Is exponentiation open?

Question

Already for $2\times 2$ matrices the exponential map is not open. However, the diagonalization trick does not work for algebras of functions. Hence the question

Is the map $f\mapsto \exp(f)$ open on the complex space $C[0,1]$?

Answer 1

Yes. Suppose $f\in C[0,1]$ and let $$C=\inf\{|\exp(f(x))|:x\in[0,1]\}.$$ Given $g\in C[0,1]$, let $H:[0,1]\times[0,1]\to\mathbb{C}$ be the linear homotopy from $\exp(f)$ to $g$ (that is, $H(x,t)=t\exp(f(x))+(1-t)g(x)$). Note that if $\|g-\exp(f)\|<C$, then the image of $H$ is contained in $\mathbb{C}\setminus\{0\}$. It follows that the homotopy $H$ lifts to the universal cover $\exp:\mathbb{C}\to\mathbb{C}\setminus\{0\}$. More precisely, there is a unique map $\tilde{H}:[0,1]\times[0,1]\to\mathbb{C}$ such that ${\exp}\circ\tilde{H}=H$ and $\tilde{H}(x,0)=f(x)$ for all $x$. Explicitly, we can compute $\tilde{H}$ using a contour integral: $$\tilde{H}(x,t)=f(x)+\int_{\exp(f(x))}^{H(x,t)}\frac{dz}{z}$$ where the integral is along the linear path. Taking $\tilde{g}(x)=H(x,1)$, we then have $\exp(\tilde{g})=g$.

To conclude that $\exp$ is open on $C[0,1]$, we now just need to show that we can ensure $\tilde{g}$ is arbitrarily close to $f$ by making $g$ sufficiently close to $\exp(f)$. This follows from the fact that $$\tilde{g}(x)-f(x)=\int_{\exp(f(x))}^{g(x)}\frac{dz}{z}$$ where the integral is computed along the linear path. If $\delta<C/2$ and $\|g-\exp(f)\|<\delta$, then $\frac{1}{z}$ is bounded by $2/C$ along this entire path and the path has length $<\delta$, and so $|\tilde{g}(x)-f(x)|<\frac{2\delta}{C}$. This goes to $0$ as $\delta$ goes to $0$, and it follows that $\exp$ is open.