Transversality theorems for the weak topology

In his 1979 paper Trotman proves, using the techniques of the Thom transversality theorem, that under some conditions on the dimensions of the manifolds under consideration, openness of the set of maps transverse to a stratification in the strong (Whitney) topology implies that the stratification is $(a)$-regular. Here we first discuss the Thom transversality theorem for the weak topology and then give a similiar kind of result for the weak topology, under very weak hypotheses. Recently several transversality theorems have been proved for complex manifolds and holomorphic maps. In view of these transversality theorems we also prove a result analogous to Trotman's result in the complex case.

What would seem to be an obvious generalization of the transversality theorem 1.1 is to replace the submanifold S by a collection of submanifolds such that their union is a closed subset, which appears as exercise 3 on page 59 in Golubitsky and Guillemin's book [4]. Unfortunately the exercise is not correct as stated and we will provide a counterexample later. A similar kind of mistake appears in exercise 8 on page 83 in Hirsch's book [6]; we will also give a counterexample to this exercise.
In fact there seem to be no complete correct statements published to date of such results and perhaps due to the mistakes in the standard textbooks, many recent papers contain errors of a similar kind. See Trotman [10] for a brief discussion about some mistakes occurring in papers of Thom, Chenciner and Wall. More recently, Loi [8] proves a transversality theorem for definable maps in o-minimal structures, but his definition of the definable jet bundle is incorrect and due to the problems with definitions his proof is not correct as stated. Also, it is generally believed that the openness of the set of maps transverse to a stratification in the strong topology implies a-regularity of the stratification without any restrictions on the dimensions of manifolds, see for example remark 1.3.3 on page 38 in Goresky and Macpherson [5], but this is not true; see remark 1.5 below. In fact Goresky and Macpherson [5] also prove a more general transversality theorem, proposition 1.3.2 on page 38 and claim that the openness of maps, restricted to a stratification in the source manifold, transverse to a stratification in the target manifold in the strong topology implies that both stratifications are a-regular and refer to Trotman [10], but Trotman does not prove such a statement in his paper.
Here we prove a proposition which is more general than the transversality theorem and also gives the correct formulation of the incorrect exercises mentioned above. We need the following definitions: A stratification Σ of a subset V of a manifold M is a locally finite partition of V into submanifolds of M . The submanifolds in the partition are called strata. By a locally finite partition we mean that each point of V has a neighbourhood meeting only finitely many strata.
Let S 1 and S 2 be two strata of Σ, S 2 is said to be a-regular over S 1 at x ∈ S 1 ∩S 2 if for every sequence of points {y i } in S 2 converging to x such that lim i→∞ T yi S 2 exists, we have A stratification is called a-regular if every pair of strata (S i , S j ) is a-regular at every point in the intersections S i ∩ S j and S j ∩ S i .
Notice that our definition of a stratification allows that none of the strata be a closed subset; see figure 1. Both S 1 and S 2 are not closed and yet their union is closed. We say that a map f is transverse to a stratification Σ at a point x if f is transverse to each stratum S i at x. Now we state the result: Proposition 1.2. Let M and N be smooth manifolds and let Σ be an a-regular stratification of a closed subset S ⊂ N . Then, The strata of a stratification need not be closed submanifolds and they can be of same dimension but the union of the strata should be closed.
The denseness of transversal maps is proved in Hirsch [6] (theorem 2.5 on page 78 and theorem 2.8 on page 80). In fact, we don't need a-regularity to prove the denseness. Here we prove results about openness of transversal maps.
The transversality theorem can be seen as a special case of our proposition, namely, when the stratification of the underlying closed set has only one stratum. Lemma 1.4. Let M and N be manifolds and let Σ be an a-regular stratification of a closed subset S of N and let f ∈ C ∞ (M, N ). If x ∈ M is such that f x Σ then there exists a coordinate chart (φ, U ) of M at x such that for each compact K ⊂ U there is a weak subbasic neighbourhood N (f, (φ, U ), (ψ, V ), K, ) * of f such that every member g of this neighbourhood satisfies g K Σ.
Proof: Without loss of generality we can assume that Σ has only two distinct strata S 1 , S 2 . We have four cases: Since S is closed, we can choose a coordinate chart (ψ, Thus for each compact set K ⊂ U we can choose > 0 small enough and take a weak subbasic neighbourhood N (f, (φ, U ), (ψ, V ), K, ) of f such that if g belongs to this neighbourhood then in particular g(K) ⊂ V ; hence g(K) ∩ S = ∅ and thus g K Σ.
Set of all maps g ∈ C ∞ (M, N ) such that g(K) ⊂ V and ||D r f φ,ψ (x) − D r g φ,ψ (x)|| < for all r ≥ 0 and all x ∈ φ(K) Suppose here that the dimensions of M and N are m and n respectively. Let the dimension of S 1 be s.
Let π : R n → R n /(R s × 0) be the natural projection map. The situation is described below: Denote by L(R m , R n /(R s × 0)) the set of linear maps between R m and R n /(R s × 0); it is a normed space.
Define a map η : U → R such that η(u) denotes the distance of the linear map Note that η is a continuous map and η(x) > 0 since f is transverse to S 1 at x and the nonsurjective linear maps L(R m , R n /(R s × 0)) form a closed set. Thus, there exists a neighbourhood U of x on which η > 0. Hence if K is any compact subset of U then η will assume a positive minimum value, say 2 , on K. Clearly then, the weak neighbourhood N (f, (φ| U , U ), (ψ, V ), K, ) is the required weak neighbourhood of f .
Using the same argument as in the case 2, we can find coordinate charts (φ , U ) around x and (ψ , V ) around f (x) such that for each compact K ⊂ U there is a weak subbasic neighbourhood N (f, (φ , U ), (ψ , V ), K, ) such that every member of this neighbourhood is transverse to S 1 on K.
to be a basis for the neighbourhoods of x. Then, for each i there is a compact set Taking limits † on both sides and using the properties of sequences of points in grassmannians we have: has the property that all its members are transverse to S 2 on all of K.
It is easy to see that for a suitable and any compact K ⊂ U , the subbasic neighbourhood and all its members are transverse to S 1 and S 2 on K.
The proof is exactly the same as the case 3 with S 1 and S 2 interchanged.
is a weak open neighbourhood of f and is contained in T K , as required. † We may assume that all of the above limits exist by taking subsequences if necessary.
The corresponding statement about the strong topology can be proved by taking a countable covering of the closed set K and then taking the countable intersection of the weak neighborhoods thus obtained.
Remark 1.5. If, in a stratification, the codimension of a stratum of the minimal dimension is greater than the dimension of M , then we don't need a-regularity to prove that the set of maps transversal to the stratification on a compact set is open.
Our result is more general than the result of exercise 15 on page 84 in [6], for we don't need any of the submanifolds of the stratification to be closed and they can be of same dimension.  In the above example S is a compact submanifold of N , but there exist maps arbitrarily close to f in the weak topology, which are not transverse to S. So, the set of maps transverse to S in the above case is not open in the weak topology. Thus, it is a counter example to the exercise 8 on page 83 in [6]. Notice that f is transverse to S 1 as well as S 2 but we can find maps arbitrarily close to f not transverse to S 1 by simply shifting f to the right side; see figure 3. Thus this is a counterexample to exercise 3 on page 59 in [4]. Now, let M and N be manifolds and Σ be a stratification of a closed subset V of N . Suppose that the codimension of a stratum of minimal dimension is less than or equal to the dimension of M . Then, in fact, we can prove the converse of the statement of our proposition. A similar kind of result is proved in [10] for the strong topology.
We suppose that the strata are of dimension at least 1. Proof: Let X and Y be two distinct strata in Σ such that Y is not a-regular over X at some x ∈ X ∩ Y . Then, there exists a sequence Let v ∈ T x X be such that v / ∈ τ and E be the one dimensional subspace of T x N spanned by v. Now, choose a basis for T x N such that we can write Let (φ, U ) be a coordinate chart around w ∈ M such that φ(w) = − → 0 ∈ R m and (ψ, V ) be a coordinate chart around x ∈ N such that ψ(x) = − → 0 ∈ R n .
Then, D x ψ : T x N → R n is a linear isomorphism of vector spaces. Under this isomorphism (1.1) and (1.2) become and Suppose that the dimension of H + τ is p. Then n − r ≤ p < n. Let {v 1 , . . . , v p } be a basis of D x ψ(H) + D x ψ(τ ) such that {v 1 , . . . , v n−r } forms a basis of H. Extend this basis to a basis {v 1 , . . . , v p , v p+1 , . . . , v n−1 , v } of R n .
Let m be the dimension of M and define a map L : R m → R n (this map is well defined because m ≥ n − r) by, L(a 1 , . . . , a m ) = a 1 v 1 + a 2 v 2 + . . . + a n−r v n−r .

By using a bump function we may construct a map
The situation is described below: This implies that, f w X and f w Y (since x / ∈ Y ).
We will now construct a sequence of smooth maps {f k } which converge to f in the weak topology such that for sufficiently large k, f k w Σ, which will be a contradiction to the hypothesis that the set First note that y k ∈ V for sufficiently large k. Thus Let H k be the subspace of D y k ψ(T y k N ) spanned by v k 1 , . . . , v k n−r . Then, by (1.5) we have lim k→∞ H k = D x ψ(H) and since the left hand side of (1.6) is spanned by a subset of {v k 1 , . . . , v k p } and p < n. Now, define the map f k ∈ C ∞ (M, N ) first on the open set U ⊂ M by the following formula for its local representative: Notice that f k φ,ψ (w) = y k and D φ(w) f k φ,ψ is the linear map L k : R m → R n defined by L k (z) = z 1 v k 1 + . . . + z n−r v k n−r . and so by (1.6), Now define f k outside U to be equal to f . This gives a smooth map f k . Notice that for sufficiently large k, by (1.6), f k w Y , which is the same as saying, f k w Σ . It is easy to see that lim k→∞ f k = f in the weak topology (in fact f k tends to f also in the strong topology), which is a contradiction to the hypothesis that T Σ = {f ∈ C ∞ (M, N ) : f w Σ} is open in the weak topology.
Remark 1.7. Our proof is inspired by the proof of the main theorem in [10] for the strong topology. We have explicitly constructed the maps to arrive at the contradiction while in [10] Trotman only mentions the existence of such maps.

Holomorphic case
Our next aim is to prove, if possible, a similar result for complex manifolds and holomorphic maps between them. We can extend the above result in the following way.
Let X be a complex analytic subvariety of N . Then, by a result of Whitney [11], X can be stratified into complex submanifolds. In fact, X can be stratified such that the strata fit together well in the sense that their tangent spaces satisfy nice regularity conditions. But we don't need these regularity conditions in the following. We have, This case turned out to be too easy to handle. Now, let M be a Stein manifold and N be any complex manifold. Let H(M, N ) denote the set of all holomorphic maps between M and N . If we regard M and N as smooth manifolds and denote C ∞ (M, N ) the set of all smooth maps, then C ∞ (M, N ) can be given two topologies, the weak topology and the strong (Whitney) topology (when M is compact, the two topologies are same), see Hirsch [6].
Since H(M, N ) ⊂ C ∞ (M, N ), we can give H(M, N ) the relative weak and strong topologies. However only the weak relative topology makes sense because the strong relative topology gives the discrete topology on H(M, N ). Thus we work with the weak topology on H(M, N ).
Many transversality theorems have been proved in this case (see Kaliman and Zaidenberg [7] and Forstnerič [3]). It is worth clarifying that in the transversality theorem (theorem 4.2) in [3] we don't need the complex analytic subvarieties to be a-regular. Also, in lemma 4.4 in [3] the complex analytic subvariety in the source manifold need not be a-regular and the one in the target manifold only has to be a-regular.
Here we give a partial converse to the lemma 4.4 in Forstnerič [3] which says:  Proof: Without loss of generality, suppose that the strata are of dimension at least 1. Since M is Stein, we can assume that M is a complex submanifold of some C p such that w = 0 ∈ C p , clearly p ≥ n − r.
We will again prove the proposition using contradiction. Suppose Σ is not aregular. Let X and Y be two distinct strata in Σ such that Y is not a-regular over X at some x ∈ X ∩ Y . Then, there exists a sequence In fact, by the curve selection lemma, we can choose y i to lie on an analytic curve.
Let v ∈ T x X be such that v / ∈ τ . Now, find a vector subspace H of C n of dimension n − r, as in proposition 1.6, such that and Let l be the dimension of D x ψ(H) + D x ψ(τ ) (n − r ≤ l < n). Now choose a basis {v 1 , . . . , v l } of D x ψ(H) + D x ψ(τ ) such that {v 1 , . . . , v n−r } forms a basis of D x ψ(H)and extend it to a basis {v 1 , . . . , v l , v l+1 , . . . , v n−1 , v } of C n where D x ψ(v) = v . Now, define a map L : C p → C n (this map is well defined because p > n − r) by, L(a 1 u 1 + · · · + a p u p ) = a 1 v 1 + a 2 v 2 + . . . + a n−r v n−r .
where v 1 , . . . , v n−r is a basis of D x ψ(H).
Notice that, L(T w M ) = D x ψ(H) and L is a holomorphic map. Let g : C p → N be defined by g = ψ −1 • L then g(w) = x, g is a holomorphic map and D w g = (D x ψ) −1 L because L is a linear map. Thus by (2.1), g w Σ.
We will now construct a sequence of holomorphic maps {g k } between C p and N which converge to g in the weak topology such that for sufficiently large k, g k w Σ.
As in proposition 1.6, first note that y k ∈ V for sufficiently large k. Thus, lim k→∞ D y k ψ(T y k Y ) = D x ψ(τ ). Now, choose a basis {v k 1 , v k 2 , . . . , v k l , v k l+1 , . . . , v k n−1 , v k } of D y k ψ(T y k N ) such that D y k ψ(T y k Y ) belongs to the span of {v k 1 , . . . , v k l } and lim Let H k be the subspace of D y k ψ(T y k N ) spanned by v k 1 , . . . , v k n−r . Then, by (2.5) we have lim k→∞ H k = D x ψ(H) and we have H k + D y k ψ(T y k Y ) = C n .
(2.6) since the left hand side of (2.6) is spanned by a subset of {v 1 , . . . , v l } and l < n. Now, define L k : C p → C n by the formula L k (a 1 u 1 + · · · + a p u p ) = ψ(y k ) + a 1 v k 1 + a 2 v k 2 + . . . + a n−r v k n−r and set g k = ψ −1 • L k . Clearly then, g k (w) = y k , by (2.5) the sequence of maps g k converges to the map g, L k (T w M ) = H k and by (2.6), g k w Σ. Now, let g| M = f and g k | M = f k , then clearly f and the f k 's are holomorphic maps between M and N , the sequence f k converges to f in the weak topology and f w Σ but for large enough k, f k w Σ, which is a contradiction to the hypothesis that the set T is open in the weak topology.