- #1
kakarotyjn
- 98
- 0
Question 1: $$0 \to A\mathop \to \limits^f B\mathop \to \limits^g C \to 0$$ is an exact short sequence,in order to prove $$\cdots \to H^q (A)\mathop \to \limits^{f^* } H^q (B)\mathop \to \limits^{g^* } H^q (C)\mathop \to \limits^{d^* } H^{q + 1} (A) \to \cdots$$ is an exact long cohomology sequence,we need to prove$${\mathop{\rm Im}\nolimits} f^* = \ker g^* $$,I can prove $$
{\mathop{\rm Im}\nolimits} f^* \subset \ker g^* $$ because $${\mathop{\rm Im}\nolimits} f = \ker g$$,but how to prove $
{\mathop{\rm Im}\nolimits} f^* \supset \ker g^* $? \\
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Question 2: What makes the cohomology groups different by digging 2 points of $R^2$ from that of $R^2$ ?How does the closed and exact differential forms change?
Thank you!
the visualized questions are also in the attach files
{\mathop{\rm Im}\nolimits} f^* \subset \ker g^* $$ because $${\mathop{\rm Im}\nolimits} f = \ker g$$,but how to prove $
{\mathop{\rm Im}\nolimits} f^* \supset \ker g^* $? \\
\\
\\
Question 2: What makes the cohomology groups different by digging 2 points of $R^2$ from that of $R^2$ ?How does the closed and exact differential forms change?
Thank you!
the visualized questions are also in the attach files