# Proof of Poincaré duality

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?