The isomorphism $H^1(X;\mathbb Z_2) \rightarrow \operatorname{Hom}(\pi_1(X),\mathbb Z_2)$ and $w_1(E)$ On page $87$ of Hatcher's book Vector Bundles and K-Theory it states that, assuming $X$ is homotopy equivalent to a CW complex ($X$ is connected), there are  isomorphisms
$$H^1(X;\mathbb Z_2) \rightarrow \operatorname{Hom}(H_1(X),\mathbb Z_2)  \rightarrow \operatorname{Hom}(\pi_1(X),\mathbb Z_2).$$
My questions:

*

*We can directly calculate the first cohomology group with $\mathbb Z_2$ coefficients from the fundamental group. A simple example is if $\pi_1(X)=\mathbb Z$, then $H^1(X;\mathbb Z_2)=\mathbb Z_2$, or if $\pi_1(X)=\mathbb Z \oplus \mathbb Z$, then $H^1(X;\mathbb Z_2)=\mathbb Z_2 \oplus \mathbb Z_2$. Are these statements correct?


*Is there any restriction to this isomorphisms? Assuming that $X$ is homotopy equivalent to a CW complex seems very general to me, i.e. any topological manifold is homotopy equivalent to a CW complex.


*Does it mean actually the first Stiefel-Whitney class $w_1(E)\in H^1(M;\mathbb Z_2)$ of a real vector bundle $E$ can always be computed (using pullback) from the $w_1(E|_\mathcal L)$ of non-contractible loops $\mathcal L \in M$ (where $M$ is the base space)? That is, if I know $w_1(E|_\mathcal L)$ of all (nonequivalent) non-contractible loops $\mathcal L$, do I know $w_1(E)$? If so, does this conclusion need isomorphism $H^1(X;\mathbb Z_2) \rightarrow \operatorname{Hom}(\pi_1(X),\mathbb Z_2)$, or do we just need to use the pullback to prove it is true?
 A: As has been pointed out, your computations are correct. As for the assumptions, I actually don't think they are necessary. The first isomorphism comes from the Universal Coefficient Theorem, and the second is a consequence of the Hurewicz Theorem. Now let me address your third question.
The isomorphism $H^1(X; \mathbb{Z}_2) \to \operatorname{Hom}(H_1(X; \mathbb{Z}), \mathbb{Z}_2)$ takes the form $[\alpha] \mapsto \Phi_{[\alpha]}$ where $\Phi_{[\alpha]} : H_1(X; \mathbb{Z}) \to \mathbb{Z}_2$ is given by $\Phi_{[\alpha]}([c]) = \langle[\alpha], [c]\rangle = \alpha(c)$.
The isomorphism $\operatorname{Hom}(H_1(X; \mathbb{Z}), \mathbb{Z}_2) \to \operatorname{Hom}(\pi_1(X), \mathbb{Z}_2)$ takes the form $\Phi \mapsto \Phi'$ where $\Phi' : \pi_1(X) \to \mathbb{Z}_2$ is given by $\Phi'([f]) = \Phi(f_*[S^1])$.
Combining, we see that the isomorphism $H^1(X; \mathbb{Z}_2) \to \operatorname{Hom}(\pi_1(X), \mathbb{Z}_2)$ takes the form $[\alpha] \mapsto \Phi_{[\alpha]}'$ where $\Phi_{[\alpha]}' : \pi_1(X) \to \mathbb{Z}_2$ is given by $\Phi_{[\alpha]}'([f]) = \Phi_{[\alpha]}(f_*[S^1]) = \langle[\alpha], f_*[S^1]\rangle = \langle f^*[\alpha], [S^1]\rangle$.
Now if $E \to X$ is a vector bundle, then $w_1(E) \in H^1(X; \mathbb{Z}_2)$ corresponds to $\Phi_{w_1(E)}' : \pi_1(X) \to \mathbb{Z}_2$ which is given by $\Phi_{w_1(E)}'([f]) = \langle f^*w_1(E), [S^1]\rangle = \langle w_1(f^*E), [S^1]\rangle$. So we can determine $\Phi_{w_1(E)}'$, and hence $w_1(E)$, from $w_1(f^*E)$ for every $[f] \in \pi_1(X)$.
Note that $w_1(f^*E) \in H^1(S^1; \mathbb{Z}_2) \cong \mathbb{Z}_2$, and since a vector bundle over $S^1$ is trivial if and only if its first Stiefel-Whitney class vanishes, we see that $\Phi_{w_1(E)}'([f]) = 0$ if and only if $f^*E$ is trivial. Therefore, $w_1(E) = 0$ if and only if $f^*E$ is trivial for all $[f] \in \pi_1(X)$, i.e. $E$ is orientable if and only if its pullback to every loop is trivial.
A: The first isomorphism $H^{1}(X,\mathbb{Z}/2) \cong \text{Hom}( H_{1}(X),\mathbb{Z}/2)$ follows by the universal coefficient theorem.
The second isomorphism follows since for any group $G$ we have $\text{Hom}(G,\mathbb{Z}/2) \cong \text{Hom}(G_{\text{ab}},\mathbb{Z}/2).$
I can't help you with part (3) as I don't know any of that theory.
