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It is easy to construct surjective locally univalent holomorphic functions $f: {\mathbb D}\to {\mathbb C}$, where ${\mathbb D}$ is the open unit disk.

I am pretty sure that the answer to the following is positive and well-known (to experts) and is somewhere in the literature:

Question. Are there (non-affine) surjective entire functions $f: {\mathbb C}\to {\mathbb C}$ without critical points?

Such functions (up to an additive constant) would have the form $f(z)=\int_0^z e^{h(w)} dw$, where $h(w)$ is another nonconstant entire function. I do not see, however, how to verify surjectivity of such $f$. Of course, it can miss at most one point in ${\mathbb C}$ and, hence, it would be astounding if indeed this was the case for arbitrary $h$. However, I do not see how to rule this out. Applying Cauchy's argument principle in this setting looks extremely messy.

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up vote 7 down vote accepted

Picards theorem will do the trick. Take $$h(z)=z\cdot e^{\int_0^zf(t)dt}$$ where $f(z)=\frac{e^z-1}{z}.$ Then, clearly $$h'(z)=e^{\int_0^zf(t)dt}(1+zf(z))=e^{\int_0^zf(t)dt+z}\ne 0$$ and $h$ is locally univalent. If $h(z)$ is not surjective, then by Picard it omits just one complex value $A$ and we can write $h(z)=e^{g(z)}+A.$ But then equality $$e^{g(z)}+A=z\cdot e^{\int_0^zf(t)dt}$$ is impossible since the LHS takes value $0$ infinitely often while the RHS does not.

In fact, the same proof works for any function of the form $h(z)=g(z)\cdot e^{\int_0^zf(t)dt}$ with $g'(z)+g(z)f(z)\ne 0$ and $g(z)$ is a polynomial.

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Very nice, thank you! – studiosus Nov 6 '13 at 15:56
Instead of writing $h(z) - A = e^{g(z)}$, you can conclude that on $\mathbb{C} - \{0\}$, $h$ misses both $0$ and $A$ (as $z = 0$ is the only value for which $f(z) = 0$), and use big Picard theorem. – xyzzyz Nov 7 '13 at 23:56

The simplest surjective entire function without critical points is the "probability integral", $\int_0^z e^{-\zeta^2}d\zeta$.

Proof. Suppose it omits $a$. As it is of order 2, it must be of the form $a+e^P$, where $P$ is a polynomial of degree $2$. But this function has critical points:-)

The argument extends to $\int e^Q$ for any polynomial of degree at least $2$.

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But how do you see that this function is surjective? – studiosus Nov 6 '13 at 21:30
I see, very nice! – studiosus Nov 6 '13 at 21:38

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