(d-1)^n$.

Since $f$ is GI, all its fibers are transversal to the hyperplane at infinity. This implies that $f$ has only isolated singularities in the affine and that the Milnor numbers $\mu_{p_i}(t)$ and $\mu_{p_i}^{{\infty}}$ are all zero, for all $t$. Then formula (\[eq:chi\]) shows that $b_{n-1}(G) = (d-1)^n$. Further, since our $f$ is a [[$\mathcal F$[[ISI]{}]{}]{}]{} polynomial, it is in particular a polynomial with at most isolated singularities at infinity, in the sense of [@ST], and therefore its general fiber is a bouquet of $(n-1)$-spheres, by [@ST Theorem 3.1].

\[l:bif\] A [[$\mathcal F$[[ISI]{}]{}]{}]{} polynomial $f : {{\mathbb C}}^n \to {{\mathbb C}}$ is general-at-infinity if and only if $b_{n-1}(G) = (d-1)^n$.

In one direction, this follows from Proposition \[p:max\]. The reciprocal is a direct consequence of the formula (\[eq:chi\]) for $b_{n-1}(G)$ in terms of $\mu_{p_i}$ and $\mu_{p_i}^{{\infty}}$.

A GI polynomial may have different configurations of isolated singularities in the affine, provided that their total Milnor number is precisely $(d-1)^n$. A large class of GI deformations is given by the following result.

\[c:nongi\] Let $P$ be a deformation of a [[$\mathcal F$[[ISI]{}]{}]{}]{} polynomial $f_0$ such that $W_0 \cap \{ \phi_d = 0\} 
=\emptyset$. Then $P$ is general-at-infinity.

By definition, $P$ is GI 