*EDIT:* *The key thing that I am wondering about is the linearity of the P2 strategy, not the constancy of P1. (The latter is straightforward.)*

**Question:** Is the following result already known? Or is it a straightforward corollary of some more general principle?

**Intuitive version** (slightly wrong): In a normal form game, adding a small cost $c$ to a particular action of a player (independently of the action of the opponent) changes the opponent's Nash equilibrium strategy linearly (in the size of $c$) and leaves the player's NE strategy constant.

**Formal version:** Let $G$ be a two-player normal-form game with actions denoted $i$, $j$ and payoffs $u_1(i,j)$, $u_2(i,j)$.
For a fixed action $i_0$ of P1, denote by $G'_c$ the game that is identical to $G$, except $u'_1(i_0, j) := u_1(i_0, j) - c$.
Then there is a finite set of exception points $-\infty = e_0 < \dots < e_n = \infty$ such that for every $c_0 \in (e_k, e_{k+1})$ and $\sigma^{c_0} \in \textrm{NE}(G'_{c_0})$,
there is a **linear** mapping $c \in [e_k, e_{k+1}] \mapsto \sigma^c_2$ such that $(\sigma^{c_0}_1, \sigma^c_2) \in \textrm{NE}(G'_c)$ (and the value for $c = c_0$ coincides with $\sigma^{c_0}_2$).

**Comments:**
(1) I know how to prove the statement "from scratch" using the LP formulation of NE (and the notion of basic solutions). But I was wondering whether there is perhaps a more general result that this follows from.
(2) It seems something similar holds if we only change a single cell in the payoff matrix. (This will cause a linear change in the *relative weights* of probabilities in $\sigma_2^{(\, \cdot \, )}$. IE, no longer linear, but still quite simple.)