Bijection of pair of top spaces

Question

I have been battling with this problem and I dont know how to prove the following:

Given a pair of topological spaces $(X,A)$, let $q : (X,A) \rightarrow (X/A,*)$ be a quotient map where * denotes $A/A$

1. show that $q$ induces, for every pointed space $(Y,y)$ a bijection: $$q^{*}:[(X/A,*),(Y,y)] \rightarrow [(X,A),(Y,y)].$$

2. Conclude that, for $n\ge 1$, there is a bijection $[(\mathbb{S^n},*),(Y,y)] \rightarrow [(\mathbb{I^n},\delta \mathbb{I}^n),(Y,y)]$, where $$\delta \mathbb{I}=\{(t_1,t_2,\cdots,t_n) \in \mathbb{I}^n| \,\,t_1\cdots t_n \cdot (1-t_1)\cdots(1-t_n)=0\}$$

Answer 1

Let $f:(X,A)\to(Y,y_0)$ be a map representing an element of $[(X,A),(Y,y_0)]$. Then, as $f(A)=y_0$ by the universal property of quotients there exists a unique map $\bar{f}:(X/A,\ast)\to(Y,y_0)$ making the diagram $$\require{AMScd} \begin{CD} (X,A) @>{p}>> (X/A,\ast)\\ @V{f}VV @VV{\bar{f}}V\\ (Y,y_0) @>>{id_Y}> (Y,y_0) \end{CD} $$ commute. That is, $p^*[\bar{f}]=[f]$, and so $p^*$ is surjective.

Now, suppose that $f_1,f_2:(X,A)\to(Y,y_0)$ are two homotopic maps with induced maps $\bar{f}_1,\bar{f}_2:(X/A,\ast)\to(Y,y_0)$ respectively. Then, as $f_1\simeq f_2\,\operatorname{rel}{A}$ there exists a homotopy $$H:X\times I\to Y,\quad H(A\times I)=y_0.$$ Since $H(A\times I)=y_0$, $H$ factors through another homotopy $\bar{H}:X/A\times I\to Y$: $$\require{AMScd} \begin{CD} X\times I @>{p\times id_I}>> X/A\times{I}\\ @V{H}VV @VV{\bar{H}}V\\ Y @>>{id_Y}> Y \end{CD} $$ which gives $\bar{f}_1\simeq\bar{f}_2$. That is, $[\bar{f}_1]=[\bar{f}_2]$, and so $p^*$ is injective.

For a reference to the above fact I used about homotopies, check Switzer's algebraic topology text, chapter $0$.