Question on homotopy equivalence of the action $\pi_1(X)$ on $\pi_n(X)$

Question

I have a question on the details of this MathOverflow question. In the second paragraph, the OP claims that there is a homotopy equivalence $$\mathbb{S}^n \to \mathbb{S}^n \vee I$$ for any $n \geq 1$ and where $I$ denotes the unit interval. This homotopy equivalence should take the basepoint of $\mathbb{S}^n$ to the endpoint of the unit interval $I$, i.e. $1$. I am failing to explicitely constructing such a homotopy equivalence. The thing which I got so far, is that we consider $(I,0)$ as a pointed space. Could anybody help me writing out this homotopy equivalence?

Answer 1

Let's take a certain model for $S^n$. Consider $I\times S^{n-1}$ and collapse each of $0\times S^{n-1}$ and $1\times S^{n-1}$ to points. Call this space $X$ and let the collapsed $1\times S^{n-1}$ be the basepoint. This $X$ is homeomorphic to $S^n$.

Now start with $I\times S^{n-1}$ but collapse $0\times S^{n-1}$ to a point and each $t\times S^{n-1}$ to a point for each $t\in[1/2,1]$. This new space $Y$ is homeomorphic to $S^n\vee I$. Again let the collapsed $1\times S^{n-1}$ be the basepoint.

The natural quotient map $f:X\to Y$ is now a basepoint-preserving homotopy equivalence.

Answer 2

Let $S\in S^n$ be the South pole and $N$, the north pole. Define $d: S^n \to I$ by $d(x) = \frac{|x-S|}{2}$. Note that $d(S) = 0$ and $d(N) = 1$. Now define $c: S^n \to S_S^n \vee S_N^n$ by collapsing the equator $S^{n-1}$, and assume the northern hemisphere maps to $S_N^n$. Finally, take the composition $S^n \to S_S^n\vee S_N^n \to S_S^n \vee I$, where the last map is given by $1\vee d$.