show/hide this revision's text 3 The answer is wrong; [made Community Wiki]

It seems like i am making

There was a dumb mistake somewhere. Until i fix it i will leave the mistaken answer. Please see the commens.

(Temporarily ignore)The fundamental error in my answeris yes for equivalent but simpler sounding reasons.

The stiefel-whitney classes are natural, this is taken as an axiom error was in Milnor and Stasheff. I recommend checking out Chapter 4 of their book. This will show you just how far one can get with misunderstanding the axioms, which do in fact hold given naturality of the construction above in algori's excellent answer. We have $w_i(f^*\tau_N)=f^*w_i(\tau_N)$ and $f^*w_i(\tau_N)=w_i(\tau_L)$ since w_i$. $f^: H^(L) f: L \to H^*(N)$ is N$ inducing an isomorphism and there are obviously bundle maps that cover f between in cohomology does not imply anything about the various induced map of the tangent bundles. I like This is where i made my fundamental error. Please see the axiomatic approach comments for details or look at earlier versions of this answer.

I had hoped that there would be a bit more , but you it really just boils down to "axiomatic" proof in the naturality sense of the stiefel whitney classesMilnor and Stasheff.

I hope i am not making some dumb mistake, If anyone comes up with one please let me know if i amfeel free to put one here.(Temporarily ignore)

Thanks to Tom and Dan for the comments!
show/hide this revision's text 2 added 172 characters in body

It seems like i am making a dumb mistake somewhere. Until i fix it i will leave the mistaken answer. Please see the commens.

(Temporarily ignore) The answer is yes for equivalent but simpler sounding reasons. The stiefel-whitney classes are natural, this is taken as an axiom in Milnor and Stasheff. I recommend checking out Chapter 4 of their book. This will show you just how far one can get with the axioms, which do in fact hold given the construction above in algori's excellent answer. We have $w_i(f^*\tau_N)=f^*w_i(\tau_N)$ and $f^*w_i(\tau_N)=w_i(\tau_L)$ since $f^: H^(L) \to H^*(N)$ is an isomorphism and there are obviously bundle maps that cover f between the various bundles. I like the axiomatic approach a bit more, but you it really just boils down to the naturality of the stiefel whitney classes.

I hope i am not making some dumb mistake, please let me know if i am. (Temporarily ignore)
show/hide this revision's text 1

The answer is yes for equivalent but simpler sounding reasons. The stiefel-whitney classes are natural, this is taken as an axiom in Milnor and Stasheff. I recommend checking out Chapter 4 of their book. This will show you just how far one can get with the axioms, which do in fact hold given the construction above in algori's excellent answer. We have $w_i(f^*\tau_N)=f^*w_i(\tau_N)$ and $f^*w_i(\tau_N)=w_i(\tau_L)$ since $f^: H^(L) \to H^*(N)$ is an isomorphism and there are obviously bundle maps that cover f between the various bundles. I like the axiomatic approach a bit more, but you it really just boils down to the naturality of the stiefel whitney classes.

I hope i am not making some dumb mistake, please let me know if i am.