I am currently trying to read Quillen's Cobordism paper. Still on the first page though so the emphasis is on the trying :P.
A map of manifolds $f: Z \to X$ can be give an complex orientation in the following way: If $\dim (f) = \dim T_z Z - \dim T_{f(z)}X$ is even then an orientation for $f$ is an equivalence class of factorizations $Z \to E \to X$ where $E$ is a complex bundle over $X$ and the map $Z \to E$ is an embedding with a complex structure on the normal bundle.
I am assuming that the definition of the normal bundle is: $$ \upsilon_{i(z)}=T_{i(z)}E/i_*(T_zZ) $$ above each point of $Z$. I am taking everything to be real the previous equation. The dimension of this is: $\dim E - \dim Z$. Considering the whole bundle over $Z \subset E$ the dimension of the normal bundle is $ \upsilon_iZ =\dim E - \dim Z + \dim Z =\dim E$.
Obviously this is where I am mistaken. This would mean that the requirement that the function be even dimensional is entirely arbitrary.
I cannot think of anything nice to fix this.
Two factorizations are equivalent if there is an isotopy between them that preserves the complex structure. This means if $E$ and $E'$ are two factorisations and $E,E'$ are subbundles of $E''$ then there is an embedding endowed with a complex structure on the normal bundle $i'': Z \times I \to E \times I$ that restricts to $i$ and $i'$ with the same complex structure on either end.
Quillen says $i'':X \times I \to E \times I$ but that does not make sense. I haven't gotten to this bit yet.
For $\dim f$ odd consider $(f, \epsilon): Z \to X \times \mathbb{R}$ where $\epsilon(z)=0$. Then for generalised maps split it up.
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This is a different way to look at it given to me by a professor in my department. I have added little bit of explanation to what he gave me.
The map $\pi$ is a submersion so we can pull back $TX$ to $TE$. This map has a kernel, lets call it $T_FE$, which is fiberwise, over $X$, isomorphic to $TF$. This gives a short exact sequence of bundles: $$ 0 \to T_FE \to TE \stackrel{T_\pi}{\to} \pi^*TX \to 0 $$ (This whole thing is a little like lifting $F \to E \to X$ to a SES of bundles in $TE$.) This SES splits $TE=\pi^*TX+T_FE$
With everything as in Quillen (just like above), now look at the normal bundle $$\begin{align} v_i & = TE - i_*TZ \\ &= \pi^*TX + T_FE - i_*TZ \end{align} $$ Which has the dimension work out in both the even and odd cases for $f$ and to me at least feels better.
