Let $R$ be a rational function of degree $d$ mapping the Riemann sphere to itself:$$R(z) = \frac{a_d z^d + a_{d-1} z^{d-1} + \cdots + a_0}{b_d z^d + b_{d-1} z^{d-1} + \cdots + b_0}$$ where $a_d$ and $b_d$ are not both zero. And suppose that a sequence of coefficients $\{(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n\}$ converges to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. Let $R_n$ be the rational function with coefficients $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$.

Is it a theorem that: The sequence $\{R_n\}$ converges uniformly to $R$ on the Riemann sphere if and only if for all $n$ sufficiently large the degree of $R_n$ matches the degree of $R$?

I ask because I am reading Beardon's *Iteration of Rational Functions* and in section 2.8 he introduces a mapping $\Psi:\mathcal{R}\to\mathbb{C}\mathbb{P}^{2d+1}$, which takes a rational function to its coefficients. Beardon omits the details, but claims that $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$, where $\mathcal{R}_d$ is the space of rational functions having degree exactly $d$.

My thought is, if for $n$ sufficiently large $R_n$ has degree $d$, then $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n \in \Psi(\mathcal{R}_d)$. Now $\Psi^{-1}$ is defined and continuous at $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$, because $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$. And (up to multiplication by a constant) $R_n = \Psi^{-1}((a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n)$. Since $\Psi^{-1}$ is continuous on the sequence of coefficients and at the limit of the sequence of coefficients, $R_n$ converges to $R$. But I would like to know:

Is the convergence always uniform under these circumstances?

Conversely, if for all $n$ there exists $R_{m>n}$ in the sequence with degree not equal to $d$, then $\{R_n\}$ cannot converge to $R$ because (Beardon also states that) the spaces $\{\mathcal{R}_0, \mathcal{R}_1, \ldots, \mathcal{R}_d\}$ are the connected components of the space of rational functions with degree at most $d$.

The reason that I think this might be a theorem is that Beardon introduces a metric $\rho(R,S) = \sup_{z\in\mathbb{C}_\infty} \sigma_0 (R(z),S(z))$, $\sigma_0$ the chordal metric on $\mathbb{C}_{\infty}$, just a couple of pages earlier than the point where he introduces the mapping $\Psi$. It is tempting to read "$\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$" as "$\Psi$ is a homeomorphism of the metric space $(\mathcal{R}_d,\rho)$ onto $\Psi(\mathcal{R}_d)$." If that is the correct reading then $R_n$ converges to $R$ under the metric $\rho$ and the convergence is clearly uniform. But I am not sure that this is the correct reading since Beardon is only explicitly treating $\mathcal{R}_d$ and $\mathbb{C}\mathbb{P}^{2d+1}$ as topological spaces at the point where he introduces $\Psi$.

Note: I have edited the question in response to Margaret Friedland's answer, in an attempt to clarify my assumptions (and hopefully my source of confusion). I am a little worried that the reason my point of confusion was unclear is that it is actually trivial. I apologize in advance if that is the case.