Exponential omits two sphere values

By [F2] the function $f$ is entire, so as a meromorphic sphere map it has no poles: $f(z)\mathrm{\ne }\mathrm{\infty }$ for every $z$, i.e. $f$ omits $\mathrm{\infty }$; and $\mathrm{\mid }f(z)\mathrm{\mid }=e^{\mathrm{Re}z}>0$, so $f$ omits $0$.

Let $a\in \mathbb{C}\setminus \{0\}$ and let $b$ be a logarithm of $a$, so $e^b=a$ by [F4]. For $z\in \mathbb{C}$, by [F3], $e^z=a$ holds if and only if $e^{z-b}=1$, if and only if $z-b\in 2\pi i\mathbb{Z}$, if and only if $z=b+2\pi ik$ for some $k\in \mathbb{Z}$. Hence the preimage of every nonzero finite value is exactly this arithmetic progression in $b$.

At a point $z=b+2\pi ik$ the derivative is $f^{\mathrm{\prime }}(z)=e^z=a\mathrm{\ne }0$ by [F2] and step 1.2; therefore $f-a$ has a simple zero there, that is, the value $a$ is attained only with multiplicity one.

Combining steps 1.1, 1.2 and 2.1: the two sphere values $0$ and $\mathrm{\infty }$ are omitted, and no other sphere value is omitted. In the notation of [F1], for $a=0,\mathrm{\infty }$ the counting function $N(r,a;f)$ vanishes identically and the whole characteristic sits in the proximity term, $m(r,a;f)=T(r,f)+C(f,a)$; the corresponding deficiency-one computation is carried out in the companion example on this page devoted to deficiencies of elementary functions.

The argument is choice-free: the only choices made are the fixed logarithm $b$ of $a$ and the integer enumeration of the progression, both of which are data of the example. Countable Choice is carried only because the surrounding Nevanlinna quantities and their exceptional-set interface are stated under it.