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The second Stiefel-Whitney class $w_2$ can be identified with a (homotopy class of) map $BSO\to B^2\mathbb Z_2$ (I write $\mathbb Z_2$ the cyclic group of order 2, and omit the dimension assumed to be high enough). It happens to have a model that is a group homomorphism (claimed in Basic Bundle Theory and K-Cohomology Invariants, Husemöller et al. section 12.3), and its homotopy fiber is the $Spin$ group (here), which allows us to write the fiber sequence $$ \mathbb Z_2\to Spin\to SO \to B\mathbb Z_2 $$ of group homomorphisms, so that we have the following fiber sequence of classifying spaces $$ SO \to B\mathbb Z_2\to BSpin\to BSO \overset{w_2}{\to} B^2\mathbb Z_2 $$ with $SO \to B\mathbb Z_2$ classifying the bundle $Spin\to SO$.

The question of lifting an oriented orthogonal structure to a $Spin$ structure is then answered by considering the homotopy class of the composed map into $B^2\mathbb Z_2\simeq K(\mathbb Z_2,2)$.

Two questions:
I assume $SO\to B\mathbb Z_2$ is the same in both fiber sequences and consider it defined by the second one.

  1. I assume that we use the Segal classifying space for $B\mathbb Z_2$ which has a group structure. Why can $SO\to B\mathbb Z_2$ be chosen as a group homomorphism?

  2. Why is $Spin\to SO$ the homotopy fiber? I read (see this answer or this comment) that a G-principal bundle $P\to B$ is a model for the homotopy fiber of its classifying map $B\overset{f}{\to} BG$, but I am unable to prove it. I get that they somehow classify the same thing: bundle morphisms to $P$ provided with a section, and maps to $B$ that have a trivializable pullback bundle. Being somewhat uncomfortable with universal properties in the homotopy category, I am looking for a more precise proof. I tried to use the fiber sequence $$\Omega BG \to P_f \to B$$ and prove that factorising $P\to P_f \to EG$ I have a weak equivalence between the homotopy fibers (which I know are weakly equivalent) but wasn't able to.

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