Milnor fibre is a bouquet of spheres

Remarks

The proof has three steps. Morse theory shows that $F$ has the homotopy type of a finite CW complex of dimension at most $n$. Milnor shows that, for an isolated singularity, $F$ is $(n-1)$-connected. A finite CW complex of dimension $n$ that is $(n-1)$-connected is homotopy equivalent to a wedge of $n$-spheres. For $n \ge 2$ this follows from the Hurewicz and Whitehead theorems, and for $n = 1$ a connected graph is homotopy equivalent to a wedge of circles. The number of spheres is the rank of $H_n(F; \Z)$, which Milnor identifies with $\mu$ through the degree of the gradient map.

Example: for $f = z_0^2 + \cdots + z_n^2$, $\mu = 1$, and the fibre $\set{z_0^2 + \cdots + z_n^2 = \delta}$ near $0$ is homotopy equivalent to the single sphere $\set{x \in \R^{n+1} : x_0^2 + \cdots + x_n^2 = \delta}$ for $\delta > 0$.

The hypothesis that the singularity is isolated is needed. For $f(z_0, z_1, z_2) = z_0 z_1$ on $\C^3$, the Milnor fibre is homotopy equivalent to a circle, not to a wedge of $2$-spheres.

Going deeper

For a student

The theorem shows that the reduced homology of the Milnor fibre is concentrated in degree $n$ and is free of rank $\mu$. So the monodromy of the Milnor fibration is described by a single $\mu \times \mu$ integer matrix, and the intersection form on $H_n(F; \Z)$ is a $\mu \times \mu$ integer form. A basis of $H_n(F; \Z)$ can be chosen of vanishing cycles, one for each nondegenerate critical point of a generic perturbation of $f$. This is the setting of Picard–Lefschetz theory.