Why if $M$ is a complex surface, $H^2(M;\mathbb Z)$ is classifying $\mathbb S^1$-bundles ? I know $H^2(M;\mathbb Z)$ does classify complex line bundle (just using the exponential sequence for continous complex-valued functions). Thanks in advance !
2026-03-25 23:34:51.1774481691
Principal $\mathbb S^1$-bundle
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$C-\{0\}$ retracts to $S^1$, thus $H^2(M,Z)$ classifies also $S^1$-bundles if it classifies $C-\{0\}$ bundle.