Equivalence relation in join of two spaces Let $X$ and $Y$ be topological spaces and let I be the unit interval. The join $X*Y$ is defined as
$$(X\times Y\times I)/\sim\,\,\mbox{with }\,\,I = [0; 1]$$
and the relation is given by $(x,y,0)\sim (x′,y,0), (x,y,1)\sim (x,y′,1)$ for $x\in X, y\in Y$.
I don't understand $\sim$ relation, i.e, $$\forall (x; y; t),\,\,(x'; y'; t')\in X\times Y\times I,\,\, (x; y; t) \sim (x'; y'; t') \leftrightarrow ??$$
Can you explain for me? Thank you very much!
 A: You need to go back to the intuitive idea of the join. 
Suppose $X,Y$ are two subsets of Euclidean space of sufficiently high dimensions that the line segments joining points $x \in X$ to points $y \in Y$ do not meet. The union of these line segments is then the join  $X * Y$. Its points are $rx + sy$ where $r,s\geqslant 0$, and $r+s=1$. Here are some pictures 
  
$$\text{Fig.5.5}$$
taken from Topology and Groupoids: pdf, Section 5.7. 
Of course  we do not want to use generally an embedding in Euclidean space, so we define $X*Y$ to consist of "formal points" $rx + sy$ with $r,s$ as above except that if $r=0$ we ignore $rx$, getting just $y,$ and if $s=0$ we ignore $sy$, getting just $x$. A useful  topology on $X*Y$ is then an initial topology  as detailed in that Section, and which  I won't labour on here. This topology makes the join associative. 
However a more commonly used topology is a final topology as an identification space of $X \times Y \times [0,1]$. 
The initial topology is good for deciding continuity of maps into the join, while the final topology is good for deciding continuity of maps out of the join. 
A: The relation says $(x, y, t) \sim (x', y', t')$ if and only if either 1) $y = y'$ and $t = t' = 0$ or 2) $x = x'$ and $t = t' = 1$
