Question on the construction of mapping from space of strategy profile into itself in Nash(1951)

Question

To appeal to Brouwer fixed point theorem, Nash(1951) constructed a continuous mapping $\operatorname{T}$ from strategy profile space into inself:

For player $i$, the probability of a pure strategy $a_{i} \in A_i$ is

$$\operatorname{T}s_i(a_i) = \frac{s_i(a_i) + \phi_{i, a_i}(s)}{1+\sum_{b_i \in A_i}\phi_{i, b_i}(s)}$$

in which, $A_i$ is the player $i$'s pure strategy space, $s_i$ is a mixed strategy that assigns each pure strategy with its probability.$s$ is a strategy profile, a vector of every player's strategy.

My question is why we need $\phi_{i, a_i}(s) = \max \{0, u_i(a_i, s_{-i})-u_i( s)\}$? Why not to replace $\phi_{i, a_i}(s)$ with $u_i(a_i, s_{-i})-u_i( s)$, which is$$\operatorname{T'}s_i(a_i) = \frac{s_i(a_i) + u_i(a_i, s_{-i})-u_i( s)}{1+\sum_{b_i \in A_i}(u_i(b_i, s_{-i})-u_i( s))}$$

It seems to me, Besides being continuous, $\operatorname{T'}$ statisfies that $\sum_{b_i \in A_i}\operatorname{T'}s_i(b_i)=1$ and that a fixed point of $\operatorname{T'}$ necessarily constitutes a nash equilibrium.

Answer 1

To appeal to the Brouwer fixed point theorem, one wants this map to go from a nonempty, compact, and convex to itself. Using the "max", we know the map goes from the compact unit simplex to itself. Your map would map oints in the unit simplex to points outside the unit simplex.