Quick definition:
Given $g$, a strategy N-tuple $u^* = (u_1^*,...,u^*_N)$ is said to be a Nash equilibrium if:
$$J_i(u_i^*, u^*_{-i}) \leq J_i(u_i, u^*_{-i}), i \in N$$
where $J$ is the cost function, $u^*_{-i}$ is strategy for all players except for $i^{th}$ player.
What $\leq$ as used in the definition? How do you compare two vectors?
For example, given matrix game (say hawk dove game)
$$M = \begin{bmatrix} (0,0) & (-5,-2) \\ (-2, -5) & (-4,-4) \end{bmatrix}$$
Claim: two pure strategy nash equilibrium are (-5,-2), (-2,-5) where each entry represents the cost = negative of gain
How do you compare $J_1 = (-5,2)$ and $J_2 = (-4,-4)$ as according to the definition? Clearly, we cannot say $J_1 \leq J_2$ if we were evaluating each term pairwise, nor can we say $J_1 \geq J_2$...how can we say that solution that gives $(-5,-2)$ a nash equilibrium?
In the position with row player 1 choosing the first option and column player 2 choosing the second option, the costs to each are represented with $(-5,-2)$, so the cost to player $1$ is $-5$ (i.e. the payoff is $+5$) and $J_1(1,2) = -5$, while the cost to player $2$ is $-2$ (i.e. the payoff is $+2$) and $J_2(1,2) = -2$.
Given that the column player 2 is choosing the second option, this is an equilibrium for row player 1, since $J_1(1,2) \le J_1(1,2)$ trivially while $J_1(1,2) \le J_1(2,2)=-2$.
Given that the row player 1 is choosing the first option, this is an equilibrium for column player 2, since $J_2(1,2) \le J_2(1,1) = 0$ while $J_1(1,2) \le J_1(1,2)$ trivially.
So this is a Nash equilibrium since neither player can reduce their individual cost (increase their individual payoff) by changing strategy.