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

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.