Is there an easy classification (and proof) of the possible branched covers between Seifert fibered *integer* homology spheres which are fiber-preserving and branched over fibers (or at least what the possible relations between the orders of the singular fibers should be)?


There exist branched covers of the form $f:\Sigma(p\_1,\ldots,p\_k) \to \Sigma(p\_1,\ldots,p\_{k-1})$ (branched cover over a regular fiber) or $\Sigma(p\_1,\ldots,p\_k \cdot n) \to \Sigma(p\_1,\ldots,p\_k)$ (branched cover over the singular fiber of order $p\_k$). I think that everything should be something like a composition of maps of this form (or at least the Seifert invariants should be related by such moves). There is a very long and tedious brute force method given by applying the results of Huang (<http://www.ams.org/journals/proc/2002-130-08/S0002-9939-02-06306-2/>) that seems like it should work, but it also feels like there should be a quick way to prove this fact (or disprove it).



Classification of fiber-preserving branched covers between Seifert fibered integer homology spheres
The class of homology sphere Seifert-fibered spaces fiber over 2-orbifolds $S^2(p\_1,\ldots,p\_k)$, where $\gcd(p\_i,p\_j)=1, i\neq j$. If one has a fiber-preserving map of one homology sphere to another, then there is a corresponding non-zero degree map between the base orbifolds obtained by quotienting the fibers. 


Suppose that the map has degree $d >1$, so we have a degree $d$ map $\phi: S^2(p\_1,\ldots,p\_k)\to S^2(q\_1,\ldots,q\_l)$. I claim then that $l\leq 3$. 


The maps between the underlying topological spaces of the orbifolds must be a branched cover $\varphi: S^2\to S^2$. Associate branching data to this map as follows. Let there be $r$ critical values of $\varphi$ (branch points). Associated to each branch point $y\_i$ is branch data $A\_i= (e\_{i1},\ldots,e\_{in\_i})$ a partition of $d$, so $e\_{i1}+\cdots+e\_{in\_i}=d$, and $n\_i$ is the cardinality of $\varphi^{-1}(y\_i)$. First, note that for each singular point of the orbifold $S^2(q\_1,\ldots, q\_l)$, the underlying point $y$ of $S^2$ must be a branch point, since otherwise $S^2(p\_1,\ldots, p\_k)$ would have at least $d$ orbifold points with branching multiple of $q\_i$ for some $i$, contradicting that it underlies a homology sphere. So we have $l\leq r$, and we may assume that the first $l$ branch points of $\varphi$ correspond to orbifold points. Now, suppose $y\_i$ is a branch point corresponding to an orbifold point of order $q\_i$. Then the order of the orbifold singularity of the preimage with branch data $e\_{ij}$ must be a multiple of $q\_i/\gcd(e\_{ij},q\_i)$. Thus, we have that $\gcd(q\_i/\gcd(e\_{ij},q\_i),q\_i/\gcd(e\_{im},q\_i))=1$ for $1\leq j\leq m\leq n\_i$. In particular, for at most one $j$ does $\gcd(e\_{ij},q\_i)=1$, and if so, all the rest must be multiples of $q\_i$. For example, we might have $A\_i=(1,q\_i,\ldots,q\_i)$, where $1+(n\_i-1)q\_i=d$. In this case, one may check either $q\_i=2$ and $n\_i\leq (d+1)/2$, or $n\_i\leq d/2$. If there are no preimage branch points of order $1$, then $e\_{ij}\geq 2$ for all $j$, and thus $n\_i\leq d/2$. We also have from the euler characteristic computation (Hurwitz formula) $\sum\_{i,j} (e\_{ij}-1) = 2d-2$. Thus, we have $\sum\_{i=1}^{l} \sum\_{j=1}^{n\_i} (e\_{ij}-1) \leq 2d-2$. Since $\sum\_j e\_{ij}=d$, we have $ld-\sum\_{i=1}^{l} n\_i \leq 2d-2$. But $n\_i\leq d/2$, except for one possible branch point which has $q\_i=2$ and $n\_i\leq (d+1)/2$. Then we get $(l/2-2)d\leq -1$, so $l \leq 3$. So if the degree is $>1$, then one gets stringent requirements on the image orbifold, namely that it has at most 3 branch points. I think some further analysis of the branch data for $l=3$ should rule out this possibility, except for the sort of example of $S^3\to \Sigma(2,3,5)$.


Otherwise, the map $\phi$ has degree 1. In this case, I think one can show that the map is a composition of the sorts of maps you suggest, since one can only decrease the orders of the orbifold points in a degree one branched cover between 2-orbifolds. 


**Addendum:** I didn't work through the case $l=3$ completely, but here's a special case that works out cleanly. We have $\frac{1}{q\_1}+\frac{1}{q\_2}+\frac{1}{q\_3}<1$ (the base orbifold can't be spherical or Euclidean). Suppose all of the branch points over these singularities have at least one branch datum of $1$. Then the branch data must look like $(1,k\_2q\_i,\ldots,k\_{n\_i}q\_i)$, so $d\geq 1+(n\_i-1)q\_i$, or $n\_i\leq 1+\frac{d-1}{q\_i}$. By the Riemann-Hurwitz formula, $$d\leq n\_1+n\_2+n\_3-2 \leq 1+ (d-1)(\frac{1}{q\_1}+\frac{1}{q\_2}+\frac{1}{q\_3}) <d,$$
a contradiction. This case must hold, for example, when all the $q\_i$ are prime powers. 


The other cases should be even easier, since then the fractions $n\_i/d$ should be much smaller, but I haven't gone through the analysis. 


