Symmetric Nash gain map ($T_i(x)=\frac{x_i+[(Px{)}_i-x^{T}Px{]}_{+}}{1+{\sum }_j[(Px{)}_j-x^{T}Px{]}_{+}}$)

The continuous gain map takes a probability simplex to itself. The Brouwer fixed-point theorem supplies a fixed point, whose positive gains would contradict the zero strategy-weighted average payoff excess. Thus all gains vanish, giving a strategy that is a best response to itself. In a symmetric bimatrix game, this produces a symmetric equilibrium.

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.