I want to construct the them spectrum $MU$ using the definition of spectrum associated to a $(B,f)$-structure. Here are the relevant definitions:
A $(B,f)$-structure is a collection of strictly commutative diagrams of the form
such that the maps $f_n$ are fibrations.
A $(B,f)$-manifold is a $n$-manifold whose classifying map for the normal bundle into $BO(n)$ lifts to $X_n$.
One then pullback the canonical bundle over $BO(n)$ via $f_n$ to a bundle over $X_n$ (call it $\gamma_n$) and then define $$MX(n):=\text{Thom}(\gamma_n)$$
I want to discuss the case of complex manifold, which according to Kochman's book it's a $(B,f)$-structure. Kochman uses a different way of indexing (more general), but I want to stick as much as it is possible to these definitions (Switzer's definitions)
The problem for an almost complex structure is that one has to deal with $odd$-indexes, and my guess would be to set $X_{2n+1}=0$ since clearly (real)-odd dimensional complex manifold don't exist. The problem is that I can't retrive the commutative diagram above, since the maps $f_n$ would be the (fibration substitute) of the inclusion $BU(n)\to BO(2n)$.
So in conclusion:
Is it possible to build $MU$ via a $(B,f)$-structure as defined in Switzer?
If you have a bunch of pointed spaces $X_0, X_2, X_4, \ldots$ with maps $f_{2n} : \Sigma^2 X_{2n} \to X_{2n+2}$ you can build a spectrum out of them by setting $X_{2n+1} = \Sigma X_{2n}$, taking the maps $\Sigma X_{2n} \to X_{2n+1}$ to be the identity and the maps $\Sigma X_{2n+1} \to X_{2n+2}$ to be $f_{2n}$.
Note that the resulting spectrum can definitely still have its odd homotopy groups be zero. The homotopy groups of a spectrum are $\pi_k(X) = \mathop{\mathrm{colim}}_{j \to \infty} \pi_{k+j}(X_j)$, which can be zero even if the individual $\pi_{k+j}(X_j)$ are not.