# De Rham cohomology of $S^1$ with compact supports (Bott/Tu) This is a question about Example 2.9, in Bott/Tu - Differential Forms in Algebraic Topology.

Consider the decomposition of $$S^1=U\cup V$$ by two open sets, as in the figure above. Then both $$U$$ and $$V$$ are diffeomorphic to $$\Bbb R$$, and the intersection $$U\cap V$$ is diffeomorphic to $$\Bbb R \coprod \Bbb R$$. We know that $$H_c^q(\Bbb R)$$ (de Rham cohomology with compact support) is $$\Bbb R$$ if $$q=1$$ and is $$0$$ otherwise. Thus to compute $$H^q_c(S^1)$$, the only interesting portion of the M-V seq. is as follows. $$0\to H_c^0(S^1)\to H_c^1 (U\cap V)=\Bbb R^2 \xrightarrow{i} H_c^1(U)\oplus H_c^1(V) =\Bbb R^2 \to H^1_c(S^1)\to 0$$

It follows that $$H_c^0(S^1)$$ and $$H_c^1(S^1)$$ is equal to $$\ker (i)$$ and $$\text{coker}(i)$$, respectively. The map $$i$$ is given by $$\omega \mapsto (\omega_U,\omega_V)$$, where, for instance, $$\omega_U$$ is the extension of $$\omega$$ (which is a $$1$$-form on $$U\cap V$$ with compact support) by zero. Then the book says that the image of $$i$$ is $$1$$-dimensional so we get $$H_c^1(S^1)=\Bbb R=H_c^0(S^1)$$. But how do we know that $$\text{image}(i)$$ is $$1$$-dimensional?

• @NoelLundström In the original de Rham cohomology, yes, but with the cohomology with compact support, the arrows are reversed May 24 '20 at 11:56
• Yes ofcourse, my bad. May 24 '20 at 12:35

They show Early in the book that $$H_c^1(\mathbb{R})=\frac{\Omega^1_c(\mathbb{R}^1)}{\ker\int_{\mathbb{R}^1}}$$ Calculate the image of the linear map $$(\int,\int)\circ i^*:H^1_c(U \cap V) \to \mathbb{R}^2$$ $$\omega \mapsto (\omega_U,\omega_V)\mapsto (\int_{U}\omega_U,\int_V \omega_V)$$ The second maps is an isomorphism. Show that $$\int_{U}\omega_U=\int_{V}\omega_V$$ to deduce the desired result.