I got a question on this proof I read in a paper.
Let $f(z)=\frac{e^z}{z+1}$. We know that $-1$ is the only pole of the function. Now we take $L_0=[-1,\infty) \cup \infty$(only real axis) and $L_{-1}$ is the preimage of $L_0$(can be anywhere), this means $L_{-1}=f^{-1}(L_0)$. Now we take $l$ a maximally connected subset of $L_{-1}$ which is bounded. Now the next step in this proof states, that $f(z):l \rightarrow L_{0} \setminus 0$ is onto. The author uses this to state, that $-1$ has to be in $l$ which i understand.What i dont understand is, why $f(z):l \rightarrow L_{0} \setminus 0$ should be surjective, and I would appreciate any help, that you can give me.