Are symmetric equilibria continuous with respect to the payoff matrix? Assume a two player symmetric game where the payoff for the row player is given by:
$$
A = \left( \begin{array}{cc}
a_{1,1} &amp; a_{1,2} &amp;\cdots &amp; a_{1,n}\\
a_{2,1} &amp; a_{2,2} &amp;\cdots &amp; a_{2,n} \\
\vdots &amp; \vdots &amp; \ddots &amp; \vdots\\
a_{n,1}&amp;a_{n,2}&amp;\cdots &amp;a_{n,n}
\end{array} \right)
$$

We denote by $\Delta$ all probability vectors over $[n]$.

A symmetric equilibrium for the game is a vector $a\in \Delta$ such that 
$$\forall x\in \Delta: x^tAa\leq a^tAa$$

Assume we increase the value of some specific coordinate $i,j$ by a variable $t$. Denote the new matrix by $A(t)$.


  Assuming $A$ is non-singular, is the change in the symmetric equilibrium continuous in $t$?




I'm aware that the question might not be enough well-defined, as there could be more than a single equilibrium for $A$, but is there necessarily a symmetric equilibrium for $A(t)$ which converges to the symmetric equilibrium for $A$ as $t\to 0$?
If I understand your question correctly then the answer is no. Consider

$$
A(h) =  \left( \begin{array}{ccc}
1 &amp; 1 &amp; 0 \\
0 &amp; 2 &amp; 0 \\
2+2h &amp; 0 &amp; 3 \end{array} \right)
$$
For the game defined by $A(0)$ the vector $a^t(0) = \left( \frac{1}{2}, \frac{1}{2}, 0\right)$ yields an equilibrium. For any $h&gt;0$ however
$$
A(h) \cdot a(0) = \left( \begin{array}{ccc}
1 \\
1 \\
1+h \end{array} \right),
$$
so $a(0)$ is no longer an equilibrium. If play around with the probabilities you will see that nearby strategies do not constitute an equilibrium either. Hence if you have a positive sequence $h_1, h_2, ...$ such that $\lim_{n\to \infty} h_n = 0$ then you cannot select a sequence of vectors $a(h_1), a(h_2), ...$ such that they are equilibrium strategies of their respective games $A(h_1), A(h_2), ...$ and
$$
\lim_{n\to \infty} a(h_n) = a(0).
$$

In this counterexample I used that an action that was never played in $a(0)$ yielded the same payoff as actions with positive supports. If this is not the case then I think the answer to your question would be affirmative. Here is a proof for the case where $a&gt;0$. 

Since $a^t$ is a best response to $a$ we have
$$
A\cdot a = \underline{1} \cdot c
$$
where $\underline{1}$ is a vector with $1$ at all coordinates and $c \in \mathbb{R}$.
If $A$ is non-singular then
$$
a = A^{-1} \cdot \underline{1} \cdot c.
$$
As $a$ is a probability vector, $c$ is uniquely determined by $A$:
$$
1 = \underline{1}^t \cdot a = \underline{1}^t \cdot A^{-1} \cdot \underline{1} \cdot c
$$
I will use $A(h)$ to denote the perturbed matrix, and $a(h)$, $c(h)$ to denote the associated vectors. The original matrix and vectors occur at $h=0$. Since matrix inversion is continuous
$$
\lim_{h \to 0} c(h) = \lim_{h \to 0} \frac{1}{\underline{1}^t \cdot A(h)^{-1} \cdot \underline{1}} = \frac{1}{\underline{1}^t \cdot A(0)^{-1} \cdot \underline{1}} = c(0).
$$
Similarly
$$
\lim_{h \to 0} a(h) = \lim_{h \to 0} A(h)^{-1} \cdot \underline{1} \cdot c(h) = A(0)^{-1} \cdot \underline{1} \cdot c(0) = a(0).
$$
This is not an exact proof. To be exact you would have to specify the norm of a matrix, and then to prove that the above limits exist you would have to have a condition like
$$
\lim_{h \to 0} ||A(h)|| = ||A(0)|| \neq 0.
$$