I'm reading a proof of Arnold's theorem about analytic linearization of analytic circle diffeomorphisms. The following result is used in the proof and I don't see why it should be true.

Let $S_\sigma= \{ z:|\Im z|<\sigma \}$. We have a function $h$ that is holomorphic on $S_{\sigma-\delta}$, and is real-valued on the real-axis. We are also give that for $z \in S_{\sigma-\delta}$ we have $|h(z)|<\delta$, and for $z \in S_{\sigma-2\delta}$ we have that $|h'(z)|<1$. Let $H$ be defined by $H(z)=z+h(z)$, so $H$ sends any point $z \in S_{\sigma-\delta}$ less than $\delta$ far away from itself.

I want to show that $H(S_{\sigma-2\delta})$ contains $S_{\sigma-3\delta}$. We have that $h$ sends the real axis to itself, so $H$ does also. As well, because $h(z)<\delta$ for $z \in S_{\sigma-\delta}$, for $\Im z=\sigma-2\delta$ we have $\Im H(z) > \sigma-3\delta$, and for $\Im z = -(\sigma-2\delta)$ we have $\Im H(z)< -(\sigma-3\delta)$. The text seems to conclude from this that $H(S_{\sigma-2\delta})$ contains $S_{\sigma-3\delta}$. This would follow from the above facts if we knew that $H(S_{\sigma-2\delta})$ had to be simply connected, but why should this be the case? (I hope I said enough to make it clear why this WOULD follow if we knew that the image were simply connected.)