I asked a similar question a couple of days ago, but I think I have since made some tangible progress that warranted a new question instead of just editing the old one. The old question is now deleted, but I'll summarise my problem here.
I'm interested in proving the seemingly well-known fact that the space of rational maps ${Rat}_d$ of degree $d$ on the Riemann sphere to itself given the topology of uniform convergence (the Riemann sphere is given the spherical metric) can be embedded as an open dense subset of the projective space $\mathbb{P}^{2d + 1}$ with the Zariski topology. (I don't think the topologies matter here too much.) I'm not very experienced with the ideas in algebraic geometry, but I have tried to use the accepted answer for this question as a means for me to not only believe that an embedding exists, but to also be certain in its veracity.
I define a map $\Psi: {Rat}_d \to \mathbb{P}^{2d + 1}$ as follows. First, if $f(z) = p(z)/q(z)$ is a rational map of degree $d$ written in lowest form, then we can homogenise $p$ and $q$ to obtain the homogeneous polynomials
$$P(x, y) = y^dp(x/y) \quad \text{and} \quad Q(x, y) = y^dq(x/y).$$
Calling $F_\mathbf{a}(x, y) = \sum\limits_{j = 0}^d a_jx^jy^{d - j}$ these homogeneous polynomials, where $\mathbf{a}$ is the $(d + 1)$-tuple $(a_0, \dots, a_d)$, I can define $\Psi$ using homogeneous coordinates as
$$\Psi([F_\mathbf{a}(x, y) : F_\mathbf{b}(x, y)]) = [\mathbf{a} : \mathbf{b}].$$
Now the claim I have is that this map is a homeomorphism onto its image. I already know that its image is open in $\mathbb{P}^{2d + 1}$ (being the complement of the zero set of the resultant, which is itself a homogeneous polynomial), and it wasn't too hard to show that its image was dense in $\mathbb{P}^{2d + 1}$.
So I have tried using the linked accepted answer above as a way to prove that $\Psi$ is actually a homeomorphism onto its image. In the accepted answer, one defines (in my notation) a map $p: \mathbb{P}^{2d + 2} \times \mathbb{P} \to \mathbb{P}$ by $p([\mathbf{a} : \mathbf{b} : x : y]) = [F_\mathbf{a}(x, y) : F_b(x, y)]$ and then shows that $p$ is continuous with respect to the Zariski topologies on $\mathbb{P}^{2d + 2} \times \mathbb{P}$ and $\mathbb{P}$. This would then imply that $\Psi^{-1}$ is continuous via a sequence argument.
But my issue with this reasoning is that wouldn't one first start with a convergent sequence in $\mathbb{P}^{2d + 1}$, then show that this map $$p' : \mathbb{P}^{2d + 1} \times \mathbb{P} \to \mathbb{P}, \quad p'([\mathbf{a} : \mathbf{b}], [x : y]) = [F_\mathbf{a}(x, y) : F_\mathbf{b}(x, y)]$$ is continuous with respect to the product topology on $\mathbb{P}^{2d + 1} \times \mathbb{P}$ (both factors have the Zariski topology individually)? Because then by the same reasoning as the accepted answer, we would conclude that a convergent sequence in $\mathbb{P}^{2d + 1}$ gives a convergent sequence in ${Rat}_d$. The reasoning in the accepted answer seems to say that if one starts with a convergent sequence in $\mathbb{P}^{2d + 3}$, then one ends up with a convergent sequence in ${Rat}_d$, which I don't think is quite right for this problem. In my experience, Zariski topologies aren't particularly nice to products (i.e. I shouldn't be able to say that $\mathbb{P}^{2d + 1} \times \mathbb{P} = \mathbb{P}^{2d + 2}$ as topological spaces).
Any thoughts and suggestions would be much appreciated!