= Gain-map proof of symmetric equilibrium
{title2=$F_i(p)=\dfrac{p_i+[e(i,p)-e(p)]_+}{1+\sum_j[e(j,p)-e(p)]_+}$}
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 $r_i=p_iS$. 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>.
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