On the Wikipedia page of Fundamental class and Stiefel-Whitney class :
If ''M'' is not orientable, $H_n(M,\mathbf{Z}) \ncong \mathbf{Z}$, and so one cannot define a fundamental class ''M'' living inside the integers. However, every closed manifold is $\mathbf{Z}_2$-orientable, and $H_n(M,\mathbf{Z}_2)=\mathbf{Z}_2$ (for ''M'' connected). Thus every closed manifold is $\mathbf{Z}_2$-oriented (not just orient''able'': there is no ambiguity in choice of orientation), and has a $\mathbf{Z}_2$-fundamental class. This $\mathbf{Z}_2$-fundamental class is used in defining Stiefel–Whitney class.
My question is that:
What distinguish the $\mathbf{Z}_2$-oriented and orientable? Am I correct to say that we can have unorientable surfaces ($\mathbb{RP}^n$ for even $n$) but they are still $\mathbf{Z}_2$-oriented? Please helps clarifying this point.
"This $\mathbf{Z}_2$-fundamental class is used in defining Stiefel–Whitney class." Isn't that all the $j$-th $\mathbf{Z}_2$-homology class give rise to all the $j$-th Stiefel–Whitney class for $1 \leq j \leq n$?
then what is the emphasis of $\mathbf{Z}_2$-fundamental class? (the $n$-th $\mathbf{Z}_2$-homology class here)
how do we know whether the $j$-th cohomology class $H^j(M,\mathbf{Z}_2) $ is the $j$-th Stiefel–Whitney class but not other cohomology classes? (arent there other possible choices?)