Dtyjlui

I'm interested in showing that $S^m * S^n approx S^{m+n+1} $, as discussed in exercise 0.18 of Hatcher's Algebraic Topology. One way to show it would be to show that $X * Y approx Sigma(X wedge Y)$. This fact shows up only in a later exercise in the chapter, so I have to think he's looking for something else.

This, this, and this question allude to answers. In particular, there is this answer, which I don't understand:

If $A$ and $B$ are subsets of $mathbb R^n$ then you can map $Atimes Itimes B$ to $mathbb R^n$ by
$(a,t,b)mapsto (1−t)a+tb$. If the resulting continuous map $A∗Bto mathbb R^n$ happens to
be one to one then you can ask whether it gives a homeomorphism to its
image. If $A$ and $B$ are compact, then the answer is necessarily yes,
since a continuous bijection from a compact space to a Hausdorff space
is always a homeomorphism.

This suffices to see that the join of spheres is a sphere.

$S^m * S^n = (S^m times S^n times [0, 1])/sim$ where $sim$ identifies the top $S^m times S^n times {0}$ to $S^m$ and the bottom $S^m times S^n times {1}$ to $S^n$. Cutting this in half gives $$S^m * S^n = (S^m times S^n times [0, 1/2])/!!sim cup_{S^m times S^n times {1/2}}; (S^m times S^n times [1/2, 1])/!!sim$$

In the first piece, $sim$ does nothing except pinching the copy of $S^m times {0}$ to a point. Thus, the first piece is homeomorphic to $C(S^m) times S^n cong D^{m+1} times S^n$. Similarly, $sim$ just pinches the copy of $S^n times {0}$ in the second piece, so that one is homeomorphic to $S^m times C(S^n) cong S^m times D^{n+1}$. So the space is homeomorphic to

$$D^{m+1} times S^n cup_{S^m times S^n} S^m times D^{n+1} cong D^{m+1} times partial(D^{n+1}) cup_partial partial(D^{m+1}) times D^{n+1} \ ;;;;;;;;;;;;;; cong partial(D^{m+1} times D^{n+1}) \ ;;;;;; cong partial(D^{m+n+2})$$

Which is just $S^{m+n+1}$ $blacksquare$

Let $S^n subset mathbb R^{n+1}$ and $S^msubset mathbb R^{m+1}$, then we can consider that they are both in $mathbb R^{n+m+2} cong mathbb R^{n+1} oplus mathbb R^{m+1}$. Then define the map

$$ phi : mathbb S^n times [0,pi/2] times mathbb S^m to S^{n+m+1} subset mathbb R^{n+m+2}$$

by $phi(a, t, b) = (cos t) a + (sin t) b$. This map descend to the quotient (still call) $phi :S^n *S^m to S^{n+m+1}$, which is bijective. Then as the domain is compact and the image is Hausdorff, $phi$ is a homeomorphism. (This is essentially the idea given in your question)

There are other ways of defining the join $X=X_1*X_2$ with a topology. In Topology and Groupoids, Chapter 5, the analogy is taken with the join of two subsets of a high dimensional $mathbb R^n$ by saying a point of the join is a formal sum $r_1x_1+r_2x_2$ where $x_i in X_i$ and $r_1+r_2=1, r_1,r_2 in [0,1]$, except that if $r_1$ or $r_2$ is $0$ then we ignore that term. There are partial functions $xi_i:X to [0,1], eta_i:X to X_i$ defined by $r_1x_1+r_2x_2 mapsto r_i, r_1x_1 +r_2x_2 mapsto x_i$, respectively. Here $eta_i$ has domain $xi_i^{-1}(0,1]$ . We give $X$ the initial topology with respect to these functions.

One advantage of using initial topologies here is that the join becomes associative.

A homeomorphism $S^p * S^q to S^{p+q+1}$ is defined by
$$rx+sy mapsto (xsin rpi /2, y sin spi /2). $$

This is related to this stackexchange answer and picture.

Consider $phi : S^n times S^m times [0,1] to S^{n+m+1}$ defined by,
$$phi((x_0,dots,x_n),(y_0,dots,y_m),t)=(sqrt{1-t})x_0,dots(sqrt{1-t})x_n,sqrt{t}y_0,dots,sqrt{t}y_m)$$

Then it is easy to check that $phi$ is Continuous , onto and fails to be one-one precisely at $S^n times S^m times {0}$ and $S^n times S^m times {1}$ .
For $t=0$, $phi((x_0,dots,x_n),(y_0,dots,y_m),0)=(x_0,dots ,x_n,0,dots,0)=phi((x_0,dots,x_n),(tilde{y_0},dots,tilde{y_m}),0)$
Thus, $S^n times S^m times {0}$ is 'collapsed' to $S^n$ and similarly, $S^n times S^m times {1}$ is 'collapsed' to $S^m$ .

Now, invoking Universal Property of Quotient, we get an onto,continuous map $ bar{phi} : S^n * S^m to S^{n+m+1}$ which is clearly one-one and thus $bar{phi}$ is a continuous bijection from a Compact space to a Hausdorff space and thus it turns out to be a homeomorphism!