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Oct 30 at 2:43 comment added Carlos Esparza I think you just need to pick a "product" coordinate neighborhood $U \cong U_1 \times U_2$ inside $\Phi^{-1}(W)$. Knowing that $\Phi^* f = 0$ in that neighborhood allows you to conclude that $f$ is zero on $\Phi(U) \cong U_2$.
Oct 20 at 11:38 history asked psl2Z CC BY-SA 4.0