# Making A Mapping Injective

*Finding a set of nearly independent objects*

Wikipedia bio source |

Giuseppe Vitali was the mathematician who famously used the Axiom of Choice, in 1905, to give the first example of a non-measurable subset of the real numbers.

Today I want to discuss another of his results that is a powerful tool.

The existence of a set that cannot properly be assigned a measure was a surprise at the time, and still is a surprise. It is a wonderful example of the power of the Axiom of Choice. See this for details.

We are interested in another of his results that is more a theorem about coverings. It is the Vitali covering theorem–see this. The theorem shows that a certain type of covering—ah, we will explain the theorem in a moment.

The power of this theorem is that it can be used to construct various objects in analysis. There are now many applications of this theorem. It is a powerful tool that can be used to prove many nice results. I do not know of any—many?—applications of the existence of a non-measurable set. Do you know any?

## Making A Mapping Injective

Let’s look at an application of the Vitali theorem that may be new. But in any case it may help explain what the Vitali theorem is all about.

Suppose that . We can make the map surjective if we restrict to be equal to . It is not so simple to make the map injective, but we can in general do that also.

Theorem 1Let be a surjective function from to . Then there is a subset of so that is injective from to .

*Proof:* For each in select one from the set and place it into . Recall is the set of so that .This of course uses the Axiom of Choice to make the choices of which to choose. Then clearly is the required set.

The difficulty with this trivial theorem is that cannot be controlled easily if it is constructed via the Axiom of Choice. It could be a very complicated set. Our goal is to see how well we can control if we assume that the mapping is smooth.

How can we do better? The answer is quite a bit better if we assume that is a “nice” function. We give up surjectivity onto but only by a null set.

Theorem 2Suppose that is a surjective smooth map from to where and are open subsets of . Also suppose that locally is invertible. Then there is a subset of so that

- The complement of is a null set.
- The map is injective from to .

That is that for all distinct points and in , . Moreover the map from to is smooth.

## Set Coverings

How can we prove this theorem? An obvious idea is to do the following. Pick an open interval in so that for an open set in and so that is injective from to . Setting to clearly works: the map is injective on . This is far from the large set that we wish to have, but it is a start. The intuition is to select another open interval that is disjoint from so that again is injective from to . We can then add to our .

We can continue in this way and collect many open sets that we add to . Can we arrange that the union of these sets yield a so that is most of ? In general the answer is no. Suppose that the intervals are the following:

for Roughly we can only get about half of the space that the intervals cover and keep the chosen intervals disjoint. If we select then we cannot select since

Vitali’s theorem comes to the rescue. It allows us to avoid his problem, by insisting that intervals have an additional property.

## The Vitali Covering Theorem

The trick is to use a refinement of a set cover that allows a disjoint cover to exist for almost all of the target set. The next definition is critical to the Vitali covering theorem.

Definition 3Let be a subset of . Let be intervals over in some index set . We say these intervals are acoverof proved is a subset of the union of all the intervals. Say the intervals also are aVitalicover of provided for all points in and all , there is an interval that contains and .

The Vitali theorem is the following:

Theorem 4Let be a subset of . Let be intervals for in some index set . Assume that the family is a Vitali cover of . Then there is a countable subfamily of disjoints intervals in the family so that they cover all of except for possibly a null set.

The Vitali theorem can be extended to any finite dimensional space . Then intervals become disks and so on.

## Open Problems

Do you see how to prove Theorem 2 from Vitali’s theorem? The insight is now one can set up a Vitali covering of the space .

A result of Poincaré says that a surjective morphism of abelian varieties can be restricted to give an isomorphism.Cf Shafarevich Basic Algebraic Geometry. Analogues will be most interesting.

«are a cover of {U} proved {U}» you probably mean «provided»