Connected component of identity of an automorphism group is a subgroup Let $D$ be a bounded domain in $\mathbb{C}^n$ and let Aut$(D)$ be the set of biholomorphic functions from $D$ to $D$. Define a metric on Aut$(D)$ by the supremum norm, $d(f,g) = \sup_{z\in D}|f(z)-g(z)|_{\mathbb{C}^n}$ and let Aut$^{Id}(D)$ to be the connected component of identity. 
I want to show that Aut$^{Id}(D)$ is a subgroup. My idea is to prove that for Aut$(D)$, connectedness is equivalent to path connectedness, then it will be done. But I'm so sure its true and I think there might be a more direct way to do this. 

Let $m:\text{Aut}(D)\times \text{Aut}(D)\to \text{Aut}(D)$ be the multiplication map. We wish to show that the image $m(\text{Aut}^{Id}(D)\times \text{Aut}^{Id}(D))$ is a subset of $\text{Aut}^{Id}(D)$. However, we know that the image is connected since $m$ is continuous. 
Does that help? Also, one needs to show that $\text{Aut}^{Id}(D)$ is closed under inverses but the idea is similar.
