I was trying to read Kervaire's 1960 paper where he first shows the existence of a manifold that does not admit differentiable structure and I got stuck. On the second page of the paper where he starts to define his invariant he writes: where $u_2 \in H^{10}(\Omega;\mathbf Z_2)$ is the reduction modulo $2$ of $e_2 \in H^{10}(\Omega)$
As I understand the construction cannot be done with coefficients other than $\mathbf Z_2$. But I don't understand why this is the case. On the same page further down he again uses reduction modulo $2$. Why is he forced to consider homology over $\mathbf Z_2$? It is not clear to me why integer coefficients would not work and why the construction is only well defined in $\mathbf Z_2$. I'd be very grateful if you could help me understand his paper by explaining to me why one needs $\mathbf Z_2$.
It is clear to me that homology over $\mathbf Z_2$ is oblivious to orientation of the manifold. However it does not explain why in this particular case other coefficients would make the whole constructions not well defined.
If you read the proof of Lemma 1.2 where he proves his invariant is well-defined, you can get a sense of why $\Bbb Z/2$ coefficients are useful. Using obstruction theory, Kervaire shows that if $f$ and $g$ are two maps such that $f^\ast(e_1) = g^\ast(e_1)$, then $$(f^\ast(u_2) - g^\ast(u_2))[s_{10}] = u_2[h\omega_{10}(f,g)[s_{10}]]$$ for any $10$-simplex $s_{10}$, where $\omega_{10}(f,g) \in C^{10}(K; \pi_{10}(\Omega S^6))$ is the obstruction cocycle, $K$ is a triangulation of $X$, and $h: \pi_{10}(\Omega S^6) \longrightarrow H_{10}(\Omega S^6)$ is the Hurewicz homomorphism.
A result of Serre implies that $u_2 [h \alpha]$ is the mod $2$ Hopf invariant of the element of $\pi_{11}(S^6)$ represented by $\alpha \in \pi_{10}(\Omega S^6)$ (recall that we have an isomorphism $\pi_{k+1}(X) \cong \pi_k(\Omega X)$). Since there are no elements in $\pi_{11}(S^6)$ with odd Hopf invariant, it follows that Kervaire's invariant is well-defined.
If instead we used $\Bbb Z$-coefficients, we would have $$(f^\ast(e_2) - g^\ast(e_2))[s_{10}] = e_2[h\omega_{10}(f,g)[s_{10}]],$$ and again by Serre we have that $e_2[h\alpha]$ is the non-reduced (integer) Hopf invariant of the class in $\pi_{11}(S^6)$ corresponding to $\alpha \in \pi_{10}(\Omega S^6)$. But in this case, for any even integer $m$, one can construct an element of $\pi_{11}(S^6)$ with Hopf invariant $m$. Hence we cannot conclude that the $\Bbb Z$-valued Kervaire invariant is well-defined. Reducing mod $2$ gets rid of this problem.