Proving that $S^{1} * S^{1}$ (join) is homotopiclly equivalent to $S^{3}$.

Question

Proving that $S^{1} * S^{1}$ is homotopiclly equivalent to $S^{3}$.

I know that the definition of homotopy equivalence from Allen Hatcher is: A map $f: X \rightarrow Y$ is called a homotopy equivalence if there is a map $g: Y \rightarrow X$ such that $f \circ g$ homotopic to $\mathbb{1_{Y}}$ and $g \circ f$ homotopic to $\mathbb{1_{X}}$.

And I have solved $a$ and $c$ in the following question:

But I still do not know how to solve this problem, will the solution be similar to one of the above problems?

Any help will be greatly appreciated!

EDIT:

Also I have this corollary (I am not so much sure that it will be helpful here):

If it is helpful here please tell me how?

Answer 1

It seems that you refer to Hatcher. In the section "Join" on p.9 Hatcher defines the join $X * Y$ and also the iterated join $X_1 * \ldots * X_n$. He proves that if each $X_i = S^0$, then the join $X_1 * \ldots * X_n$ is homeomorphic to $S^{n-1}$. In particular $S^0 * S^0 \approx S^1$ which implies $$S^1 * S^1 \approx S^0 * S^0 * S^0 * S^0 \approx S^3. $$

Answer 2

The bottom line is that to prove two spaces $X,Y$ are homotopy equivalent you must produce a homotopy equivalence $f : X \to Y$, i.e. you must write down a formula for an appropriate function $f : X \to Y$ and then prove that the function you wrote down is a homotopy equivalence. This will require you to write down three more formulas: one for an appropriate function $g : X \to Y$ to be a homotopy inverse of $f$; one for a homotopy $H : X \times [0,1] \to X$ between $g \circ f$ and the identity; and one for a homotopy $H' : Y \to [0,1] \to Y$ between $f \circ g$ and the identity.

In this case, as noted in a comment, you have the option of proving the stronger statement that $S^1 * S^1$ and $S^3$ are homeomorphic. This still requires you to produce a homeomorphism $f : S^1 * S^1 \to S^3$, this time by writing down a formula for an appropriate function $f$ and then proving that the function you wrote down is a homeomorphism.

If you're lucky, someone else will have written the formula down for you, and that appears to be the case here: you can just read p. 9 of Hatcher to find the formula, according to the answer of @PaulFrost.

But if you're not so lucky, then your job might be hard. You have to use your mathematical experience/imagination/expertise, or whatever else you might have at your disposal, to figure out the correct formula.