Gain-map proof of symmetric equilibrium ($F_i(p)=\frac{p_i+[e(i,p)-e(p){]}_{+}}{1+{\sum }_j[e(j,p)-e(p){]}_{+}}$)

For a symmetric finite game, pure-action expected payoffs $e(i,p)$ against independent identical opposing mixed strategies are continuous. The displayed map takes the probability simplex continuously into itself. At a Brouwer fixed-point theorem fixed point, put $r_i=[e(i,p)-e(p){]}_{+}$ and $S={\sum }_i{r}_i$. The fixed-point equations give $r_i=p_{i}S$. If $S>0$, every supported action has strictly positive excess payoff, contradicting ${\sum }_i{p}_i(e(i,p)-e(p))=0$. Hence $S=0$ and every supported action is a best response, proving a symmetric Nash equilibrium.