Vtyjkyuk

Assume that $E$ is a (reduced) extraordinary multiplicative cohomology theory. Then it is called complex-oriented if there is a class $c^E_1 in E^2(mathbbCP^infty)$, such that the map $i^*: E^2(mathbbCP^infty) to E^2(mathbbCP^1) cong E^*(pt.)$, induced by the inclusion $i : mathbbCP^1 to mathbbCP^infty$, sends $i^*(c^E_1)=1 in E^*(pt.)$. However, this probably is the short story of the real meaning behind the complex-oriented idea. I found somewhere that this definition, implies the existence of a Thom class for $E$. In other words, for a complex orientation of $E$ suffices to specify a Thom class and vice versa. How do we prove that these two facts are equivalent?

The statement to show is:

$E$ is complex-oriented iff there is a Thom class $u_V in E^2k(X^V)$ for any complex vector bundle of rank $k$.

The $(Leftarrow)$ direction follows from choosing the tautological line bundle $L$ over $mathbbCP^infty$. We get a Thom class $u_L in E^2((mathbbCP^infty)^L) cong E^2(mathbbCP^infty)$ since $(mathbbCP^infty)^L simeq mathbbCP^infty$. By definition/by the Thom isomorphism, the restriction of the Thom class $u_L$ to a fiber gives the generator of $E^2(S^2)$.

For the $(Rightarrow)$ direction, if $V$ is a line bundle, we are done by pulling back the first Chern class $c_1^E$ along the map on Thom complexes induced classifying map for $V$. We can reduce the general case to the rank 1 case by the splitting principle.

Here are some more details. Suppose $V$ is a line bundle over $X$. There is a pullback of line bundles
beginarrayccc
V & rightarrow & L \
downarrow & & downarrow \
X & rightarrow & mathbbCP^infty.
endarray

There is an induced map on Thom complexes $X^V to (mathbbCP^infty)^L simeq mathbbCP^infty$. Since $E$ is complex-oriented, there is a first Chern class $c_1^E in E^2(mathbbCP^infty)$, which we can then pullback along this map to get an element $u_V in E^2(X^V)$. To show that $u_V$ is indeed a Thom class, we observe that it's true by complex-orientation $c_1^E$ begets a generator when you restrict it to a fiber of the Thom complex (which is $S^2$). Then we use the above pullback square to deduce that the fibers of the Thom complexes are identified with each other, so $u_V$ also restricts to a generator along a fiber.

Now suppose $V downarrow X$ is an arbitrary complex vector bundle of rank $k$, and the argument is to use the splitting principle to reduce this to the previous case. As you say, there is a little bit of work needed to extend this to complex-oriented cohomology, but the argument is essentially the same. Recall how the splitting principle for ordinary cohomology is proved. We need to show:

1. Pulling back $V$ along the map $mathbbPV to X$ yields a vector bundle that splits off a line bundle.


2. The induced map $E^*(X) to E^*(mathbbPV)$ is monic.

(1) is just geometry, and nothing changes from the original case. (2) is more interesting. We claim a fortiori that $E^* mathbbPV cong E^* X [t]/c_t(V)$. This follows from the Leray-Hirsch theorem for $E$-cohomology, and the essential ingredient of that theorem is the collapse of the Atiyah-Hirzebruch-Serre spectral sequence for $mathbbCP^k-1 to mathbbPV to X$. Using complex-orientability again, we see that the element $c_1^E in E^2 mathbbCP^k-1$ in the spectral sequence must be a permanent cycle, and using the ring structure we see that all of its powers are permanent cycles too. This verifies the condition of the Leray-Hirsch theorem, and the claim is proved.

Returning to the main argument, we can repeatedly apply the splitting principle to get some pullback
beginarrayccc
L_1 oplus cdots oplus L_k & rightarrow & V \
downarrow & & downarrow \
F & rightarrow & X.
endarray

We also get Thom classes $u_L_i in E^2(F^L_i)$. Using the external cup product and $F^L_1 oplus cdots oplus L_k cong F^L_1 wedge cdots wedge F^L_k$, we get a Thom class $u_L_1 cdots u_L_k in E^2k(F^L_1 oplus cdots oplus L_k)$. Moreover, this is in the image of $E^2k(X^V)$, and so specifies a well-defined Thom class for $X^V$.

Spelled out in this detail, it might perhaps be easier to use the complex orientation to compute $E^*(X^V)$ and directly establish a Thom isomorphism, and thus a Thom class, since this is essentially the argument for the Leray-Hirsch theorem above.

There was also a question in the comments about $k$ vs. $2k$. Just consider ordinary homology: $H^textodd(mathbbCP^infty) = 0$, so we really do want $H^2k$. This is an artifact of the fact that there are real spheres of odd dimension that complex geometry cannot see.