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On perturbations of continuous maps

We give sufficient conditions for the following problem: given a topological space X, a metric space Y, a subspace Z of Y, and a continuous map f from X to Y, is it possible, by applying to f an arbitrarily small perturbation, to ensure that f(X) does not meet Z? We also give a relative variant: if f(X') does not meet Z for a certain subset X' of X, then we may keep f unchanged on X'. We also develop a variant for continuous sections of fibrations, and discuss some applications to matrix perturbation theory.

It is worth noting that being X-avoidable is a property of the pair (Z, Y ) not an intrinsic property of Z; nevertheless, when the context does not make this ambiguous, we will let Y be implicit and refer to Z alone.
Thus our main problem can be rephrased as follows.
Problem 1.2 Given a topological space X and a metric space Y , which subsets of Y are X-avoidable?
To fix the terminology, let us take this definition.

Definition 1.3
A manifold of dimension n ∈ N is a Hausdorff space where every point has a neighborhood that is homeomorphic to R n .
Here is our main result.
Theorem (See Theorem 3.4) Let n, q ∈ N with n < q. Let X be a normal space of Lebesgue covering dimension n. Let Y be a metric topological manifold. Let Z be a subset of Y of Lipschitz codimension (see Definition 3.2) at least q. Then Z is X-avoidable.
Notice that here, while Y is required to be a metric manifold, much greater generality is allowed for X.
We also give results showing that unions of X-avoidable sets are still X-avoidable: first a general result on finite unions (Proposition 3.5) and then, in the locally compact case, a result on countably infinite unions (Theorem 3.6). We then obtain a relative variant (Proposition 4.1): if the restriction of the map f to a certain closed subset C of X already avoids Z, then we may perturb f outside of C to make it avoid Z everywhere, without modifying f on C. We also prove a variant for continuous sections of locally trivial fibrations (Theorem 5.1) under the assumption that the base space is paracompact. Finally, in Section 6 we discuss some applications to matrix perturbation theory that motivated this work.

Review of Some Dimension Theory
Here we review just a few selected notions and results, with no aim to offer a general review of this topic. See [2] or [3] for the general theory. Let us first recall the classical notion of Lebesgue covering dimension, which we will just call dimension. Given a topological space X, a refinement of an open covering (U i ) i∈I of X is an open covering (V j ) j∈ J of X such that for all j ∈ J, there exists i ∈ I such that V j ⊂ U i . [2] or [3]) Let d ∈ N. A topological space X is said to have dimension at most d if any open covering of it has a refinement (V j ) j∈ J such that for all x ∈ X, the set { j ∈ J, x ∈ V j } has at most d + 1 elements.

Definition 2.1 (See
Obviously, the dimension of X is then defined as the smallest d such that X has dimension at most d, or ∞ if no such d exists. It is true that R d has dimension d.

Definition 2.2
For any topological space X, we will let dim X denote its (Lebesgue covering) dimension.

B. Jacob
Let us also recall the notion of a normal space.

Definition 2.3
A topological space is said to be normal if any two disjoint closed subsets have disjoint neighborhoods.
We will need some variants and refinements of the implication (i)⇒(ii) in the above theorem. In order to obtain them, we will simply adapt the classical proof of that theorem. The following technical lemmas are used in that proof. Here, by ϕ(F) = λ we mean that ϕ(x) = λ for all x ∈ F.

Perturbations of Continuous Maps
The proof of the following lemma follows very closely the classical proof of Theorem 2.4. Our only change is to replace the uniform topology by the source limitation topology, i.e., replace the constant ε by a function on X. This is a straightforward adaptation, but it will allow us to do much more than the classical theorem allowed. Lemma 3.1 Let n ∈ N. Let X be a normal space such that dim X n. For any f ∈ C(X, R n+1 ) and for any ε ∈ C(X, R >0 ), there exists g ∈ C(X, R n+1 ) such that for all x ∈ X, g(x) − f (x) < ε(x) and g(x) = 0.
Since X is normal, by Urysohn's Lemma 2.5, for all i there exists a continuous func- We may therefore define a continuous function g i : X → R by letting, for all x ∈ X, for all x and all i, and therefore It remains to show that g(x) = 0 for all x ∈ X. Suppose that g(x) = 0 for some x ∈ X. Then g i (x) = 0 for i = 1, . . . , n + 1. It follows that Let us now introduce the notion of "Lipschitz codimension".
Let Y be a metric topological manifold of dimension p. Let Z be a subset of Y . One says that Z has Lipschitz codimension at least q in Y if there exists an open covering (U i ) i∈I of Y and, for all i ∈ I, a homeomorphism ϕ i from U i to an open subset of R p , such that: is a subset of R p−q seen as the vector subspace of R p consisting of vectors ending with q zeros.
The next step is to prove that we can locally avoid Z, that is, separately in each open set U i as in Definition 3.2: Lemma 3.3 Let n, p, q ∈ N with q > n, let X be a normal space of dimension n, let Y be a metric topological manifold of dimension p, with metric denoted by d, let Z be a subset of Y of Lipschitz codimension at least q, let (U i ) i∈I be an open covering of Y , and let (ϕ i ) i∈I be a family as in Definition By Lemma 3.1 applied to α and the function ε ′ , there exists a continuous map β from and β(x) = 0. Let ρ : R p → R p−q be the map discarding the q last components. Define a map γ : Also notice that for all x ∈ V i , γ(x) does not belong to the closure of ϕ i (Z), since the q last components of γ(x) are β(x) = 0, and by Definition 3.2 we know that any vector in ϕ i (Z) has its last q components equal to 0.
Since γ is a continuous map from V i to ϕ i (U i ), we may let . Also note that for all x ∈ V i , since γ(x) does not belong to the closure of ϕ i (Z), it follows that g i (x) does not belong to the closure of Z. Finally, by definition of the function ε ′ , we may extend g i to a continuous map from X to Y by letting Let us now glue the local charts together to prove the main theorem of this section.

Theorem 3.4
Let n, q ∈ N with q > n, let X be a normal space of dimension n, let Y be a metric topological manifold, and let Z be a subset of Y of Lipschitz codimension at least q. Then Z is X-avoidable.
Proof Let d be the metric on Y , let ε ∈ C(X, R >0 ), and let f ∈ C(X, Y ). We have to show that there exists g ∈ C(X, Y ) such that for all x ∈ X, d( f (x), g(x)) < ε(x) and g(x) does not belong to the closure of Z. Let (U i ) i∈I be an open covering of Y and (ϕ i )) i∈I be a family as given by Definition 3.2, as Z has codimension at least q in Y .
Since Y is metrizable, it is paracompact, and therefore we may assume without loss of generality that the covering (U i ) i∈I is locally finite -indeed, the existence of the corresponding family (ϕ i )) i∈I in Definition 3.2 passes to refinements. Again since Y is paracompact, there is a partition of unity (u i ) i∈I subordinate to (U i ) i∈I , and we may replace U i by the support of u i so that (u i ) i∈I is precisely subordinate to (U i ) i∈I . For i ∈ I, let V i = f −1 (U i ), and let v i = u i • f , so that (V i ) i∈I is a locally finite open covering of X and (v i ) i∈I is a partition of unity precisely subordinate to it. Let be a well-ordering on I. We assume without loss of generality that (I, ) is an ordinal. For x ∈ X and i ∈ I, let Let us prove by transfinite induction that for any ordinal i with i I, letting Zero case: The case i = 0 follows immediately from Lemma 3.3, as we have X 0 = V 0 .
Successor/limit case: Suppose now that the induction hypothesis is known to hold for all j < i where i is a fixed ordinal, 0 < i I. Let us show that it also holds for i. For all j < i we have a map h j as given by the induction hypothesis. Recall that (V j ) j∈I is a locally finite covering of X, so in particular, for all x ∈ X, the set of all j ∈ I such that x ∈ V j is finite. For all x ∈ X, let j x ∈ I be the greatest j < i such that x ∈ V j , and let h i (x) = g jx (x). This defines a map h i from X to Y . Let us check that it is continuous. Again by local finiteness of the covering (V j ) j∈I , for any j < i, for any x ∈ V j , there is a neighborhood W of x that intersects only finitely many of the V k , for k ∈ I. Letting k be the greatest among these finitely many indices, we see that h i agrees with g k on W , hence is continuous at x. Thus the map h i is continuous. Apply Lemma 3.3 to the map h i , the local chart (U i , ϕ i ), and the epsilon-function v i ε. Call g i the resulting map. It is then immediate to check that g i has the desired properties, showing that our induction hypothesis holds for i.
The following auxiliary results allow us to show that certain unions of avoidable subsets are avoidable. They illustrate again how the source limitation topology is more suitable than the uniform topology here.
Let us start with finite unions.

Proposition 3.5 Let X be a normal space, let Y be a metric topological manifold, let n ∈ N * , and let (Z i ) 1 i n be a family of X-avoidable closed subsets of Y . Then their union 1 i n Z i is X-avoidable.
Proof By induction on n, the case n = 1 is vacuously true. Suppose that the results holds for a fixed n and let us establish it for n+1. Let f ∈ C(X, Y ) and ε ∈ C(X, R >0 ).
We may assume that f already avoids Z 1 , . . . , Z n . Let us show that it avoids Z n+1 . Define a function η on X by

B. Jacob
Notice that η(x) > 0 for all x ∈ X because the Z i are closed. Apply Theorem 3.4 to the map f , the subset Z n+1 , and the function η. The resulting map g avoids Z n+1 by construction, and it still avoids Z 1 , . . . , Z n because of our particular choice of η.
Let us now prove that, when X is locally compact, we may actually avoid the union of countably many closed subsets. Theorem 3.6 Let X be a normal, locally compact space, let Y be a metric topological manifold, and let (Z i ) i∈N be a family of X-avoidable closed subsets of Y . Then their union i∈N Z i is X-avoidable.
Proof Let d be the metric on Y . For k ∈ N, let W k = i k Z i . Let f ∈ C(X, Y ) and ε ∈ C(X, R >0 ). Let us construct a sequence (g k ) k∈N of continuous maps from X to Y such that: for all x ∈ X.
The existence of g 0 follows from the X-avoidability of Z 0 . Let us now suppose that g 0 , . . . , g k is already constructed for some fixed k, and let us construct g k+1 . Notice that η k (x) > 0 because the Z i are closed. Apply Theorem 3.4 to the map g k , the function η k , and the subset Z k+1 . The resulting map g k+1 has the desired properties. This completes the construction of the advertised sequence (g k ) k∈N . It follows from the above-listed properties of that sequence that it converges uniformly on every compact subset of X (since the sequence (2 −k−2 ) k∈N is summable). Let g be its limit. Since X is locally compact, it follows that g is a continuous map from X to Y . Notice that for all x ∈ X, we have d( f (x), g(x)) < ε(x) (since 1/2 + k∈N 2 −k−2 = 1). It remains to show that g actually avoids i∈N Z i . Let x ∈ X and i ∈ N. We want to show that g(x) ∈ Z i . We have g i (x) ∈ Z i . For j i, let δ j = d(g j (x), Z i ). We have δ j > 0, since Z j is closed, and by construction we know that δ j+1

Relative Variant
One of the benefits of the choice of the source limitation topology is that we can easily give a relative version of the above results, i.e., if f already avoids Z on a certain closed subset of X, we can choose g to agree with f on that subset. However, we will have to make the further assumption that the base space is perfectly normal.
Recall that a space X is perfectly normal if for any disjoint closed subsets E, F of X, there exists a continuous function ϕ from X to [0; 1] such that (ϕ(x) = 0 ⇔ x ∈ E) and (ϕ(x) = 1 ⇔ x ∈ F). By contrast, if X is only normal, then the function ϕ given by Urysohn's lemma may take the values 0 or 1 outside of E and F. For instance, metrizable spaces are perfectly normal. Proposition 4.1 Let X be a perfectly normal space, let C be a closed subset of X, let X ′ be the complement of C in X, let Y be a metric topological manifold, with metric d, let Z be a X ′ -avoidable subset of Y , let f be a continuous map from X to Y , and let ε be a continuous function from X to R >0 . Suppose that Then there exists a continuous map g from X to Y such that for all x ∈ C, g(x) = f (x) and for all x ∈ X, g(x) ∈ Z and d( f (x), g(x)) < ε(x).
Proof Since X is perfectly normal and C is closed, there exists a continuous function η from X to [0; 1] vanishing exactly on C. Let ε ′ be the restriction of ηε to X ′ , and let f ′ be the restriction of f to X ′ . Since Z is X ′ -avoidable, there exists a continuous map g ′ from X ′ to Y such that for all x ∈ X ′ , g ′ (x) ∈ Z and d( f ′ (x), g ′ (x)) < ε ′ (x). Finally, extend g ′ into a continuous map g from X to Y by letting g(x) = f (x) for all x ∈ X ′ .

Variant for Locally Trivial Fibrations
In this section, we extend our results to the case of locally trivial fibrations under the additional assumption that the base space X is paracompact. Recall that paracompact spaces are normal (Dieudonné's theorem). The proof is similar to that of Theorem 3.4, except that we now work on the base space X of the fibration instead of working on Y (which is now the fibre of the fibration), whence the need to make the paracompactness assumption on X.
Theorem 5.1 Let X be a paracompact space, let Y be a metric topological manifold, with metric d, let Z be a subset of Y that is U -avoidable for all open subsets U of X, let A be a locally trivial fibration over X with fibre Y , let B be a locally trivial sub-fibration of A over X with fibre Z, let f be a continuous section of A over X, and let ε be a continuous function from X to R >0 . Then • there exists a continuous section g of A over X such that for all x ∈ X, g(x) ∈ B x and d( f (x), g(x)) < ε(x); • if, moreover, X is perfectly normal and C is a closed subset such that for all x ∈ C, f (x) ∈ B x , then the map g may be taken so that for all x ∈ C, g(x) = f (x).
Proof Once the first statement is proved, the proof of the second statement is similar to that of Proposition 4.1, so let us only prove the first statement here. Let ε ∈ C(X, R >0 ), and let f be a continuous section of A over X. We have to show that there exists a continuous section g of A over X such that for all x ∈ X, d( f (x), g(x)) < ε(x) and g(x) does not belong to the B x . Let (U i ) i∈I be an open covering of X trivializing A and B, and let (ϕ i )) i∈I be a family of maps as in Definition 3.2 applied to Z. Since X is paracompact, we may assume without loss of generality that the covering (U i ) i∈I is locally finite -indeed, the existence of the corresponding family (ϕ i )) i∈I in Definition 3.2 passes to refinements. Again since X is paracompact, there is a partition