# Pullback bundle and Chern classes

The following is from Bott and Tu. If $$f : M \to N$$ is a map between two manifolds and $$E$$ is a complex bundle over $$N$$, then the pullback $$f^{-1}E$$ is a bundle over $$M$$. If the Chern classes of $$E$$ vanish, then the naturality property implies that so do those of $$f^{-1}E$$. Taking the Chern classes to be a measure of twisting of a bundle, we may assert that pulling back dilutes a bundle, i.e. makes it less twisted. One extreme is when $$f$$ is constant; in this case $$f^{-1}E$$ is trivial.

One may wonder if there exists a bundle which is so twisted that every bundle is a pullback of this universal bundle. Such a bundle indeed exists, at least for manifolds of finite type; it is the universal quotient bundle on the Grassmannian $$G_k(\Bbb C^n)$$ for $$n$$ sufficiently large.

I have some naive questions about this phrase. First, is there some simple example that would illustrate this "diluting" via $$f$$? If I consider for example $$f:S^1 \to S^1$$ and give the codomain the Möbius bundle, can I choose $$f$$ such that the pullback bundle $$f^{-1}E$$ untwists the Möbius bundle to $$S^1 \times \Bbb R$$ or $$S^1 \times \Bbb C$$?

Next, can I use this "twistness" to classify bundles? I.e. are bundles with the same twistness isomorphic? I guess that twistness means at least in the complex setting that the Chern classes are equal.

Yes, before discussing how to "untwist" the Möbius bundle, just recall that $$S^1 \simeq \mathbb R P^1$$ and that line bundles over $$M$$ are classified by the homotopy classes of maps $$[M, \mathbb R P^\infty] \simeq H^1(M,\mathbb Z_2)$$. So there are only two line bundles over $$S^1$$ up to isomorphism. As you can check that the Möbius band is not the trivial bundle, then any line bundle over $$S^1$$ must be trivial, or isomorphic to the Möbius band.
As is commonly done in algebraic geometry to denote line bundles over $$\mathbb C P^n$$, denote the trivial real line bundle on $$S^1$$ as $$\mathcal O(0)$$ and the Möbius band as $$\mathcal O(1)$$. As an exercise, you may check that, if $$f : S^1 \to S^1$$ is a smooth map of topological degree $$\deg f$$, then $$f^*\mathcal O(1) \simeq \mathcal O(1)^{\otimes \deg f} =: \mathcal O(\deg f)$$ Moreover, you can check that $$\mathcal O(n) \simeq \mathcal O(m)$$ if and only if $$m \equiv n$$ mod $$2$$, thus $$\mathcal O(n)$$ is trivial when $$n$$ is even and isomorphic to the Möbius band when $$n$$ is odd.
In general it's difficult to classify the bundle "twistness" via algebraic invariants. You have characteristic classes that can be helpful to distinguish non-isomorphic bundles, but having the converse is false in general. For example, the tangent bundle of $$S^2$$ has trivial total Stiefel-Whitney class but it is not the trivial bundle.