$\DeclareMathOperator{\co}{\operatorname{H}}$Fix a commutative ring $R$. Let $X$ be a connected topological space $X$ which is "$R$-Poincaré of dimension $n$", that is, there exists a (fundamental) class $[X] \in \co_n(X;R)$ such that for every $j$ the map $$- \cap [X] \colon \co^j(X;R) \to \co_{n-j}(X;R)$$ of $R$-modules is an isomorphism. Then the square $X^2$ is connected and $R$-Poincaré of dimension $2n$ with fundamental class as the homology cross product $[X] \times [X] \in \co_{2n}(X^2 ; R)$. Consider the following cohomology classes:
- Let $\mu \in \co^n(X;R)$ be the unique class that satisfies $\mu \cap [X] = 1 \in R = \co_0(X;R)$.
- Writing $D \colon X \to X^2$ for the diagonal map, let $\Delta \in \co^n(X^2;R)$ be the unique class that satisfies $$\Delta \cap ([X] \times [X]) = D_*[X] \in \co_n(X^2;R) \, .$$
Fix $x_0 \in X$ and consider the map \begin{align*} f \colon X &\to X^2 \\ x &\mapsto (x_0,x) \, . \end{align*}
Question: Does it follow that $f^*(\Delta) = \mu$?
The answer is yes when $R=k$ is a field because we can select a homogenous $k$-basis $B$ of $\co^*(X;k)$ with $1 \in B$ and dual basis $B^\sharp$ (so that $1^\sharp = \mu$) and establish the formula (Theorem 11.11 in Milnor--Stasheff) $$\Delta = \sum_{\beta \in B} (-1)^{|\beta|} \beta \times \beta^\sharp \, .$$
