I am working through a proof of Poincaré duality. I don't understand the one step marked in bold.
Let $R$ be a ring. Pick an $R$-orientation $(o_x; x\in\mathbb{R}^m)$ of $\mathbb{R}^m$. Pick $r\in \mathbb{N}$ and let $o_{B_r}$ be the unique $R$-orientation of $\mathbb{R}^m$ along $B_r=\{x\in \mathbb{R}^m \vert \ \ \vert x \vert \leq r\}$. This exists since $B_r$ is compact. We want to show that the map $$(-)\cap o_{B_r}: H^m(\mathbb{R}^m, \mathbb{R}^m\setminus B_r;R)\rightarrow H_0(\mathbb{R}^m;R)$$ is an isomorphism. Recall that for $\alpha \in H^p(X),\beta \in H^q(X), c\in H^{p+q}(X)$ we have $\langle \alpha \cup \beta, c\rangle=\langle \alpha, \beta \cap c\rangle$. In particular $\langle \beta, o_{B_r}\rangle=\langle 1, \beta \cap o_{B_r}\rangle$ for all $\beta \in H^m(\mathbb{R}^m, \mathbb{R}^m\setminus B_r;R)$. By the Universal Coefficient theorem the map $\kappa: H^m(\mathbb{R}^m, \mathbb{R}^m\setminus B_r;R)\rightarrow \operatorname{Hom}(H_m(\mathbb{R}^m, \mathbb{R}^m\setminus B_r;R),R), \beta \mapsto \langle \beta,-\rangle$ is an isomorphism (the Ext term vanishes by the reduced long exact sequence of the pair $(\mathbb{R}^m, \mathbb{R}^m\setminus B_r;R)$). We conclude that the map $(-)\cap o_{B_r}$ is an isomorphism.
Why does the bijectivity of $\kappa$ together with $\langle \beta, o_{B_r}\rangle=\langle 1, \beta \cap o_{B_r}\rangle$ imply that $(-)\cap o_{B_r}$ is an isomorphism?
