The Unapologetic Mathematician

Mathematics for the interested outsider

Exact sequences split

Now that we know the splitting lemma, we can show that every short exact sequence of vector spaces splits!

To see this, we’ll need to refine an earlier result. Remember how we showed that every vector space has a basis. We looked for maximal linearly independent sets and used Zorn’s lemma to assert that they existed.

Here’s how we’re going to refine this result: start with a collection S of linearly independent vectors. Then we don’t just look for any maximal collection, but specifically for a maximal collection containing S. Clearly if we have a linearly independent set B containing S which is maximal among such sets, it is also maximal among all linearly independent sets — it is a basis. On the other hand, the previous argument (with Zorn’s lemma) says that such a maximal linearly independent set must exist.

What does this mean? It says that any linearly independent set can be completed to a basis. If we start with the empty set (which is trivially linearly independent) then we get a basis, just like before. So we recover the same old result as before.

But look what we can do now! Take a short exact sequence

\mathbf{0}\rightarrow U\xrightarrow{S}V\xrightarrow{T}W\rightarrow\mathbf{0}

and pick any basis \left\{e_i\right\}_{i\in\mathcal{I}} of U (notice that we’re using a generic, possibly infinite, index set). Now hit this basis with S to get \left\{f_i\right\}_{i\in\mathcal{I}} with f_i=S(e_i). I say that this is a linearly independent set in V.

Why is this? Well, let’s say that there’s a linear combination r^if_i=0 (only finitely many of the r^i can be nonzero here). This linear combination is in the image of S, since we can write it as r^iS(e_i)=S(r^ie_i). But exactness tells us that S is injective, and so we have r^ie_i=0. But then all the r^i have to vanish, since the e_i form a basis!

So we’ve got a linearly independent set \left\{f_i\right\}_{i\in\mathcal{I}} in V. We now complete this to a maximal linearly independent set \left\{f_i\right\}_{i\in\mathcal{I}'} with \mathcal{I}\subseteq\mathcal{I}'. This is now a basis for V, which contains the image of all the basis elements from U.

Now turn this around and define a linear transformation S':V\rightarrow U by specifying its values on this basis. Set S'(f_i)=e_i for i\in\mathcal{I} and S'(f_i)=0 for i\in\mathcal{I}'\setminus\mathcal{I}. Then the composition S'\circ S is the identity on U (check it on our basis), and so the sequence splits, as we said.

Notice here that all the Zorniness only matters for infinite-dimensional vector spaces. Everything we’ve done here works in \mathbf{FinVec}(\mathbb{F}) without ever having to worry about such set-theoretic problems. However, given my politics I have no problem with using the Axiom of Choice when push comes to shove. I’m just pointing this out in case you’re the squeamish sort.

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June 26, 2008 - Posted by | Algebra, Linear Algebra

11 Comments »

  1. [...] since every short exact sequence splits we have an isomorphism . This is the content of the rank-nullity [...]

    Pingback by The Rank-Nullity Theorem « The Unapologetic Mathematician | June 27, 2008 | Reply

  2. [...] Today I just want to point out a dual proposition to the one I refined last week. At that time we stated that any linearly independent set can be expanded to a basis. This followed [...]

    Pingback by Spanning sets « The Unapologetic Mathematician | June 30, 2008 | Reply

  3. [...] the system is inhomogenous in general, and as such it might not have any solutions. Since every short exact sequence of vector spaces splits we can write . Then the vector will have some component in the image of , and some component in [...]

    Pingback by The Index of a Linear Map « The Unapologetic Mathematician | July 22, 2008 | Reply

  4. [...] the map allows us to break up as (since short exact sequences split). On the other hand, considering the map allows us to break up as . Exactness tells us that , [...]

    Pingback by The Euler Characteristic of an Exact Sequence Vanishes « The Unapologetic Mathematician | July 23, 2008 | Reply

  5. [...] say we’re given a list of linearly independent vectors in . They must be a basis, since any linearly independent set can be completed to a basis, and a basis of must have exactly elements, which we already have. Then we can use the as the [...]

    Pingback by The General Linear Groups « The Unapologetic Mathematician | October 20, 2008 | Reply

  6. [...] we know that some of the most important exact sequences are short exact sequences. We also saw that every short exact sequence of vector spaces splits. So does the same hold for representations? It turns out that no, they don’t always, and [...]

    Pingback by Do Short Exact Sequences of Representations Split? « The Unapologetic Mathematician | December 17, 2008 | Reply

  7. [...] take a basis of . Since this is a linearly independent set spanning a subspace of , it can be completed to a basis for all of . Now we can use this basis of to write out the matrix of and use our formula from [...]

    Pingback by The Determinant of a Noninvertible Transformation « The Unapologetic Mathematician | January 14, 2009 | Reply

  8. [...] also know that any linearly independent set can be expanded to a basis. In fact, we can also extend any orthonormal collection of vectors to an orthonormal basis. Indeed, [...]

    Pingback by Orthonormal Bases « The Unapologetic Mathematician | April 29, 2009 | Reply

  9. [...] An important fact about the category of vector spaces is that all exact sequences split. That is, if we have a short exact [...]

    Pingback by Orthogonal Complements « The Unapologetic Mathematician | May 4, 2009 | Reply

  10. [...] when we needed to show that every vector space has a basis, or Tychonoff’s theorem, or that exact sequences of vector spaces split. So it’s sort of a mixed bag. In practice, most working mathematicians seem to be willing to [...]

    Pingback by Non-Lebesgue Measurable Sets « The Unapologetic Mathematician | April 24, 2010 | Reply

  11. [...] submodule — and — then we can pick a basis of . And then we know that we can extend this to a basis for all of : [...]

    Pingback by Reducibility « The Unapologetic Mathematician | September 23, 2010 | Reply


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