The claim I want to prove is that:
Let $n \geq 1$. $H_n(B^n,\mathbb{S}^{n-1})$ is generated by $[\iota_n+S_n(\mathbb{S}^{n-1})]$.
Here is my work:
Since $\Delta^n \cong B^n$ and $\partial\Delta^n \cong \mathbb{S}^{n-1}$, we identify them.
$\bullet$ We proceed by induction. Let $n \geq 1$.We have the following relationship (the double arrow indicates isomorphism): $$H_n(B^n,\mathbb{S}^{n-1}) \leftrightarrow H_n(B^n/\mathbb{S}^{n-1},*)=\tilde{H}_n(\mathbb{S}^n)=H_n(\mathbb{S}^n) \leftrightarrow H_{n+1}(B^{n+1},\mathbb{S}^n)$$ When $n=1$, $H_1(B^1,\mathbb{S}^0)\leftrightarrow \tilde{H}_0(\mathbb{S}^{0})=\mathbb{Z}$. Recall that $\tilde{H}_0(\mathbb{S}^0)=\text{ker}H_n(f)$, where $$H_n(f): H_0(\mathbb{S}^0) \to H_n(*), [ac^0_1+bc^0_{-1}] \mapsto [(a+b)*]$$ From this, $\tilde{H}_0(\mathbb{S}^0)$ is generated by $[c^0_1-c^0_{-1}]$. Since $\partial$ is an isomorphism, an element $[\sigma+S_1(\mathbb{S}^0)] \in H_1(B^1,\mathbb{S}^0)$ is a generator of $H_1(B^1,\mathbb{S}^0)$ iff $[d\sigma]=\partial[\sigma+S_1(\mathbb{S}^0)]=[c^0_1-c^0_{-1}]$. In fact, $d\iota_1=c^0_1-c^0_{-1}$. So $[\iota_1+S_1(\mathbb{S}^0)]$ generates $H_n(B^n,\mathbb{S}^{n-1})$.
$\bullet$ Suppose the statement holds for $n=k \geq 1$. Let $n=k+1$. Note that \begin{align*} q_*[\iota_n+S_n(\mathbb{S}^{n-1})] &=[q_*(\iota_n)+S_n(*)]\\ &=[q] \quad(\text{By identification of }H_n(B^n/\mathbb{S}^{n-1},*) \text{ and } H_n(\mathbb{S}^n)) \end{align*} \begin{align*} \partial[\iota_{n+1}+S_{n+1}(\mathbb{S}^n)] &=[d\iota_{n+1}] \end{align*}
My question is: how do we show that $[q]=[d\iota_{n+1}]$?