214 D. SALAMON, FLOER HOMOLOGY
Remark 5.7. Let (M,>..) be a branched manifold with charts {(Mi, >..i,cpi)}iEJ·
Then the set
Mreg = {x EM : x E M i====? x E int(Mi)}
is open and dense in M (see Lemma 5.10 below). The definition shows that M has
a well defined tangent space Tx M for x E Mreg· If x tJ. Mreg, then there is a tangent
space Tx Mi for every i E I with x E Mi and these tangent spaces of the different
branches need not be naturally isomorphic. 0
Exercise 5 .8. Let (M, >..)be a branched manifold with charts {(Mi, >..i, cpi)}iEJ and
suppose that each set J(j) in the partition of the index set consists of a single point.
Prove t hat M is a smooth manifold and >.. : M ___, Q is constant on the components
of M. Hint: Each Mi is an open subset of M. D
Exercise 5.9. Define M = D x {±1}/ ,..__,where D C C denotes t he closed unit
disc and t he equivalence relation is given by (z, -1) ,..__, (z^2 , 1) for lz l = 1. Define
>.. : M ___, Q by >..([z, -1]) = 1 /2 for lz l < 1 and >..([z, 1]) = 1 for lz l ::::; l. Prove
that (M, >..) admits the structure of an oriented branched 2-manifold. Note that
this space cannot be expressed as a union of closed 2-manifolds. D
Lemma 5.10. Suppose that M is a metric space with a covering {Mi}iEI and a
decomposition I= LJi l(j) of the index set into disjoint finite subsets l(j) such that
(i) for every j , M(j) = LJiEI(j) Mi is an open subset of M,
(ii) for every i E J(j), M i is a relatively closed subset of M(j),
(iii) for all i, i' E J, intM,, (Mi n Mi') = intM, (Min Mi').Then Mreg = {x EM : x E Mi====? x E int(Mi)} is a dense open subset of M.
Proof. The proof is in three steps.
Step 1: Mreg is open.
We prove t hat Mi' n int(Mi) C int(Mi') for all i , i' E J. To see this note that t he
set M i' n int(Mi) is open relative to Mi' and hence
Mi' n int(Mi ) c intM,, (Min Mi') = intM, (Min M i').
Now let x E Mi' n int (Mi). By what we have just proved, x E int M; (Mi n Mi' ), and
hence there exists an open neighbourhood Uc M of x such that Un Mi C Min Mi'.
Hence Un int(Mi) is an open set in M which contains x and is contained in M i'·
Hence x E int( M i'). Thus we have proved t hat
for all i, i' E J. This implies
and hence Mreg is open.
Step 2: For every i E J(j),
Mi' n int(Mi) c int(Mi')
Mreg = LJ int(Mi ),
iEJM i - int(Mi ) = u (Min Mi' - intM,, (Mi n Mi')).
i'EJ(j}-{i}