To elaborate off Martin's comment, one can generally say a number of interesting things. Consider a fixed, finite set of players $N$ and, for each, a finite set of actions $S_i$ for $i \in N$. Then the space of all payoff functions for agent $i$ with these agents/actions can be identified with $\Gamma_i = \mathbb{R}^{\prod_i |S_i|}$, and the space of games with $\prod_i \Gamma_i$. Let $\Sigma = \prod_i \Delta(S_i)$, and define:
$$E= \{(g,\sigma) \in \Gamma \times \Sigma : \textrm{$\sigma$ is a Nash equilibrium of $g$} \}.$$ Finally, for any locally compact space $K$, denote its one-point compactification as $\bar{K}$.

Now, results-wise, firstly it is straightforward from the definition of Nash equilibrium that $E$ is the graph of the 'equilibrium correspondence' $\phi: \Gamma \to \Sigma$, which is upper-hemicontinous in $g$. It is not, however, everywhere lower-hemicontinuous.$^1$

The result Martin cites may be formally stated as:

**Theorem**: (Kohlberg & Mertens 1986 *Econometrica*) Let $\bar{p}: \bar{E} \to \bar{\Gamma}$ denote the continuous extension of the projection $p:E \to \Gamma$ with $p(\infty) = \infty$. Then $\bar{p}$ is homotopic to a homeomorphism (under a homotopy taking $\infty$ to $\infty$ and $E$ to $\Gamma$).

This structure theorem tells us a great deal. For starters, there necessarily always exists a subset of Nash equilibria which are robust to any perturbation of payoffs as $\bar{E}$ is basically a deformed rubber sphere above/enclosing the sphere of games (in the sense that a nearby game has a nearby equilibrium).

Moreover, we have a particularly nice structure to work with: by the definition of von Neumann-Morgenstern utilities, the graph of $\phi$ may be defined as a finite collection of polynomial equalities and inequalities, and hence $\phi$ is a semi-algebraic correspondence (correspondence whose graph is a semi-algebraic set). In particular, for every game $g$, $\phi(g)$ admits a finite triangulation, i.e. there are finitely many connected components of equilibria for any game.

In light of this, at any point $g$ of lower-hemicontinuity of $\phi$, $\bar{p}^{-1}(g)$ is finite and there exists a neighborhood of $g$, $U$, for which $\bar{p}^{-1}(U)$ is a finite disjoint union of open balls. As deg$(\bar{p})=1$, then one obtains that at any point of lower-hemicontinuity of $\phi$, the set of Nash equilibria is finite and odd in number.

Furthermore, a result of Blume and Zame$^2$, the semi-algebraicity of $\phi$ implies that the set on which $\phi$ fails to be lower-hemicontinuous is strictly lower dimensional and hence of measure zero (note that we cannot just appeal to Sard's theorem to get this result as $E$ is *not* a smooth manifold). Hence the number of equilibria of a finite normal form game is generically finite and odd.

Thus to conclude, for generic games, not much interesting topologically happens (though to reach this point, in the spirit of your question, at least a little topology in the form of Brouwer degree rears its head). But for more interesting games, plenty can happen, within the purviews outlined above.

$^1$ For an example of this, it suffices to consider a 2x2 game with upper-left payoff pair $(1,1)$, lower-right $(t,t)$ for $t < 0$, and off-diagonal payoffs $(0,0)$. There is a unique Nash equilibrium, but at $t=0$ there are two.

$^2$ A lemma on page 3 of this paper.

extensive form gamessolely in terms of the topology of the equilibrium manifold at their normal form representation. $\endgroup$ – Pete Caradonna Mar 5 '17 at 22:31