Let $f : \mathbb{D} \rightarrow \mathbb{D}$ be an holomorphic function in $\mathbb{D}$ with $f(0)=0$. Let $a_1, a_2,\ldots,a_n$ be $n$ different points in $\mathbb{D}$ with $f(a_j)=b \; \forall \; j =1,\ldots,n$ with $b \in \mathbb{D} \backslash \{0\}$.
We define $\varphi_{a_j}(z)=\frac{z-a_j}{1-\bar a_jz}$, $h(z)= \prod_{j=1}^{n} \varphi_{a_j}(z)$ and $g(z) = \frac{\varphi_b (f(z))}{h(z)}$.
Prove that $|g(z)|\leq 1$, for all $z \in \mathbb{D}$.
Now I show my attemp:
First I've already prove some properties of the function $\varphi_a$ which are $\varphi_a (a) = 0$, $\varphi_a (0) = -a$, $\varphi_a \; o \; \varphi_{-a} (z) = z \; \forall a,z\in \mathbb{D}$ and that $\varphi_a$ is a biyection between $\mathbb{D}$ and itself.
To prove that $|g(z)|\leq 1$ for all $z \in \mathbb{D}$ I do it by induction.
First I show that $|\frac{\varphi_b (f(z))}{\varphi_{a_1}(z)}|\leq 1$.
I do it by considering $\psi (w)= \varphi_b \; o \; f \; o \;\varphi_{-a}(w)$, which verifyes that $|\psi(w)| \leq 1 \; \forall w \in \mathbb{D}$ and $\psi (0) = 0$.
$\psi$ is also holomorphic in $\mathbb{D}$ because it is composition of holomorphic functions.
Now I can use Schwarz Lemma to conclude that $|\psi (w)| \leq |w|$. Now considering $w = \varphi_{a_1}(z)$ we get by the definition of $\psi$ that $|\varphi_b (f(z))| \leq |\varphi_{a_1}(z)|$, hence $|\frac{\varphi_b (f(z))}{\varphi_{a_1}(z)}|\leq 1$.
Now I assumed that $g_{n-1}(z) := \frac{\varphi_b (f(z))}{\prod_{j=1}^{n-1} \varphi_{a_i}(z)}$ satisfyes that $|g_{n-1}(z)| \leq 1$ for all $z \in \mathbb{D}$.
Now considering $\phi (w):= g_{n-1} \; o \; \varphi_{-a_n}(w)$, then $|\phi (w)| \leq 1$ in $\mathbb{D}$ and $\phi(0)=0$. Like before this function is holomorphic because it is composition of holomorphic functions.
Using Schwarz Lemma and considering $w = \varphi_{a_n}$ I get $|g_{n-1}(z)| \leq |\varphi_{a_n}|$, and so it is prove that $|g(z)|\leq 1$ in $\mathbb{D}$ for all $n\in \mathbb{N}$.
Is this proof correct? Is there any step which is not clear or not correct? Someone sees another way easier or not to make the proof?
Thanks in advance