admissible covers vs. stable maps to P^1

Question

The paper Stable maps and tautological classes of Faber-Pandharipande shows that the Gromov-Witten classes on spaces of relative stable maps of the projective line push forward to tautological classes on moduli spaces of stable curves.

It is claimed on p. 6 that $\overline{M}_{g,0}(\mu_1,\ldots,\mu_m)=\overline{H}_g(\mu_1,\ldots,\mu_m)$, that is, the space of relative stable maps with ramification profiles $\mu_i$ over marked points of $\mathbb{P}^1$ is equal to the space of admissible covers of a genus 0 curve. However, I am confused as to how this can be correct: if the degree is 1, then I believe the space on the right is empty, but the space on the left has many components in general, coming from starting with the identity map on $\mathbb{P}^1$ and attaching ghost components of positive genus to the source. More generally, I don't see how the presence of ghost components allow one to identify a stable map with an admissible cover.

Later Proposition 1 it is stated as a corollary the fundamental class of $\overline{H}_g$ pushes forward to a tautological class on $\overline{M}_{g,n}$, but because of the discrepancy above I don't see why this is immediate.

Am I missing something?

Answer 1

The claim is that these moduli spaces coincide under the numerical condition that $$ 2g−2 + 2d=\sum_{i=1}^m (d−\ell(\mu_i)).$$ In this case there will be no ramification points outside the fibers $f^{-1}(q_i)$ and no contracted components on the domain curve. Indeed any contracted component must have positive genus by stability since $n=0$, so any additional ramification or contracted component would force the genus of the domain curve to be strictly greater than $g$ (since $g$ is the genus of an admissible cover with this profile by Riemann-Hurwitz).

In the example $d=1$ that you mentioned the numerical condition is only satisfied when $g=0$, since the right hand side must be zero.