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?
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.