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Let X be a complex normal variety and U a subvariety that is open in the analytic topology. Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
edited to include complex
Let X be a normal variety and U a subvariety that is open in the analytic topology. Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
Let X be a complex normal variety and U a subvariety that is open in the analytic topology. Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
Let X be a normal variety and U a subvariety that is open in the analytic topology. Then the map $\pi_1(U) \to \pi_1(X)$ coming from the map $U \subset V$ is surjective - why is this?
