Let $\mathfrak{A}$ be a unital $C^*$-Algebra and let $\delta: \mathfrak{A} \rightarrow \mathfrak{A}$ be a linear map that is not constantly zero and satisfies, for every $A, B\in\mathfrak{A}$, $\delta(AB) = \delta(A)B + A\delta(B)$. It is proved here by Sakai that $\delta$ (called derivation) is always coutinuous. Then define:
$$ e^{\delta}(A) = \sum_{n\geq 0}\frac{\delta^n(A)}{n!} $$
where $\delta^0(A) = A$ for each $A\in\mathfrak{A}$. For every $A, B\in\mathfrak{A}$, since one can prove that for each $N > 0$, $\delta^N(AB) = \sum_{i\leq N}\binom{N}{i}\delta^i(A)\delta^{N-i}(B)$, one can check that $e^{\delta}$ is an endormophism. Also notice that for any $\lambda\in\mathbb{C}$, we have $\lambda\delta$ defined by $(\lambda\delta)(A) = \lambda\delta(A)$ is also a derivation and given $\lambda, \mu\in\mathbb{C}$, one can show $e^{\lambda\delta}e^{\mu\delta}(A) = e^{\mu\delta}e^{\lambda\delta}(A) = e^{(\lambda+\mu)\delta}(A)$ (the same way we prove the equation of $\delta^N(AB)$). Hence $e^{\delta}$ is an automorphism with inverse $e^{-\delta}$.
My question is: is it true that, because for each $A\in\mathfrak{A}$, $e^{\delta}(A)$ is homotopic to $0$ through the path $e^{t\delta}(A)$ where $t\in[0, 1]$, then, in the context of $K$-theory, $K_0(e^{\delta})$ is the zero mapping from $K_0(\mathfrak{A})$ to $K_0(\mathfrak{A})$? I believe the path $[0, 1]\rightarrow\mathfrak{A}, t\mapsto e^{t\delta}(A)$ is continuous for each $A\in\mathfrak{A}$ because $\delta$ is bounded, but find it hard for me to believe the conclusion is true. Moreover, anyone who can provide any more info about the automorphism $e^{\delta}$ where $\delta$ is a derivation will be highly appreciated. Definitions you may need for $K$-theory can be found on this wiki page.
