This is question is regarding Proposition 5.1 on page 22-23 of Elementary Theory of Analytic Functions of One or Several Complex Variables by Cartan.
Here is the proposition:
Proposition 5.1. Suppose $\displaystyle T(X)=\sum_{n\geq 1}b_{n}X^{n}$. If the radii of convergence $\rho{(S)}$ and $\rho{(T)}$ are $\neq 0$, then the radius of convergence of $U=S\circ T$ is also $\neq 0$. To be precise, there exists an $r>0$ such that $\displaystyle\sum_{n\geq 1}|b_{n}|r^{n}<\rho{(S)}$; the radius of convergence of $U$ is $\geq r$, and, for any $z$ such that $|z|\leq r$, we have $$|T(z)|<\rho{(S)}$$ and
(5.1) $$S(T(z))=U(z)$$
I understand most of the proof. However, I don't understand how the equality $U_{n}(z)=S_{n}(T(z))$ for $|z|\leq r$ is determined.
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Relation (5.1) remains to be proved. Put $\displaystyle S_{n}(X)=\sum_{0\leq k\leq n}a_{k}X^{k}$ and let $S_{n}\circ T=U_{n}$. For $|z|\leq r$, we have $$U_{n}(z)=S_{n}(T(z))\text{,}$$ since the mapping $T\to T(z)$ is a ring homomorphism and $S_{n}$ is a polynomial.
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I have not been able to work out how the existence of the ring homomorphism $T\to T(z)$ or that $S_{n}$ is a polynomial is used to deduce the result above.
Any help to explain how this is deduced is greatly appreciated.
I think you might have already cleared the problem, but anyway I answer for people suffering from it.
Let $A_r$ be the set of the formal power series which converge on $\{ z \in K | |z| \leq r \}$. We fix $z$ such that $|z| \leq r$. Then, substituting $z$ gives a map $F \mapsto F(z)$ from $A_r$ to $K$, which we call $f_z$. Note that $T \in A_r$ and $f_z$ is denoted as $T \mapsto T(z)$ in the book. The facts that $A_r$ is a ring and that $f_z$ is a ring-homomorphism can be verified by Proposition 4.1 of the book.
Since $S_n$ is a polynomial, we can put $S_n(X) = \sum_{0 \leq k \leq n} a_k X^k$. Then, by the definition of composition, $U_n = S_n \circ T$ can be written as (a finite sum) $\sum_{0 \leq k \leq n} a_k T^k$.
Now, we can prove our assertion by $$ U_n(z) = f_z(U_n) = f_z(S_n \circ T) = f_z(\sum_{0 \leq k \leq n} a_k T^k) = \sum_{0 \leq k \leq n} a_k f_z(T)^k = \sum_{0 \leq k \leq n} a_k T(z)^k = S_n(T(z)). $$
Note that it is the fourth equality $f_z(\sum_{0 \leq k \leq n} a_k T^k) = \sum_{0 \leq k \leq n} a_k f_z(T)^k$ that is non-trivial and justified by the facts that $\sum_{0 \leq k \leq n} a_k T^k$ is a finite sum and that $f_z$ is a (K-linear) ring-homomorphism.