Consider the following theorem and proof:
Proposition: For any two player game, there exists a Nash stable point.
2-D Proof. First we embedd the square in $\mathbb{R}^4$ by $$ \Gamma=\left\{\left(p_1, p_2, q_1, q_2\right) \in \mathbb{R}^4: p_1, p_2, q_1, q_2 \geq 0, p_1+p_2=q_1+q_2=1\right\} . $$ Then define $u_i: \mathbb{R}^4 \rightarrow \mathbb{R}$ and $v_i: \mathbb{R}^4 \rightarrow \mathbb{R}$ by $$ \begin{aligned} & u_1(\mathbf{p}, \mathbf{q})=\max \{A((1,0), \mathbf{q})-A(\mathbf{p}, \mathbf{q}), 0\} \\ & u_2(\mathbf{p}, \mathbf{q})=\max \{A((0,1), \mathbf{q})-A(\mathbf{p}, \mathbf{q}), 0\} \\ & v_1(\mathbf{p}, \mathbf{q})=\max \{B(\mathbf{p},(1,0))-B(\mathbf{p}, \mathbf{q}), 0\} \\ & v_2(\mathbf{p}, \mathbf{q})=\max \{B(\mathbf{p}(0,1))-B(\mathbf{p}, \mathbf{q}), 0\} . \end{aligned} $$ Through these we define $$ \begin{aligned} F: \mathbb{R}^4 & \rightarrow \mathbb{R}^4 \\ (\mathbf{p}, \mathbf{q}) & \mapsto\left(\frac{\mathbf{p}+\mathbf{u}(\mathbf{p}, \mathbf{q})}{1+u_1(\mathbf{p}, \mathbf{q})+u_2(\mathbf{p}, \mathbf{q})}, \frac{\mathbf{q}+\mathbf{v}(\mathbf{p}, \mathbf{q})}{1+v_1(\mathbf{p}, \mathbf{q})+v_2(\mathbf{p}, \mathbf{q})} .\right) \end{aligned} $$ In particular, we note that $F(\Gamma) \subset \Gamma$. Moreover, $F$ is continuous. So by the Brouwer Fixed Point The because the square is homeomorphic to the unit disc, there exists a fixed point $\left(p^*, q^*\right)$.
So far so good. What I do not understand is the follwoing:
Claim $\left(\mathbf{p}^*, \mathbf{q}^*\right)$ is a Nash stable point.
Proof of Claim. Since $u_1, u_2$ cannot both be non-zero, wlog $u_2=0$. Then at $\left(p^*, q^*\right)$, $$ \frac{p_1+u_1}{1+u_1}=p_1 \text {. } $$ So either $u_1=0$ or $p_1=1$.
Issues: I really can not follow one bit of this. In particular, why can't we have both $u_1$ and $u_2$ non zero? This would amount to
$$ \begin{aligned} & A\left((1,0), \mathbf{q}^*\right)>A\left(\mathbf{p}^*, \mathbf{q}^*\right) \\ & A\left((0,1), \mathbf{q}^*\right)>A\left(\mathbf{p}^*, \mathbf{q}^*\right) \end{aligned} $$ which I see no problem with.
Question: Could someone formally help me understand what is going on?
Relative Definitions: Here are the definitions that have been used in my course
Defn (Two-Player game) A two-player game, or bimatrix game, is given by two matrices $P, Q \in \mathbb{R}^{m \times n}$. Player 1 , or the row player, chooses a row $i \in\{1, \cdots, m\}$, while player 2 , the column player, chooses a column $j \in\{1, \cdots, n\}$. These are selected without knowledge of the other player's decisions. The two players then get payoffs $P_{i j}$ and $Q_{i j}$ respectively.
Defn (Strategy) Players are allowed to play randomly. The set of strategies the row player can have is $$ X=\left\{x \in \mathbb{R}^m: x \geq 0, \sum x_i=1\right\} $$ and the column player has strategies $$ Y=\left\{y \in \mathbb{R}^n: y \geq 0, \sum y_i=1\right\} $$ Each vector corresponds to the probabilities of selecting each row or column. A strategy profile $(x, y) \in X \times Y$ induces a lottery, and we write $p(x, y)=x^T P y$ for the expected payoff of the row player. If $x_i=1$ for some $i$, i.e. we always pick $i$, we call $x$ a pure strategy.
Defn (Nash stable point) A Nash stable point is a pair $\left(p^*, q^*\right)$ such that $$ \begin{array}{ll} A\left(p^*, q^*\right) \geq A\left(p, q^*\right) & \forall p \in X \\ B\left(p^*, q^*\right) \geq B\left(p^*, q\right) & \forall q \in Y . \end{array} $$
You didn’t introduce $A$ and $B$, but from your definition of a Nash equilibrium it seems that they’re the payoff functions for players $1$ and $2$, respectively.
If so, the reason the inequlities in your “issues” section can’t both be fulfilled is that the payoff is linear in each argument, i.e.
\begin{eqnarray*} A\left(\mathbf p^*,\mathbf q^*\right) &=& p_1^*A\left((1,0),\mathbf q^*\right)+p_2^*A\left((0,1),\mathbf q^*\right) \\ &=& p_1^*A\left((1,0),\mathbf q^*\right)+(1-p_1^*)A\left((0,1),\mathbf q^*\right)\;, \end{eqnarray*}
so $A\left(\mathbf p^*,\mathbf q^*\right)$ is a convex combination of $A\left((1,0),\mathbf q^*\right)$ and $A\left((0,1),\mathbf q^*\right)$ and thus must lie between them.