Prove that $S^n /\{p,q\}$ is homotopy equivalent to $S^n \vee S^1$.

Question

Prove that $S^n /\{p, q\}$ homotopy equivalent to $S^n ∨ S^1$

My attempt:

I can see the picture quite clearly:

But how to write explicit homotopies ?

Thanks in Advance for help!

Answer 1

We use the fact that if $(X,A)$ is a CW pair and $A$ is contractible, then $X\simeq X/A$. Now consider the space $Y$ which is a sphere with an additional edge connecting $p,q$. Let $A$ be this edge. Then $Y\simeq Y/A,$ and $Y/A$ is homeomorphic to a sphere with $p,q$ identified. Now let $B$ be an arc connecting $p,q$ on the surface of the sphere. Then $Y\simeq Y/B$, and $Y/B$ is homeomorphic to $S^n\vee S^1$. (Contracting $B$ moves the endpoints of the added edge to coincide.) Now by the transitivity of $\simeq$ you are done.

Answer 2

For the inspiration, you should see Example 0.8 in Hatcher's book.

Drawing the same picture, he showed that $S^2 / S^0$ is homotopy equivalent to $S^2 \vee S^1$.

You can generalize his argument using $CW$-complex structure of $S^n$.