Abelianization of limit groups Let $G_1$ and $G_2$ be limit groups, and let $C_1$ and $C_2$ be cyclic subgroups of $G_1$ and $G_2$, respectively. 
Question: 
If $G$ is the amalgamated product of $G_1$ and $G_2$ with amalgamated subgroups $C_1$
and $C_2$, is it true that $d(G) = d(G/G^{'})$?
Notation: $d(G)$ is the minimum number of generators for the group $G$.
 A: Here's a silly counter-example (though I doubt this is what you have in mind). Take $G_1=G_2=\mathbb{Z}$ with $C_1=2\mathbb{Z}$ and $C_2=3\mathbb{Z}$.  Certainly $G_1,G_2$ are limit groups. Then 
$G=\mathbb{Z}*_{2\mathbb{Z}=3\mathbb{Z}}\mathbb{Z}$
the fundamental group of the trefoil knot complement. This is easily seen to have abelianization $\mathbb{Z}$, but of course is not itself cyclic.
In this example, $G$ is not itself a limit group.  There are also counterexamples in which it is. For instance, consider $G_1=G_2=F_2$, the free group of rank two, and let $C_1=C_2=\langle w\rangle$ be any maximal cyclic subgroup which is not a free factor but is non-trivial in homology (for instance $\langle a^2ba^{-1}b^{-1}\rangle$).  Then the corresponding double
$G=G_1*_{C_1=C_2} G_2$ 
has abelianization of rank three, but is itself of rank four. This latter estimate is tricky to prove, however, and I'm not sure if it can be deduced from anything in the literature. You should look at the oeuvres of Louder and Weidmann for an idea of how to prove it.
Added:
In an e-mail, Richard Weidmann has pointed out the following easy way to prove that $d(G)=4$: quotient out by $w$ to note that $G$ surjects the free product
$F_2/\langle\langle w\rangle\rangle*F_2/\langle\langle w\rangle\rangle$ ; 
this has rank 4 by Grushko's theorem, so $G$ does too.
