Let $\mathbb C$ be the complex plane, $H(\mathbb C)$ the set of all entire functions, and $D(\mathbb C)$ the set
of all non-negative divisors in $\mathbb C$.


Consider the map $Z:H(\mathbb C)\to D(\mathbb C)$ which to every entire function $f\in H(\mathbb C)$ puts into correspondence its divisor of zeros. 
There are natural topologies which make this map continuous: on $H(\mathbb C)$ this is uniform convergence on compact subsets, and on $D(\mathbb C)$ the induced topology of weak convergence of measures. (A divisor can be thought of as a discrete measure that takes integer values on all bounded sets).


The Weierstrass factoriation theorem says that this map is surjective, and he actually constructed a right inverse $W$, so that $Z\circ W=id\_{D(\mathbb C)}$.


Question: Does there exist a CONTINUOUS right inverse?


Weierstrass map is evidently not continuous. (His map depends on the ordering of zeros by absolute value, and this ordering can change when the divisor varies continuously). I can prove that there is no analytic right inverse, with the natural analytic structures on $H(\mathbb C)$ and $D(\mathbb C)$. I can also prove that there is no multiplicative continuous right inverse (multiplicative means that the sum of divisors corresponds to the product of functions). 


EDIT. Let me add for completeness a brief explanation why there is no analytic $W$.
In MR2280501, Michigan Math. J. 54 (2006), no. 3, 687–696, for every compact $E\in U$ of zero log capacity in the unit disk $U$, 
I constructed an holomrphic function $F(z,w)$ of two variables
$(z,w)\in\mathbb C\times U$, with the property that $F(z,w)\neq 0$ when $w\in E, z\in\mathbb C$ but $z\mapsto F(z,w)$ has zeros for $w\in U\backslash E$. Take some uncountable $E$. Let $D(w)$ be the divisor of the entire function $f\_w=F(.,w)$ in the $z$-plane.
Assuming that an analytic Weierstrass map exists, denote $g=W(\emptyset)$.
Then the set $\{ w:W(D(w))=g\}$ must be analytic (that is either discrete of the whole $w$-disk, but this is not so because it contains $F$ and is not equal to the whole disk.



Continuous Weierstrass map
The Weierstrass product has the form $W(z)=\prod E\_{N(a)}(z/a)$, where the product is over the set of the desired zeros $a$, and the integers $N(a)$ can be chosen freely; they only need to be large enough asymptotically to ensure convergence.


To avoid the problem you mentioned, we must make sure that $N(a)$ depends continuously on the divisor. In particular, we will also need $E\_N(z)$ for non-integer $N$, but this can be done by just interpolating in a straightforward way: if $N=n+d$ with $n\in\mathbb N$, $0\le d<1$, then we set
$$
E\_N(z) = (1-z)\exp \left( z + \frac{z^2}{2} + \ldots + \frac{z^n}{n} + d\frac{z^{n+1}}{n+1}\right) .
$$


Fix a continuous, compactly supported $F: \mathbb R\to [0,1]$ with $F(x)=1$ for $|x|\le 2$. I claim that taking
$$
N(a) = 2+|a|+ \sum\_b F(|b|-|a|)
$$
will give us continuous dependence of $W$ on the zero set. (The sum is over the zeros, with multiple zeros contributing the corresponding number of summands.)


A divisor $D'$ is close to the given divisor if it has almost exactly the same zeros $|a'|\le R$, and $D'$ can do anything whatsoever on $|z|>R$. To show that $W, W'$ are close in the desired topology, we must compare them on a fixed compact set, say $|z|\le r$, and of course we will take $R\gg r$.


Obviously, the finitely many factors corresponding to the $|a'|\le R$ are almost the same as before. So we need to show that for any (discrete) set of zeros $|a'|>R$, the corresponding part of the Weierstrass product is almost $1$ on $|z|\le r$ (provided $R$ was chosen large enough to start with).


Let's look at the contribution from the $a'$ with $n\le |a'|<n+1$. If there are $K$ of these, then they all contribute to my modified counting function $F$, so $N(a')\ge K+2$. Recall that $|E\_N(z)-1|\le |z|^{-N-1}$ for $|z|\le 1$. So the total contribution to $\sum\log |E\_{N(a')}(z/a')-1|$ from these $a'$ is $\lesssim K n^{-K-3} \le n^{-4}$ (because this function is decreasing in $K\ge 1$). This is summable, so the claim follows.


