I asked this https://math.stackexchange.com/q/1694046/309968 question already on MSE, but received no answer and I hope it's ok if I ask here for once.
Let $R$ be commutative ring with $1_R$
Lemma: Let $B\subseteq \mathbb{R}^n$ be a compact ball and let $\mu_B\in H_n(\mathbb{R}^n,\mathbb{R}^n\setminus B;R)$ be a generator. Then $$\cap \mu_B:H^k(\mathbb{R}^n,\mathbb{R}^n\setminus B;R)\to H_{n-k}(\mathbb{R}^n;R)$$is an isomorphism.
Here denotes $H^*$ singular cohomology, $\cap$ is the cap product https://en.wikipedia.org/wiki/Cap_product .
Proof: It is sufficient to consider $k=n$, otherwise both modules are trivial.
We already know that the evaluation map $$H^n(\mathbb{R}^n,\mathbb{R}^n\setminus B;R)\to \operatorname{Hom}_R(H_n(\mathbb{R}^n,\mathbb{R}^n\setminus B;R),R)$$ $$[\xi]\mapsto (\;[\sigma ]\mapsto \xi (\sigma ) \;)$$ is surjective. Therefore there exists a $\eta\in H^n(\mathbb{R}^n,\mathbb{R}^n\setminus B;R)$ such that $\eta (\mu_B )=1$. Let $$\epsilon :H_0(X;R)\to R,\; [\sum_{\sigma}r_{\sigma}\sigma ]\mapsto \sum_{\sigma}r_{\sigma}$$be the augmentation map. It is $\epsilon(\eta\cap \mu_B)=1$, hence $ \eta\cap \mu_B$ is a generator of $H_0(\mathbb{R}^n;R)$ and $\cap\mu_B$ is an isomorphism.
My question:
I understand that the proof shows that $\cap\mu_B$ is surjective. But why should $\cap\mu_B$ be injective, maybe one could choose an other element $\eta '\in H^n(\mathbb{R}^n,\mathbb{R}^n\setminus B;R)$ such that $\eta '(\mu_B )=1$?
Or do we need some extra conditions on $R$?
I appreciate your help. Best
