Yet Another Schwarz Lemma Question This is an exercise about the Schwarz lemma, to which the internet gave me no hint so far.

Let $\mathbb{D}=\{z : |z| < 1\}$, and let $f,g:\mathbb{D}\to\mathbb{D}$ be analytic, $1-1$ functions satisfying
  $$f(0)=g(0), \ f'(0)=g'(0).$$
  Prove that $f \equiv g$ on $\mathbb{D}$.

My INCORRECT try: the Schwarz lemma applied to $f-g$, $f'-g'$ yields 
$|f(z)-g(z)|\leq|z|, |f'(0)-g'(0)|\leq 1$,
$|f'(z)-g'(z)|\leq|z|, |f''(0)-g''(0)|\leq 1$.
But I don't know how to continue.
Why incorrect? Because I cannot apply the Schwarz lemma to $f-g$, since not necessarily $\forall z : |f(z)-g(z)| \leq 1$, and similarly not necessarily $\forall z : |f'(z)-g'(z)| \leq 1$.
 A: This is false as stated. Start with $$F(z)=z,\quad G(z)=z+z^2.$$
Since $G'(0)\ne0$ there exists $\delta>0$ such that $G$ is injective in the disk $|z|<\delta$. Choose $R$ so that $|F(z)|<R$ and $|G(z)|<R$ for $|z|<\delta$ and set $$f(z)=F(\delta z)/R,\quad g(z)=G(\delta z)/R.$$
It's true if you assume in addition that $g$ is surjective; in that case you can apply the Schwarz Lemma to the function $f\circ g^{-1}$. (In fact it's enough to assume that $f(D)\subset g(D)$; then you can apply Schwarz to $g^{-1}\circ f$. In fact in that case you don't even need to assume that $g$ is injective; if you assume just $g'\ne0$ then analytic continuation gives the moral equivalent of $g^{-1}\circ f$, namely a function $h:D\to D$ with $h(0)=0$ and $g\circ h=f$.)
Edit The fact that $G'(0)\ne0$ implies that $G$ is injective in some disk $|z|<\delta$ is a well known fact from elementary complex analysis. We can give an ad hoc proof here without needing any complex analysis, showing that in fact we can take $\delta=1/2$. Suppose that $|z|<1/2$, $|w|<1/2$, $z\ne w$ and $G(z)=G(w)$. Then $$z-w=w^2-z^2=(w-z)(w+z).$$Since $z\ne w$ this implies that $w+z=-1$, and now the triangle inequality implies that $1\le|z|+|w|<1$.
Knowing that $\delta=1/2$ works allows us to give an explicit counterexample: $$f(z)=z/2,\quad g(z)=z/2+z^2/4.$$(There's a simple algebraic proof above that $g$ is injective in the unit disk. This fact will be obvious to readers who know a little complex analysis, if they note that $\Re g'(z)>0$ for $|z|<1$.)
