For a real analytic function $f:(a,b)\to\Bbb R$ we have that if $f'(x)\neq 0$ for all $x$ in the interval $(a,b)$, then $f$ is injective. My question is whether similar holds if $f$ is a holomorphic function $$f:G\to\Bbb C$$ where $G\subseteq\Bbb C$ is a simply connected, open area and $f'(z)\neq 0$.
A simple counter example is $f:z\mapsto z^2$ if $G$ is simply connected such that $-1,1\in G$ but $0\notin G$. Then $f$ is not injective even though $f'(z)\neq 0$ everywhere. So the idea is to fix the preconditions somehow. A second take:
Question: Let $f:G\to S$ be holomorphic with $f'(z)\neq0$ and where $G,S\subseteq\Bbb C$ are open and simply connected areas with $f(G) = S$. Does this imply that $f$ is injective?
It seems to fix the case $f:z\mapsto z^2$ because if $-1,1\in G$, there is some path $\gamma$ that connects $-1$ and $1$. But then $f(\gamma)$ runs around $0$, and because $S=f(G)$ is simply connected, there must be $0\in S$. This means $0\in G$ and thus $f'(z)\neq0$ does not hold and the conjectured theorem does not apply.
While there is a negative answer to the OP question in Conformal map from punctured disc to disc, there seems to be some confusion about it so I will expand my comment above a bit. First it is very easy to show that there is $f: D^{*} \to D$ conformal and surjective if and only if there is $g:D\to D$ conformal, surjective and non injective as the latter implies that there is $g(z_1)=g(z_2)$ so $g$ restricted to $\mathbb D-z_1$ is still surjective, while if there is $f$ conformal, surjective and with an isolated singularity at $0$ say, then $\exp \circ f$ maps the left half plane $\Re w <0$ conformally, surjectively and non-injectively to the unit disc and hence pulling that back to the unit disc by a Mobius transform we get our $g$
Now the answer linked above sketches how to construct such conformal, surjective but non-injective map between simply connected domains, but a very easy construction is in https://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=complex;action=display;num=1234362106 the third post for example; later posts in the same forum give more examples and references to literature where such examples were given in print
So the answer to the OP question is negative but non-trivially so. Note that there are tons of surjective holomorphic maps from the unit disc to itself (eg Blaschke products, singular inner functions, etc), so without conformality the result is negative trivially