202 D. SALAMON, FLOER HOMOLOGY
is independent of the choice of this a.
The existence of a is obvious from Step 1. Moreover, if the points Zai, Zaj, Zak, Zae
are all distinct, then a is unique. If not then w(zai, Zaj, Zak, Zae) E {O, 1, oo} and a
simple combinatorial argument shows that this number is independent of a.
Step 3: Two stable Riemann surfaces z and z of genus zero with n marked points
are equivalent if and only if w1(z) = w1(z) for all I= (i, j, k, f) E In.
This is proved by induction over the number of vertices. The induction step is to
remove an endpoint from the tree. A set of indices I C { 1, ... , n}, corresponding
to the marked points on a given endpoint of the tree, can be characterized by the
conditions
(45)(46)i,i' EI, j,j' ~I
i,i^1 ,i^11 EI, j ~I
Wii'jj' = 00,Wii'i"j =/=-00.Given any such set one can reduce the number of vertices by replacing { 1, ... , n}
with { 1, ... , n} - I and reordering.
Step 4: Let w = { Wij kd E ( S^2 )1n be given. Then there exists a stable Riemann
surface z of genus zero with n marked points such that Wijke = Wijke(z) for alli, j, k, e if and only if w satisfies (42) and (43).
This is again proved by induction over the number of vertices. The key point in the
induction is to prove that, for every tuple w = { Wijke} which satisfies ( 42) and ( 43),
there exists a nontrivial subset I C {1, ... , n} which satisfies ( 45) and ( 46). Here
nontrivial means that I=/=- 0 and I=/=-{1,... , n}. For more details see [20]. D
Proof of Proposition 4.19: For any sequence z,, of stable Riemann surfaces
of genus zero with n marked points and any z one proves that the following are
equivalent.
(a) z,, DM-co nverges to z.
(b) For 11 sufficiently large there exists a surjective tree homomorphisms f" :T---+ T v and Mobius transformations <p~ E G such that f"(ai) = ai and
Zai = 1/-->00 lim (<p~)-^1 (zjv(a)i)
for a E T and i E { 1, ... , n}.
(c) For any four distinct integers i,j,k,f E {l, ... ,n}, Wijke(z)
lim,,__,oo Wijke(z,,).The proof of (a) {===} (b) ===} (c) is fairly straight forward. The hard part is to
show that ( c) implies (b). The key point here is the observation that, for any two
stable Riemann surfaces z and z' of genus zero with n marked points, there exists
a surjective tree homomorphism f : T---+ T ' with f(ai) =a~ for all i if and only if
Wijke(z') = oo ===} Wijke(z) = oo
for all i,j,k,f E {1,... , n}. For more details see [20]. D
Proof of Proposition 4.20: The proof is based on an explicit construction of
coordinate charts. Given a point [z] E Mo,n we must pick out n - 3 of the cross
ratios from which all the others can be reconstructed (in a neighbourhood of the