Cochains and Cocycles. I'm trying to prove that: 1. The cup product of two cocycles is a cocycle; 2. The cup product of a cocycle with a coboundary is a coboundary. Here's my attempt: 1. Suppose $u$ and $v$ are cocycles, i.e. $\partial(u) = \partial(v) = 0$. Then by the Leibniz rule: \begin{equation*} \partial(u \smile v) = \partial(u) \smile v + (-1)^{|u|} \ u \smile \partial (v) = 0 \end{equation*} Since $\partial(u \smile v) = 0$ we have that $u \smile v$ is a cocycle. 2. Let $u$ be a cocycle and $v$ a coboundary. Then $\partial(u) = 0$ and $v = \partial(w)$. Then $\partial(w \smile u) = \partial(w) \smile u + (-1)^{|w|} \ w \smile \partial(u) = \partial(w) \smile u = v \smile u$. Therefore $v \smile u$ is a coboundary. Similarly, $\partial(u \smile w) = (-1)^{|u|} \ u \smile v = \pm u \smile v$. I'm not quite sure if that $-u \smile v$ implies that $u \smile v$ is a coboundary. Thanks for the help :)

If $\partial(u \smile w) = - u \smile v$, then $\partial(-u \smile w) = u \smile v$.

Therefore the cup product of a cocycle and a coboundary, in either order, is a coboundary.