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This data is taken up to cobordism in the obvious way. Note that the spectrum $MTO(n)$ is not connective, which corresponds to the fact that the above makes sense for negative $k$. … with a proper map $\pi : E \to X$, an $n$-dimensional vector bundle $V \to E$, a stable isomorphism $\varphi : TE \oplus \epsilon^\ell \cong V \oplus \pi^*(TX) \oplus \epsilon^\ell$, again taken up to cobordism …

I think this theory is the same as unoriented bordism, when one tries to make sense of it. As it stands I don't think it makes sense, because I think that the expression "$\mathbb{Z}^w$" does not de …

The case $G=SO$ and $X=*$ is considered in Neumann, Walter D., Fibering over the circle within a bordism class. Math. Ann. 192 1971 191–192. where it is shown that a bordism class fibres over the c …

For a $G$-manifold $M$, taking the signature of the fixed points $M^G$ defines a homomorphism $\phi : \Omega_n^G \to \mathbb{Z}$, as if $W : M_0 \leadsto M_1$ is a cobordism then so is $W^G : M_0^G \leadsto …

This is cobordism-invariant by the usual argument. Now $\rho_i(id_M) = p_i(TM)$, and if $f : N \to M$ is a homeomorphism then $\rho_i(f) = (f^{-1})^*(p_i(TN))$. … So if the cobordism fundamental class were homeomorphism invariant, the integral Pontrjagin classes would be. This is false cf. chapter 4.4 of the Novikov conjecrure book by Kreck and Lueck. …

No: for example, there is no 1-manifold with Stiefel--Whitney number for $w_1$ equal to 1.