The strategies yielding a zero-sum game's value

Let $m, n \in \mathbb{N}_1 := \{1, 2, \dots\}$ and let $\mathbf{A} \in \mathbb{R}_{m \times n}$ be a real, $m \times n$ matrix. Denote by $\Gamma$ the two-players, zero-sum game, whose payoff matrix is $\mathbf{A}$, to be interpreted from the point of view of the row player. It is known that there is a number $v \in \mathbb{R}$, called $\Gamma$'s value, a strategy $\mathbf{x}$ for the row player and a strategy $\mathbf{y}$ for the column player, such that $v$ is $\Gamma$'s expected payoff given $\mathbf{x}$ and $\mathbf{y}$, $v = p_\Gamma(\mathbf{x}, \mathbf{y})$, and such that for any strategy $\mathbf{x}'$ of the row player and for any strategy $\mathbf{y}'$ of the column player we have $$p_\Gamma(\mathbf{x}', \mathbf{y}) \leq v \leq p_\Gamma(\mathbf{x}, \mathbf{y}')$$

Now let $\mathbf{a}$ be some strategy for the row player and let $\mathbf{b}$ be some strategy for the column player, such that $p_\Gamma(\mathbf{a}, \mathbf{b}) = v$. Is it necessarily the case that for any strategy $\mathbf{a}'$ of the row player and for any strategy $\mathbf{b}'$ of the column player the following holds? $$p_\Gamma(\mathbf{a}', \mathbf{b}) \leq v \leq p_\Gamma(\mathbf{a}, \mathbf{b}')$$

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You are basically asking whether every strategy profile that yields the value of the game is a Nash equilibrium. This is not true. Consider the following normal-form game: $$\begin{array}{cccc}&\text{L}&\text{R}\\\text{T}&1&0\\\text{B}&0&0\end{array}$$ where the row player's payoffs are shown. Consider the pure strategy $(\text{T},\text{R})$. It is easy to see that it is a minmax strategy (its payoff is maximal in the column and minimal in the row), so the value of the game vanishes: $v=0$. Also, $(\text B,\text L)$ gives the same payoff, but it is clearly not a Nash equilibrium, since the row player could just go to T and increase the payoff.