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Let $X$ be a path-connected manifold. Then by the classification of vector bundles, the collection of all (real) line bundles on $X$ is $$ \text{Vect}^1(X)\cong [X,BO(1)]=[X,\mathbb{R}P^\infty]=[X,B\mathbb{Z}_2] $$ where $\mathbb{Z}_2=\{\pm 1\}$ under multiplication.

Question. Are there any results/references giving that

$$ \text{Vect}^1(X)\cong \text{Hom}(\pi_1(X), \mathbb{Z}_2)? $$ Whether can we just take loop space $$ [X,B\mathbb{Z}_2]=[\Omega X,\Omega B\mathbb{Z}_2] =[\pi_1(X), \mathbb{Z}_2] $$ or not?

Michael Albanese
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Shiquan
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  • See http://math.stackexchange.com/questions/1277823/tautological-line-bundle-over-mathbbrpn-isomorphic-to-normal-bundle-also/1277969#1277969 and note that $H^1(X;A)= Hom(\pi_1 X,A)$ if $A$ is abelian. – Daniel Valenzuela Mar 10 '16 at 19:45

1 Answers1

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In general,

$$[X, K(G, n)] \cong H^n(X; G)$$

where $K(G, n)$ is an Eilenbeg-MacLane space (see Theorem $4.57$ of Hatcher's Algebraic Topology).

In particular, if $G$ is equipped with the discrete topology, then $BG$ is a $K(G, 1)$ so

$$[X, BG] \cong [X, K(G, 1)] \cong H^1(X; G).$$

On the other hand, for an abelian group $G$,

$$H^1(X; G) \cong \operatorname{Hom}(\pi_1(X), G).$$

This isomorphism follows from the Universal Coefficient Theorem for cohomology (see the argument given in the first paragraph of this answer which uses $G = \mathbb{Z}$, but will work for any abelian group $G$).

Therefore,

$$\operatorname{Vect}^1(X) \cong [X, B\mathbb{Z}_2] \cong H^1(X; \mathbb{Z}_2) \cong \operatorname{Hom}(\pi_1(X), \mathbb{Z}_2).$$

Note, none of the above required $X$ to be a manifold.

Michael Albanese
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