The top Stiefel-Whitney class of an orientable manifold is the Euler characteristic mod 2, since it is the mod 2 reduction of the Euler class. Does this still hold for an unorientable manifold?
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2$\begingroup$ The Euler class of (the tangent bundle of) a manifold only makes sense for an oriented manifold. So, does what still hold? $\endgroup$– Bill KronholmCommented May 4, 2011 at 3:52
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7$\begingroup$ @Bill: the question "does the top Stiefel-Whitney class coincide with the Euler characteristic mod 2?" makes sense for any closed manifold. $\endgroup$– Tom ChurchCommented May 4, 2011 at 4:54
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Every manifold is $Z_2$-orientable, so the Euler class (with $Z_2$-coefficients) is defined and coincides with the top Stiefel-Whitney class.
Back to your question: Yes, the top Stiefel-Whitney class evaluated on the fundamental class of the manifold is equal to the Euler characteristic mod 2 - regardless of the $Z$-orientability of the manifold.
This is Corollary 11.12 in Milnor, Stasheff: Characteristic Classes.
