Dual Steenrod squares

Question

Fix the ground ring $\mathbb{F}_2$ and let $X$ be a space with finite homology. Then we have an isomorphism $\Phi^i_X:H_i(X)\to H^i(X)^*,a\mapsto \langle-,a\rangle$ which allows us to define the dual Steenrod squares $$\mathrm{Sq}_r:H_i(X)\stackrel{\Phi_X^i}{\to} H^i(X)^*\stackrel{(\mathrm{Sq}^r)^*}{\to} H^{i-r}(X)^*\stackrel{(\Phi^{i-r}_X)^{-1}}{\to} H_{i-r}(X).$$ I want to understand which formulae they satisfy. Many things seem easy: For example, the “dual” Cartan formula needs the comultiplication $\Delta:H_i(X)\to \bigoplus_{p+q=i}H_p(X)\otimes H_q(X)$ which is dual to the cup product, and one should get $$\Delta_{i-r}\circ \mathrm{Sq}_r = \sum_{p+q=r}(\mathrm{Sq}_p\otimes\mathrm{Sq}_q)\circ \Delta_i.$$ What I am now trying to dualise is $$\mathrm{Sq}^nx = x\smile x ~~~\text{ for }|x|=n.$$ The right hand side is obviously the cup product precomposed with the diagonal $\nabla:H^n(X)\to H^n(X)^{\otimes 2}\subseteq \bigoplus_{p+q=2n}H^p(X)\otimes H^q(X)$. We can find a map which is dual to this diagonal and of the form $$D:\bigoplus_{p+q=2n}H_p(X)\otimes H_q(X)\to H_n(X),a\otimes b\mapsto a\,\# \,b.$$ We then should get for $a\in H_{2n}(X)$ the statement $$\mathrm{Sq}_n(a) = (D\circ \Delta)(a).$$ However, this does not look very enlightening. Does this “wrongly graded product” $D$ have any meaning? At least, it satisfies the following property for all $\alpha\in H^n(X)$ $$\langle \alpha,a\,\#\,b\rangle = \langle\alpha^{\times 2},a\times b\rangle.$$