Are branch cuts always ‘cancellable’? Many functions can be analytically continued to $\mathbb C$ except the branch cut. 
However, it appears to me that for every function $f(z)$ that has a branch cut, there always exists a non-constant meromorphic/entire function $g(z)$ such that $f(g(z))$ can be analytically continued to the whole $\mathbb C$.
For example,
$$\sqrt {x^2}=x$$
$$\ln e^x=x$$
$$\arccos\cos x =x$$
$$\ln\ln e^{e^x}=x$$
$$\operatorname{W}(xe^x)=x$$
More complicated examples:
$$f(x)=\sqrt{(x+1)(x+3)}=\sqrt{(x+2)^2-1}$$
$$g(x)=-2+\cosh x$$
$$f(x)=x^\alpha\qquad{\alpha\in\mathbb C}$$
$$g(x)=e^x$$
Is this true?
 A: There is indeed always a function such as you describe. 
First, pick an open disk in $\mathbb{C}$ where $f$ is analytic. Since $f$ is non-constant, there is a point $a$ in this region where $f'(a)\neq 0$. Thus, by the Lagrange Inversion Theorem, there is a local inverse $g(z)$ which is analytic in a neighborhood of $f(a)$. In this neighborhood, $f(g(z))=z$, and the identity function $z$ can of course be extended to the complex plane.
Now, as you likely noticed, this restriction of the domain to an open disk before analytic continuation is necessary. Any $g(z)$ such that $f(g(z))$ has no branch cut must avoid the cut of $f$. By Picard's theorem there is no entire function on $\mathbb{C}$ which can avoid more than one point, and no meromorphic function on $\mathbb{C}$ which can avoid more than two. Thus for any entire or meromorphic $g$, either $g$ is constant or the preimage of the branch cut under $g$ must be a branch cut for $g$. The second case contradicts the meromorphicity of $g$. Thus there is no non-constant meromorphic $g$ such that $f(g(z))$ is meromorphic on all of $\mathbb{C}$.
