Hatcher Algebraic Topology 0.24

Let's show the reduced versions are homeomorphic, which will show the originals are homotopic (they are not equal in general).

The join can be thought of as "lines" from $X$ to $Y$, with some collapsing. The relations are: $$ (x_1,y,0)\sim (x_2,y,0);$$ $$ (x,y_1,1)\sim (x,y_2,1).$$

The reduced version also collapses $x_0\ast Y$ and $X\ast y_0$. So the additional relations are: $$ (x_0,y,t)\sim (x_0,y_0,0);$$ $$ (x,y_0,t)\sim (x_0,y_0,0).$$

We can derive further relations too: $$(x,y,0)\sim(x_0,y,0)\sim(x_0,y_0,0);$$ $$(x,y,1)\sim (x,y_0,1)\sim(x_0,y_0,0).$$

The smash product is gotten from $X\times Y$ by collapsing $X\times y_0$ and $x_0\times Y$. The suspension of that can be thought of as $X\times Y\times I$, with the relations: $$ (x,y_0,t)\sim (x_0,y_0,t);$$ $$ (x_0,y,t)\sim (x_0,y_0,t);$$ $$ (x,y,1)\sim (x_0,y_0,1);$$ $$ (x,y,0)\sim (x_0,y_0,0).$$

The reduced suspension adds the relation $$ (x_0,y_0,t)\sim (x_0,y_0,0).$$

Now it is not hard to see you are quotienting out by the same relations for both constructions. Namely,

$$ (x,y_0,t)\sim (x_0,y_0,0);$$ $$ (x_0,y,t)\sim (x_0,y_0,0);$$ $$ (x,y,0)\sim (x_0,y_0,0);$$ $$ (x,y,1)\sim (x_0,y_0,0).$$

Some pictures from "Topology and Groupoids", Chapter 5, which you may find helpful are the join $X * Y$ as

and the subspace to be collapsed to a point to give the suspension of the smash product is where the two vertices on the mid line are the base points.