ast}(X) \otimes C_{\ast}(X)$ given by Alexander-Whitney and its "flip" (with the signs introduced to make it a chain map) agree only up to chain homotopy.

The commutative cochain problem attempts to fix this: that is, to find a graded-commutative differential graded algebra $C_1^*(X)$ associated functorially to $X$ (which may be restricted to be a simplicial complex) which is chain-equivalent to the usual (noncommutative) algebra $C^{\ast}(X)$ of singular cochains.

In *Rational homotopy theory and differential forms,* Griffiths and Morgan mention briefly that there is no way to make the cup-product skew-commutative on cochains (that is, to solve the commutative cochain problem) with $\mathbb{Z}$-coefficients, and that this is because of the existence of cohomology operations. It is also asserted that these cohomology operations don't exist over $\mathbb{Q}$ (presumably excluding the power operations). Could someone explain what this means?

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