198 D. SALAMON, FLOER HOMOLOGY
Corollary 4.11. Mo,k,A(M, J) is a compact metrizable space.
Proof. By Theorem 4.10, each point in Mo,k,A(M, J) has a countable neighbour-
hood basis and, by Theorem 4.8, limits are unique. Hence Mo,k,A(M, J) is a
Hausdorff space.^2 By Theorem 4.9, the space is sequentially compact. Now every
sequentially compact topological space with a countable basis is compact.^3 Hence
Mo,k,A(M, J) is a compact Hausdorff space. Hence the result follows from the
fact that a compact Hausdorff space is metrizable iff it has a countable basis (cf.
Kelley [24]). D
The proof of Theorem 4.9 follows essentially from the arguments in section 4 .1
which, in a slightly different form are also contained in [19, 31, 37]. Full details
of the proofs of all three theorems 4.8, 4.9, and 4. 10 can be found in [21]. It
is interesting to discuss the special case, where the target space M is a singlepoint, in more detail. In this case Mo,k,A(M, J) = Mo,k is the Deligne-Mumford
compactification of the moduli space of k distinct marked points on the 2-sphere.
In this situation Corollary 4.11 can be strengthened to the assertion that Mo,k
naturally admits the structure of a compact smooth manifold. This will be discussed
in the next section.Exercise 4.12. Consider the target space M = 82 with the standard complex
structure and denote by L E H^2 (8^2 ; Z) the positive generator. Prove that the
moduli space M 0 , 0 ,L(8^2 , i) is a single point. Prove that there is a bijectionMo,o,2L ( 82 , i) ~ <CP^2.
Hint: The singular stratum (consisting of equivalence classes of stable maps mod-
elled over a tree with 2 vertices) can be naturally identified with <CP^1 via the image
of the unique double point under the stable map. The open stratum is the space of
equivalence classes of rational maps u : 82 ---t 82 of degree 2 under the equivalence
relationProve that this open stratum can be identified with <C^2. Namely, every rational
function of degree 2 has two distinct critical values, and this pair of critical values
(up to reordering) determines the equivalence class. DExercise 4.13. As in the previous exercise consider the target space M = 82 with
the standard complex structure. Examine the limit behaviour of the sequence of
rational mapsu (z) - (z - 1 + n .l )^2 (z + 1 + .l n )^2 (z - n .l + ~) n (z + .l n + --\) nn - (z-1)2 (z+1)^2 (z-) (z+)
(^2) Let x and y b e distinct points in a topological space with countable neighbourhood bases and
unique limits. Let {Un}n and {Vn } n b e countable neighbourhood bases for x and y, respectively.
If Un n Vn =I 0 for all n then any sequence Zn E Un n Vn converges to both x and y. Hence
uniqueness of limits implies that Un n Vn = 0 for some n.
(^3) Every open cover has a countable refinement {U,.,.}nEN• consisting of all those elements from the
basis which are contained in some element of the cover. Suppose that this refinement does not
have a finite subcover. The n there exists a sequence Xn E Un such that Xn 'i Um for m < n.
Choose a convergent subsequence Xn; --> x. Then x E Um for some m. Hence there exists an
io EN s uch that Xn; E Um for i 2: io. Hence Xn; E Um for some n ; > m, a contradiction.