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How can we show that given the manifold $M=\mathbb{S}^2$ and $U=\mathbb{S}^2\setminus \{A\}$ and $V=\mathbb{S}^2 \setminus \{B\}$, then:

$$H_{dR}^1(U)\oplus H_{dR}^1(V)\rightarrow H_{dR}^1(U\cap V) $$ cannot be surjective. The reasoning in my book says that $ H_{dR}^1(U)=H_{dR}^1(V)=\{[0]\}$ and $H_{dR}^1(\mathbb{S}^1)=H_{dR}^1(U\cap V)=\mathbb{R}$, hence they cannot be surjective. But I am having trouble seeing why this is the case, as in why the above is equal to $\mathbb{R}$ and $\{[0]\}$.

Can anyone help me understand my confusion? Thank You

Aurora Borealis
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  • Note that $U, V$ are diffeomorphic to $\mathbb R^2$ and $U\cap V$ is diffeomorphic to $\mathbb S^1 \times \mathbb R$. –  May 13 '18 at 06:23
  • @I am sorry I might need alittle more than this. – Aurora Borealis May 13 '18 at 06:24
  • Um.... do you know what is $H^1_{dR} (\mathbb R^2)$? –  May 13 '18 at 06:26
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    I know the theorem on diffeomorphic invariance of $H_{dR}^k$, and the Poincare's Lemma, $H_{dR}^1(\mathbb{R}^2)$ are all exact closed 1-forms on $\mathbb{R}^2? $ which is ${[0]}$? – Aurora Borealis May 13 '18 at 06:29
  • Poincare lemma tells you that all closed forms in $\mathbb R^2$ are exact, so $H^1_{dR}(\mathbb R^2) = {0}$. –  May 13 '18 at 06:31
  • Ok in my book it says for any start shaped open set $U \in \mathbb{R}^n $, we have $ H_{dR}^1(U)={[0]},$ so if $H_{dR}^1( \mathbb{R}^2)={[0]}$, then is $\mathbb{R}^2$ a star shaped open set in $\mathbb{R}^n?$ – Aurora Borealis May 13 '18 at 06:35
  • Well, is $\mathbb R^2$ a star-shaped open subset in $\mathbb R^2$? –  May 13 '18 at 06:35
  • That is what I am confused about the result seems like it is, but I do not know why it is. – Aurora Borealis May 13 '18 at 06:37
  • Can you state the definition of a star-shaped subset? –  May 13 '18 at 06:37
  • Ok nvm it is a star shaped, but then it raises the next question how can you see that it is diffeomorphic to $U$ and $ V$ because this doesnt seem obvious. and this part is the real confusing part for me; $H^1_{dR}(U∩V)=ℝ $ I do not see how we can have this part to hold. – Aurora Borealis May 13 '18 at 06:40
  • Do you know stereographic projection? –  May 13 '18 at 06:42
  • Yes I have learned that in complex analysis. – Aurora Borealis May 13 '18 at 06:43
  • That's exactly a diffeomorphism from $\mathbb R^2$ to $\mathbb S^2\setminus{N}$. –  May 13 '18 at 06:44
  • Hmm ok I will try to see how that holds, but how about $H^1_{dR}(U∩V)=ℝ $? – Aurora Borealis May 13 '18 at 06:45
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    What does $(U \cap V)$ 'look like'? Note that it is $S^2$ with two points removes. If you remove one point, you have $S^2 - {pt}$ which by stereographic projection is diffeomorphic to $\mathbb{R}^2$. Now if I remove one more point, say the origin, I have $\mathbb{R}^2 - {(0,0)]|$, i.e. the punctured plane. Maybe you know that the form $d\theta = \frac{-y}{x^2 + y^2} dx + \frac{x}{x^2+y^2} dy$ generates the 1st cohomology of the punctured plane. In which case we get $H_{dR}^1(U\cap V) = \mathbb{R}$. – Osama Ghani May 13 '18 at 08:53

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