= Solution
Write $MU^*=\Omega_U^*$ for <complex cobordism>. A <complex orientation of a smooth map> is chosen stable complex normal data, not a complex structure on either manifold separately. More concretely, replace the continuous map by a homotopic smooth map $f_0$, choose an embedding
$$
j=(f_0,e):X\hookrightarrow Y\times\mathbb R^N,
$$
and give its <normal bundle> $\nu_j$ a <stable complex structure on a real vector bundle>. The auxiliary dimension can be chosen so the normal rank is even; further stabilization makes the <normal bundle> actually complex. Equivalence means stabilization by trivial complex summands and homotopy of the oriented embedding data. The stable identity
$$
\nu_j\oplus TX\cong f_0^*TY\oplus\mathbb R^N
$$
identifies this as orientation of the relative stable <normal bundle>. For odd relative codimension one uses an odd auxiliary suspension coordinate; the degree formula below is unchanged. This is the geometric factorization model for complex-oriented maps used in https://msp.org/agt/2018/18-7/agt-v18-n7-p.pdf[Algebraic and Geometric Topology 18, geometric cobordism discussion].
Let $d=\dim Y-\dim X$ and $r=\operatorname{rank}_{\mathbb R}\nu_j=d+N$. The chosen normal orientation gives a <Thom class> $u_{\nu_j}$ of degree $r$ and the <Thom isomorphism in complex cobordism>. For compact manifolds the <Pontryagin-Thom collapse> of a <tubular neighborhood> to its <Thom space> is
$$
c_j:Y_+\wedge S^N\longrightarrow\operatorname{Th}(\nu_j).
$$
For $a\in MU^q(X)$, define the <Gysin map in complex cobordism> by the composite
$$
MU^q(X)\xrightarrow{\ a\mapsto\pi^*a\,u_{\nu_j}\ }
\widetilde{MU}^{q+r}(\operatorname{Th}(\nu_j))
\xrightarrow{\ c_j^*\ }
\widetilde{MU}^{q+r}(Y_+\wedge S^N)
\xrightarrow{\ \text{desuspension}\ }
MU^{q+d}(Y).
$$
Thus \b[$f_!$ raises degree by the real relative codimension $d$]. This definition depends on the specified orientation and is invariant under the stated equivalences of its normal data. The plus sign on a space denotes adjoining a disjoint basepoint.
For the embedding $i:L\hookrightarrow M$ with normal rank $2n$, let $c:M_+\to\operatorname{Th}(\nu)$ be its tubular collapse. The Thom construction gives $i_!(1)=c^*u_\nu$. Restricted to $L$, this collapse is exactly the <zero section> $z:L\to\operatorname{Th}(\nu)$. Therefore
$$
\boxed{i^*i_!(1)=z^*u_\nu=e_{MU}(\nu)\in MU^{2n}(L).}
$$
This is the <self-intersection formula in complex cobordism>. For an actual complex rank-$n$ <normal bundle> its top <Chern class in complex cobordism> is the zero-section pullback of its <Thom class>, so the right side is $c_n^{MU}(\nu)$. Equivalently, under the <splitting principle for complex vector bundles> both are the product of the first <Chern classes> of the line summands. The same argument gives $i^*i_!(a)=a\,e_{MU}(\nu)$ for every $a$.
There is a real distinction if the printed word “stable” is interpreted literally. The <Euler class> of a stably complex real bundle need not equal $c_n^{MU}$ of its stable complex class. For the diagonal embedding $S^4\hookrightarrow S^4\times S^4$, the <normal bundle> is $TS^4$. It has a stable complex structure with
$$
TS^4\oplus\mathbb R^2\cong\mathbb R^6\cong\mathbb C^3,
$$
so its stable complex rank-two class is trivial and has second <Chern class> zero. However the ordinary integral image of $i^*i_!(1)$ is $e(TS^4)$, which evaluates to the <Euler characteristic> $2$. Thus the always-valid answer is the cobordism <Euler class>. The requested top-Chern identification is valid for an actual complex <normal bundle>, or if “top <Chern class>” here is explicitly being used as the Euler-class terminology for the chosen normal orientation.
Back to article page
