The top StiefelWhitney 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?

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 KronholmMay 4, 2011 at 3:52

7$\begingroup$ @Bill: the question "does the top StiefelWhitney class coincide with the Euler characteristic mod 2?" makes sense for any closed manifold. $\endgroup$– Tom ChurchMay 4, 2011 at 4:54
1 Answer
Every manifold is $Z_2$orientable, so the Euler class (with $Z_2$coefficients) is defined and coincides with the top StiefelWhitney class.
Back to your question: Yes, the top StiefelWhitney 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.