Chern Classes via Grassmannians

Question

Let $X$ be some smooth manifold and let $p:E\to X$ be a complex vector bundle of finite rank. Our goal is to classify $E$ up to bundle-isomorphisms.

According to Atiyah's K-Theory book, there is a a bijection between the isomorphism classes $Vect_n(X)$ of rank $n$ vector bundles over $X$ and $$[X\to G_n(\mathbb{C}^\infty)]$$ the homotopy classes of maps $f:X\to G_n(\mathbb{C}^\infty)$ (the infinite Grassmannian).

Thus, for each $E$ there is a unique homotopy class $[f_E:X\to G_n(\mathbb{C}^\infty)]$ determined by $E$.

This in turn induces a map on any contravariant functor $\mathcal{F}$ from the homotopy category to an algebraic category $$ \mathcal{F}([f_E]): \mathcal{F}(G_n(\mathbb{C}^\infty))\to \mathcal{F}(X) $$

Then $\mathcal{F}([f_E]) \neq \mathcal{F}([f_\tilde{E}])$ implies $E \ncong\tilde{E}$, because $E\cong\tilde{E}$ implies $[f_E]=[f_\tilde{E}]$.

Question 1: Is there any functor $\mathcal{F}$ such that $\mathcal{F}([f_E]) = \mathcal{F}([f_\tilde{E}])$ implies $E\cong \tilde{E}$? (perhaps for sufficient conditions on $X$?)

Question 2: Is it the case that the $k$th Chern class of $E$ is given by $\mathcal{F}([f_E])(c_k)$ where $c_k\in\mathcal{F}(G_n(\mathbb{C}^\infty))$, and $\mathcal{F}$ is either the de Rham or the singular $k$th cohomology functor? If yes, is there a difference between using de Rham cohomology or singular cohomology? Also, how to determine $c_k$ concretely? I realize that $c_k$ is defined as the $k$th Chern class of the classifying bundle (the bundle quotient $(G_n(\mathbb{C}^m)\times \mathbb{C}^m)/F$ where $F$ is the tautological $n$-plane bundle), but is there a way to explain the definition of $c_k$ using this language without reverting to other constructions such as the Euler characteristic or invariant polynomials?

Question 3: What could one obtain by using other functors? For instance, the singular homology functor? The $n$th homotopy group functor? Why does one never hear about these? Also, why is it that one only uses the even-degree cohomology functor in this context?

Is there a textbook that is very similar to Chern's original 1945 paper, but which uses regular vector bundles instead of sphere bundles?

Answer 1

1. It depends on what you mean by an algebraic category. It's certainly not enough to take anything that looks like cohomology, since there are nontrivial spaces which have no nontrivial cohomology (with respect to any generalized cohomology theory).

2. Yes for singular cohomology. For de Rham it's unclear what one might to mean by the de Rham cohomology of an infinite-dimensional thing.

3. Homology gives you the same information as cohomology. Homotopy is hard to compute with. The infinite Grassmannian has no odd-dimensional cohomology. Look up K-theory.