Let $G=\mathbb{Z}/2\mathbb{Z}$. Let $f\colon L \to N$ be a smooth map of connected smooth **closed** $n$-dimensional manifolds such that the induced map


$$f^\* \colon H^\*(N,G) \to H^\*(L,G)$$


is an isomorphism.


**Question**: Are the pull back of the Stiefel-Whitney classes of the tangent bundle of $N$ the Stiefel-Whitney classes of the tangent bundle of $L$?.


This is in fact true for the first Stiefel-Whitney class by considering coverings and degrees, but what about the higher degree classes?


**Motivation**: This came up because (relative) spin is important in defining Floer homology with $\mathbb{Z}$ coefficients. So I am in fact mostly interested in the following sub-question.


**Question**: In particular what about the second Stiefel-Whitney class in the case where both $N$ and $L$ are also assumed to be oriented? and if the answer is negative: what extra conditions do I need to make it positive?


The idea is that I apriori have to use $G$ coefficients, but can prove that it is a $G$-cohomology equivalence, and want to use that to start the argument over again with other coefficients, but for that I need this property of the second Stiefel-Whitney class. 


This sub-question and the relation to Floer homology is related to orientations in real $K$-theory and delooping in the following sense: take a map $h\colon X \to U/O$ by delooping we get a map $\Omega h \colon \Omega X \to \Omega U/O \simeq \mathbb{Z}\times BO$ which classifies a virtual bundle over the loop space of $X$. This bundle is oriented iff the original map composed with the canonical map $U/O \to BO$ classified a virtual $0$-dimensional bundle with vanishing second Stiefel-Whitney class. This is the main point of why orientations in Floer homology is initimitely linked with spin! In the case of a Lagrangian sub-manifold $L\subset T^\*N$ the difference of the tangent bundles precisely defines such a map $L \to U/O$ ($U(n)/O(n)$ classifies Lagrangians in $\mathbb{R}^{2n}$) such that the composition to $BO$ classifies the virtual bundle $TN-TL$. So in fact you may add this lifting property as an extra condition to the sub-question if you like, and then I would lose no generality. I believe that this condition implies that all the relative Prontryagin classes vanishes, which may be helpfull.


ADDED: in light of the answer, all this motivation and these extra possible assumptions are not important nor relevant for the actual question.



Are the stiefel-Whitney classes of the tangent bundle determined by the mod 2 cohomology?
The answer to the question is positive, due to Wu's formula. See e.g. Milnor-Stasheff, Characteristic classes, lemma 11.13 and theorem 11.14. In fact, all one needs to compute the Stiefel-Whitney classes of a smooth compact manifold (orientable or not) is the cohomology mod 2 (as an algebra) and the action of the Steenrod algebra on it. Both structures are preserved under cohomology isomorphisms induced by continuous maps.


