Orientability of a submanifold which is a preimage of a submanifold I am reading Milnor & Stacheff, Characteristic Classes, Chapter 18. There is a short review of smooth manifolds, and there is a following statement:
Suppose $f:M\to N$ is a smooth map between smooth manifolds, and suppose $f$ is transverse to a submanifold $Y\subset N$ (so that $f^{-1}(Y)$ is a submanifold of $M$). If $\nu^k$ is the normal bundle of $Y$ in $N$, then the induce bundle over $f^{-1}(Y)$ from $\nu^k$ by $f$ can be idenfied with the normal bundle of $f^{-1}(Y)$ in $M$. In particular, if $\nu^k$ is an oriented vector bundle, and if $M$ is an oriented manifold, then $f^{-1}(Y)$ is also an oriented manifold.
I see that (assuming that $M,N$ are Riemannian) the normal bundle of $f^{-1}(Y)$ can be identified with the induced bundle $f^*\nu^k$, but I can't see how the last statement follows. Am I missing a elementary theorem or something?
 A: The point is that a "difference" of two oriented bundles is oriented. On the level of vector spaces, if $W=V\oplus U$ we can say that the basis $b_V$ is positively oriented (for $W$) if when combined with any positively oriented basis $b_U$ for $U$ one gets positively oriented basis $b_W=(b_V, b_U)$ for $W$ (you can check that this is well-defined, basically because determinant of of a block diagonal matrix is the product of the determinants of the two blocks).
In fancier language, $\Lambda^{\dim W} W \equiv \Lambda^{\dim V} V \otimes \Lambda^{\dim U} U  $, so that orientation of any two of the vector spaces (identification of the top exterior power with $\mathbb{R}$) produces orientation of the remaining one. The same is then true for bundles, by repeating  this same construction over every point in the base.
In your case $W=TM$, $U=f^*\nu^k$, and $V=T f^{-1}(Y)$.
A warning is in order: One could insert various factors of $(-1)^{\dim X}$ or $(-1)^{\dim X+1}$ into the above formulas. Picking the "right" factors in a way that makes things "nice" downstream is quite non-trivial.  So while everyone will agree that $f^{-1}(Y)$ is orientable, the actual choice of an orientation may be subject to differing conventions.
