Entire functions inducing surjective maps on some spaces of holomorphic functions I was working on entire functions that induce surjective maps on Banach algebras, and I get stuck to the following question : "Let $f$ be an entire function. Assume that, for any $r>0$, there exists a holomorphic map $g_r: D(0,r)\longrightarrow \mathbb{C}$ such that $f\circ g_r(z)=z$ for any $z\in D(0,r)$. Does that imply that $f$ is an analytic automorphism of $\mathbb{C}$?"
In fact, this holds if there exists a point $z_0$ of $\mathbb{C}$ whose preimage by $f$ is finite. Indeed, in this case, one can find a sequence $(r_i)$ of real numbers tending to $+\infty$ such that $g_{r_i}(z_0)=g_{r_j}(z_0)$ for any $i,j$. Since $f\circ g_{r_i}(z)=z$ for any $z\in D(0,r_i)$, it is easy to check that two functions $g_{r_i}$ and $g_{r_j}$ coincide on $D(0,r_j)\cap D(0,r_j)$. Define the map $g:\mathbb{C}\longrightarrow \mathbb{C}$ by setting $g(z)=g_{r_i}(z)$ for any $i$ such that $z\in D(0,r_i)$. Then the function $g$ is well-defined and holomorphic on all of $\mathbb{C}$, and moreover $f\circ g(z)=z$ for any $z\in \mathbb{C}$. In particular, $g$ is entire and injective, and so it is an analytic automorphism of $\mathbb{C}$. Since $f\circ g(z)=z$ for any $z\in \mathbb{C}$, this implies that $f$ is also an automorphism. 
Therefore, what happens in the general case? If you have got any suggestions or ideas, they are welcome.
 A: In fact, the answer to the question is negative, and one can produce a counter-example. Here I would like to thank Pr. Lukas Geyer from Montana, who gave me a counter-example to another question that I asked, and it turns out that this counter-example also works for this question.
Consider the entire function $f:z\longmapsto z\sin(z)$. Let $r$ be any positive real number and fix a positive integer $k$ such that $k\pi -\frac{\pi}{2}>r$. Consider the rectangle of the complex plane delimited by the lines $\mathfrak{Re}(z)=k\pi -\frac{\pi}{2}$, $\mathfrak{Re}(z)=k\pi +\frac{\pi}{2}$, $\mathfrak{Im}(z)=k$ and $\mathfrak{Im}(z)=k$. Since $\cosh^2=1+\sinh^2$, it is easy to check that, for any complex number $z=x+iy$ with $x,y\in \mathbb{R}$:
$$
\vert z\sin(z)\vert=\vert z\vert \sqrt{\sin^2(x)+\sinh^2(y)}.
$$
For any $z$ such that $\mathfrak{Re}(z)=x=k\pi -\frac{\pi}{2}$, we get that $\vert f(z)\vert\geq \vert z\vert\geq x>r$. In the same way, one can check that $\vert f(z)\vert>r$ for any $z$ such that $\mathfrak{Re}(z)=k\pi+\frac{\pi}{2}$. Moreover, for any $z$ such that $\mathfrak{Re}(z)=x\geq k\pi -\frac{\pi}{2}$ and $\mathfrak{Im}(z)=y=k$, we get that $\vert f(z)\vert\geq \vert z\vert . \vert \sinh(y)\vert \geq \vert x\vert . \vert y\vert >r$. In the same way, one can check that $\vert f(z)\vert>r$ for any $z$ such that $\mathfrak{Re}(z)\geq k\pi-\frac{\pi}{2}$ and $\mathfrak{Im}(z)=-k$. In other words, the function $f$ has modulus $>r$ on the whole boundary of $R_k$. At this point, denote by $\gamma$ a simple positively oriented path that runs through the whole boundary of $R_k$, and set for any $w\in D(0,r)$ :
$$
N(w)=\int_{\gamma} \frac{f'(z)}{f(z)-w}dz.
$$
By construction, this integral is well-defined and the function $N$ est continuous on the disc $D(0,r)$. By the argument principle, $N(w)$ is the number of solutions (counted with multiplicities) of the equation $f(z)=w$ on the interior $R_{k,0}$ of the rectangle $R_k$. Since $D(0,r)$ is connected, the function $N$ is constant on $D(0,r)$. Moreover, the equation $f(z)=z\sin(z)=0$ has $z=k\pi$ as a unique solution on $R_{k,0}$. Since $f'(k\pi)=k\pi\cos(k\pi)+\sin(k\pi)\not=0$, we have $N(0)=1$, and so $N(w)=1$ for any $w\in D(0,r)$. In other words, the equation $f(z)=w$ has a unique solution in $R_{k,0}$ for any $w\in D(0,r)$. 
Now set $U_r=f^{-1}(D(0,r))\cap R_{k,0}$. Then the restriction map $f_{|U_r}:U_r\longrightarrow D(0,r)$ is one-to-one. Since $f$ is holomorphic, $f_{|U_r}$ is a biholomorphism from $U_r$ to $D(0,r)$. If $g_r$ denotes the reciprocal of $f_{|U_r}$, then $g_r$ is holomorphic from $D(0,r)$ to $U_r$, and $f\circ g_r(z)=z$ for any $z\in D(0,r)$. However, the function $f$ is not an analytic automorphism of $\mathbb{C}$. 
