Proof. Using Theorem 37.3.1 and after possibly shrinking $H$, assume without loss of generality that $I \in A$.

After choosing and extending a branch of the complex logarithm, let $\phi: \complex \to \complex$ be a Borel measurable function such that:

1. (i)

   $e^{\phi(z)}= z$ for all $z \in \complex \setminus \bracs{0}$.

2. (ii)

   For each $0 < r < R$, $\phi$ is bounded on the annulus $\ol{B(0, R)}\setminus B(0, r)$.

(1): Let $T \in G(A)$, then there exists $0 < r < R$ such that $\sigma_{A}(T) \subset \ol{B(0, R)}\setminus B(0, r)$. In which case, $\phi$ is a bounded Borel measurable function on $\sigma_{A}(T)$. By the Borel functional calculus, $\phi(T) \in A$ with $T = e^{\phi(T)}$. In which case, the path $t \mapsto e^{t\phi(T)}$ is a norm-continuous path in $G(A)$ from $I$ to $T$.

(2): In particular, as $e^{t\phi}(\partial B(0, 1)) \subset \partial B(0, 1)$, the spectrum of $e^{t\phi}$ as an element in the domain of the $L^{\infty}$ functional calculus, is contained in $\partial B(0, 1)$. Thus if $T$ is unitary, then the path $t \mapsto e^{t\phi(T)}$ lies in the unitary group of $A$ by Corollary 36.6.3.$\square$