Above is a way to find the cohomology groups of $S^1$. But what I do not get is how they concluded that $\delta$ is surjective. Can someone explain this to me?
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$\begingroup$Restrict yourself to : $$ ... H^0(U \cap V) \xrightarrow{\delta} H^1(S^1) \xrightarrow{i} H^1(U) \oplus H^1(V) ... $$
By exactness at $H^1(S^1)$, we have that $\ker i = \operatorname{im} \delta$. However, note that $H^1(U) \oplus H^1(V) = \mathbf 0$, so that $i$ is taking every element of $H^1(S^1)$ to $0$. Hence, the kernel of $i$ is the whole of $H^1(S^1)$, so the image of $\delta$ is also $H^1(S^1)$, which is the whole space, so $\delta$ is surjective.
A dual result is also true : If we have an exact sequence (with groups as objects and homomorphisms between them) like : $$ \mathbf 0 \xrightarrow{i} I \xrightarrow{\delta} J ... $$
Then by exactness at $I$, you can prove that $\delta$ is injective,since $\ker \delta = \operatorname{im} {i} = \mathbf 0$. You will need to apply these facts many times in your study of cohomology, I can tell you that much from a brief study of cohomology from Hatcher.
$\endgroup$ 3 $\begingroup$$\ldots \to X \underbrace{\to}_{\delta} H^1(S^1) \to 0$ means that the image of $X$ is the same as the kernel of the map out of $H^1(S^1)$, which being the zero map is the whole $H^1(S^1)$
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