Let $Rat_d$ be the space of degree $d$ rational maps $f(z) = \frac{p(z)}{q(z)}$, where $p(z) = a_d z^d + \cdots + a_0, q(z) = b_d z^d + \cdots + b_0$ are complex polynomials with no common roots and leading coefficients $a_d, b_d$ not both 0. Consider the evaluation map at $\infty$, i.e. $$Rat_d \to \mathbb{C}\mathbb{P}^1$$ where $$\frac{p(z)}{q(z)} = \frac{a_d z^d + \cdots + a_0}{b_d z^d + \cdots + b_0} \to \frac{a_d}{b_d}$$
Let $Rat_d^0$ be the fiber at 1.
Question Why is this a fibration? $$Rat_d^0 \to Rat_d \to \mathbb{C}\mathbb{P}^1$$ (This is from Milnor, p.40 Appendix B)
It is not clear if this is a fiber bundle ($Rat_d$ is not compact, so doens't look like I can apply Ehresmann's theorem directly; it's clear that this map is not proper anyway). I'm not sure how people show something is a fibration (Hurewicz? Serre?) in general.
It wasn't clear to me why the fibers should be diffeomorphic; this actually comes from bihomogenity of resultant. For closure I'm posting the details here.
Without loss of generality let's look at pre-image near 1. By taking a small enough neighborhood we can assume $b_d \neq 0$, and so in $Rat_d$, we can focus on the open set $b_d \neq 0$ as well. Abusing the notation to focus on the affine piece $b_d \neq 0$ (so really in the following it's $\frac{a_i}{b_d}, \frac{b_i}{b_d}$) we can assume that $b_d = 1$, and we are looking at the map from affine space $$H:[a_d, \cdots, a_0, 1, b_{d-1}, \cdots, b_0] \to a_d$$ where $res(a_d z^d + \cdots + a_0, z^d + b_{d-1} z^{d-1} + \cdots + b_0) \neq 0$.
Take a small neighborhood $U$ around 1 that misses 0. Note that we have diffeomorphism $$i: H^{-1} (U) \to U \times H^{-1}(1)$$ that commutes with projections, where $$i([a_d, \cdots, a_0, 1, b_{d-1}, \cdots, b_0]) = (a_d, [1, \frac{a_{d-1}}{a_d}, \cdots, \frac{a_0}{a_d}, 1, b_{d-1}, \cdots, b_0])$$ The only point here is why $$[1, \frac{a_{d-1}}{a_d}, \cdots, \frac{a_0}{a_d}, 1, b_{d-1}, \cdots, b_0] \in H^{-1} (1) $$ since the diffeomorphism part is clear afterwards. It suffices to show $$res(\frac{a_d z^d + \cdots + a_0}{a_d}, z^d + b_{d-1} z^{d-1} + \cdots + b_0) \neq 0$$ But this is because of bihomogeity of resultant, so that $$res(\frac{a_d z^d + \cdots + a_0}{a_d}, z^d + b_{d-1} z^{d-1} + \cdots + b_0) = \frac{1}{a_d^d} res(a_d z^d + \cdots + a_0, z^d + b_{d-1} z^{d-1} + \cdots + b_0) \neq 0$$