Show that \((\alpha \frown \varphi) \frown \psi = \alpha \frown (\varphi \smile \psi)\) for all \(\alpha \in C_k(X; R)\), \(\varphi \in C^\ell(X; R)\), and \(\psi \in C^m(X; R)\). Deduce that cap product makes \(H_*(X; R)\) a right \(H^*(X; R)\) module.
Solution
It is enough to check the identity on a singular \(k\)-simplex \[ \sigma=[v_0,\ldots,v_k]. \] With Hatcher’s conventions, for \(\varphi\in C^\ell(X;R)\), \[ \sigma\frown\varphi = \varphi([v_0,\ldots,v_\ell]) [v_\ell,\ldots,v_k]. \] Applying \(\frown\psi\) with \(\psi\in C^m(X;R)\) gives \[ (\sigma\frown\varphi)\frown\psi = \varphi([v_0,\ldots,v_\ell]) \psi([v_\ell,\ldots,v_{\ell+m}]) [v_{\ell+m},\ldots,v_k]. \] On the other hand, the cup product is \[ (\varphi\smile\psi)([v_0,\ldots,v_{\ell+m}]) = \varphi([v_0,\ldots,v_\ell]) \psi([v_\ell,\ldots,v_{\ell+m}]), \] so \[ \sigma\frown(\varphi\smile\psi) = \varphi([v_0,\ldots,v_\ell]) \psi([v_\ell,\ldots,v_{\ell+m}]) [v_{\ell+m},\ldots,v_k]. \] The two expressions are identical. By linearity, \[ \boxed{(\alpha\frown\varphi)\frown\psi =\alpha\frown(\varphi\smile\psi)} \] for every chain \(\alpha\).
The boundary formula for cap product implies that a cycle capped with a cocycle is a cycle, and changing either by a boundary or coboundary changes the cap product by a boundary. Hence cap product descends to \[ H_k(X;R)\times H^\ell(X;R)\longrightarrow H_{k-\ell}(X;R). \] The chain-level identity then gives on homology \[ ([\alpha]\frown[\varphi])\frown[\psi] =[\alpha]\frown([\varphi]\smile[\psi]), \] and the unit in \(H^0(X;R)\) acts as the identity. Therefore \(H_*(X;R)\) is a right module over the graded ring \(H^*(X;R)\).