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If $M$ is a smooth manifold, the total Stiefel-Whitney class of $M$ is defined to by the total Stiefel-Whitney class of the tangent bundle, i.e. $w(M) := w(TM)$.

If $M$ is a closed smooth manifold, then $w(M) = \operatorname{Sq}(\nu)$ where $\operatorname{Sq}$ is the total Steenrod square and $\nu$ is the total Wu class. One consequence of this fact is that $w(M)$ depends only on $H^{\bullet}(M; \mathbb{Z}_2)$ as a graded module over the Steenrod algebra. In particular, if $f : M \to N$ is a homotopy equivalence of closed smooth manifolds, then $f^*w(N) = w(M)$. So, for example, any exotic sphere has total Stiefel-Whitney class $1$ (because it is homotopy equivalent, in fact homeomorphic, to the standard smooth sphere).

Note that the expression $\operatorname{Sq}(\nu)$ does not depend on the smooth structure, so one could define the total Stiefel-Whitney class of a closed topological manifold by this expression, despite the fact such a manifold has no natural vector bundle.

Has this definition of Stiefel-Whitney classes been used to deduce any information about closed topological manifolds which admit no smooth structure?

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    $\begingroup$ One example of how this definition can be used is to deduce topological non-embedding theorems of some manifolds into some other manifolds (regardless of wether they admit smooth structures). For details see the algebraic topology texts by Spanier and by Dold (search the index of Stiefel-Whitney and Euler class). $\endgroup$ Jan 24, 2016 at 15:15

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