Here $S^n\vee S^1$ denotes the wedge of $S^n$ and $S^1$ with base points $*=(1,0,\ldots,0)$ and $*'=(1,0)$ respectively.
$$S^n\vee S^1:=S^n\sqcup S^1/*\sim *'$$
WLOG I assume $q=*$
My idea: I will construct a continuous map $f: S^n\to S^n\vee S^1$ such that $f(p)=f(q)=[*]=[*']$. Then this will give rise to a continuous map $\tilde{f}: S^n/\{p,q\}\to S^n\vee S^1$.
Similarly, I will define $g_1:S^n\to S^n$ such that $g(*)=p$ (this is easy to construct, we will consider an orthogonal linear operator on $\Bbb{R}^n$ with maps $*$ to $p$ and then restrict it to $S^n$). Then define $g_2:S^1\to S^n$ such that $g(*')=q$. (Here we could take $g$ to be the inclusion $S^1\hookrightarrow S^n$, $(x_1,x_2)\mapsto (x_1,x_2,0,\ldots,0)\in S^n$). Then using the universal property of disjoint union we have a continuous map $g:S^n\sqcup S^1\to S^n$, composing it with the quotient map $\pi:S^n\to S^n/\{p,q\}$ we get a map $\pi\circ g:S^n\sqcup S^1\to S^n/\{p,q\}$.
Then $\pi\circ g(*)=[p]$ and $\pi\circ g(*')=[q]$. Hence, this map gives rise to another continuous map $\tilde{g}:S^n\sqcup S^1/*\sim *'\to S^n/\{p,q\}$.
Then finally we have to argue $\tilde{f}\circ\tilde{g}\simeq 1_{S^n\vee S^1}$ and $\tilde{g}\circ\tilde{f}=1_{S^n/\{p,q\}}$
So this is my idea. But I'm unable to construct the function $f$. Can anyone help to complete the proof? Thanks for help in advance.
Best Answer
Hint: $S^n/ \{p,q\}$ homotopy equivalent to $(S^{n} \coprod I)/ \sim$ ,where $p \sim 0$ and $q \sim 1$.
The homotopy equivalence can be seen by taking a quotient map collapsing the whisker joining $p,q$.