Recall that the Stiefel-Whitney classes of a smooth manifold are defined to be those of its tangent bundle - this definition doesn't extend to topological manifolds as they don't have a tangent bundle. Wu's theorem states that for a closed smooth manifold, $w = \operatorname{Sq}(\nu)$. The expression $\operatorname{Sq}(\nu)$ makes sense for a closed topological manifold and therefore serves as a definition for the Stiefel-Whitney classes on such a manifold.


Recall that if $M$ is a closed smooth $n$-dimensional manifold, then $w\_n(M)$ is equal to the mod $2$ reduction of $e(M)$, see Corollary 11.12 of Milnor and Stasheff's *Characteristic Classes*. In particular, the Stiefel-Whitney number $\langle w\_n(M), [M]\rangle$ is the mod $2$ reduction of the Euler characteristic. Is this still true for closed topological manifolds?



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> Let $M$ be a closed topological $n$-dimensional manifold. If $w\_n(M)$ is the top Stiefel-Whitney class of $M$, as defined above, is the Stiefel-Whitney number $\langle w\_n(M), [M]\rangle$ the mod $2$ reduction of $\chi(M)$?
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Is the top Stiefel-Whitney number of a topological manifold the Euler characteristic mod two?
As you say, we define $w\_n = \sum \text{Sq}^i \nu\_{n - i}$, where $\nu\_{n-i}$ is the Wu class, the class such that $\nu\_{n-i} \cup c = \text{Sq}^{n-i} c$ for $c \in H^{i}$. So as a corollary we have $\text{Sq}^i \nu\_{n - i} = \nu\_i \cup \nu\_{n-i}$. 


Because $\nu\_j$ vanishes for $j > n/2$, the sum over $i$ is only the term $\nu\_{n/2}^2$ when $n$ is even, and $0$ when $n$ is odd. As the Euler characteristic of an odd-dimensional closed manifold vanishes, this gives the odd-dimensional case. 


Now when $n = 2k$, using Poincare duality mod 2 we see that $\chi(M) = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$ So the claim is that $\langle \nu\_k^2, [M] \rangle = \text{rk } H^k(M;\Bbb Z/2) \pmod 2.$ 


This is because $\nu\_k$ is a characteristic vector for the symmetric bilinear cup-product form on $H^k(M;\Bbb Z/2)$; in fact, for any 'characteristic vector' $y$ for a nondegenerate symmetric bilinear form over a $\Bbb Z/2$-vector space $V$, meaning that $y \cdot x = x^2$ for all $x$, we have $\text{rk } V = y^2 \pmod 2$. 


The most obvious way for me to see this is to classify nondegenerate symmetric bilinear forms over $\Bbb Z/2$ vector spaces: they are all sums of copies of $\begin{pmatrix}1\end{pmatrix}$ and $\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}$, for which the respective characteristic vectors are $(1)$ and $(0,0)$.


