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Edit: (1) below is now clear, but I'm still looking for an answer for (2).

There are two issues I have when trying to understand the classifying spaces of (oriented) vector bundles:

(1) Let $E \to X$ be an $n$-dim vector bundle with classifying map $f: X \to BO(n)$. The space $BSO(n)$ is the classifying space of oriented vector bundles and at the same time the universal cover of $BO(n)$. Then $E$ is orientable if and only if $f$ lifts to a map $\tilde{f}: X \to BSO(n)$.

Question 1: Why doesn't it contradict the fact that $f$ admits a lift if and only if $f_* (\pi_1 (X,x_0)) \subseteq p_* (\pi_1 (BSO(n),y_0))=0$? This is the lifting criterion for covering spaces, and since $BSO(n)$ is simply connected then for a lift to exist $X$ should be simply connected as well, but this is definitely not the case.

(2) I'm trying to show that $BSO(n)$ is simply connected without using that $\pi_1 (BSO(n)) = \pi_0 (SO(n))=0$, but rather using that $BSO(n)$ is the classifying space for oriented $n$-dimensional real vector bundles. Since there is a unique vector bundle over the one-point space, it is easy to see that $BSO(n)$ is path-connected.

However if I want to extend the argument for $\pi_1$, I bump into the following issue regarding basepoints:

$$ \{ S^1 \times \mathbb{R}^n \} = \{ n-\text{dim oriented vector bundles over }S^1 \} = [ S^1, BSO(n) ] \leftarrow \pi_1 (BSO(n)) $$

Question 2: The last map is surjective since $BSO(n)$ is path-connected, but how to show that it is bijective? This is true since $BSO(n)$ is simply connected, but I don't know that a priori (that's exactly what I am trying to show).

This approach is hinted in Hatcher's Vector bundles and $K$-theory, page 31, but without further comment.

Minkowski
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    For 1) you don't need that $X$ is simply connected, only that the map $f_*$ maps all the non-zero elements to zero. Geometrically what this means is that along every loop in your manifold, the restriction of the bundle should be orientable. For example the tangent bundle of the circle satisfies this property, but the mobius bundle over the circle does not. – Thomas Rot Feb 05 '21 at 10:03
  • You are absolutely right about (1) @ThomasRot – Minkowski Feb 05 '21 at 10:07
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    Do you know about the clutching construction? This allows you to classify orientable vector bundles over the circle in terms of maps from $S^0$ into $SO(n)$. This should answer 2). But this amounts to showing that $\pi_1(BSO(n))=\pi_0(SO(n))$. – Thomas Rot Feb 05 '21 at 10:08
  • But this is more basic than $\Omega BSO(n)=SO(n)$ – Thomas Rot Feb 05 '21 at 10:10
  • Yes, I know about clutching functions Then your argument yields a bijection $[S^0, SO(n)] = [S^1, BSO(n)]$. Does this bijection directly imply that $\pi_1(BSO(n))=\pi_0(SO(n))$ by taking basepoint-preserving homotopy clases of maps or do I need an extra argument to conclude? – Minkowski Feb 05 '21 at 10:44
  • I mean, I see how to conclude if $[X_1, Y_1] = [X_2, Y_2]$ implies $[(X_1,x_1), (Y_1,y_1)]* = [(X_2,x_2), (Y_2, y_2)]*$ but I'm not sure this is the case. – Minkowski Feb 05 '21 at 10:49
  • $X$ does not have to be simply connected for a lift to exist in question 1. A lift exists if $X$ is simply connected but $X$ doesn't have to be. – Noel Lundström Feb 05 '21 at 11:38
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    It's been awhile since I've thought about it, but I believe that for path connected $X$, that $[S^1,X]$ is in bijection with conjugacy classes of $\pi_1(X,\ast)$. In particular $[S^1, X]$ has one element iff $\pi_1$ has one element. – Jason DeVito - on hiatus Feb 09 '21 at 18:15
  • Jason DeVito's comment is correct. See this question for example. – Michael Albanese Feb 12 '21 at 13:18
  • I'm astonished by the easy argument @JasonDeVito gave. Indeed, the observation that a group is trivial iff there is a single conjugacy class concludes. I was aware of this bijection with the conjugacy classes of elements in $\pi_1$, but I didn't realise that was enough using that observation. Many thanks! – Minkowski Feb 12 '21 at 19:24

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