Smash product of two pointed spaces In the topology book I'm reading I found the following statement:
Let $(X, x_0), (Y, y_0)\in Top_*.$ The "smash product" of $(X, x_0), (Y, y_0)$ is defined as
$$X\wedge Y := X\times Y/X\vee Y,$$
where $$X\vee Y:=  X \bigsqcup Y/\sim = X\bigsqcup Y/\{x_0, y_0\}$$
My question is why $$X\wedge Y = (X\times Y)/(X\times \{y_0\})\cup (\{x_0\}\times Y)$$
I don't know how to prove it and I hope, that someone can help.
 A: Your definition of $X\wedge Y$ is an abuse of notation (albeit a common one), and the actual definition is that
\begin{equation}
X\wedge Y = (X\times Y)/(X\times \{y_0\})\cup (\{x_0\}\times Y).\tag{1}
\end{equation}
Indeed, the "definition" 
\begin{equation}
X\wedge Y := X\times Y/X\vee Y \tag{2}
\end{equation}
 is strictly speaking meaningless since $X\vee Y$ is not a subset of $X\times Y$.  Statement (2) is actually just an informal shorthand for statement (1), since $X\vee Y$ is homeomorphic to the subset $(X\times \{y_0\})\cup (\{x_0\}\times Y)\subseteq X\times Y$ (the map $f:X\coprod Y\to(X\times \{y_0\})\cup (\{x_0\}\times Y)$ given by $f(x)=(x,y_0)$ for $x\in X$ and $f(y)=(x_0,y)$ for $y\in Y$ satisfies $f(x_0)=f(y_0)$ and hence induces a map $g:X\vee Y\to (X\times \{y_0\})\cup (\{x_0\}\times Y)$, which you can show is a homeomorphism).  So when your book writes (2), what it actually literally means is (1).
A: Although the question is old now, let me remark that the wedge $(X,x_0) \vee (Y,y_0)$  is the sum, or coproduct, of $(X, x_0), (Y, y_0)$ in the category $Top_*$. Recall that a categorical sum of objects $A_1,A_2$ consists of an object $A_1 + A_2$ and two morphisms $i_k : A_k \to A_1 + A_2$ with a certain universal property which determines the "sum triple" $(A_1 + A_2,i_1,i_2)$  uniquely up to isomorphism.
There are various concrete constructions of the wedge $\vee$. Let us denote by $\vee_{prod}$ the following:
$$(X,x_0) \vee_{prod} (Y,y_0) = X\times \{y_0\}\cup \{x_0\}\times Y$$ with the subspace topology from $X \times Y$ and basepoint $(x_0,y_0)$. The map $i_X : (X,x_0) \to (X,x_0) \vee_{prod} (Y,y_0)$ is given by $i_X(x) = (x,y_0)$, similarly $i_Y$.
We have a map 
$$j_X : (X,x_0) \to (X,x_0) \times(Y,y_0) = (X \times Y, (x_0,y_0)), j_X(x) = (x,y_0) ,$$
similarly $j_Y$. Given any wedge construction $\vee$, the universal property of the sum produces a unique map $j : (X,x_0) \vee (Y,y_0) \to (X,x_0) \times(Y,y_0)$ such that $j \circ i_X = j_X, j \circ i_Y = j_Y$.
The smash product is then defined as the pushout of the pair of maps $(j,c)$, where $c : (X,x_0) \vee (Y,y_0) \to \ast$ = one-point space. The map at the opposite side of $c$ is a quotient map $\hat{c} : (X,x_0) \times(Y,y_0) \to (X,x_0) \wedge (Y,y_0)$.
If you look at the standard construction of the pushout as a quotient of the product $(X,x_0) \times(Y,y_0) \times \ast$ and take $\vee_{prod}$ as wedge construction, you will see it results precisely in your concrete definition which we denote for the moment by $(X,x_0) \wedge_{prod} (Y,y_0)$.
You can alternatively take any other wedge $\vee$ and prove that the quotient map $p : (X,x_0) \times(Y,y_0) \to (X,x_0) \wedge_{prod} (Y,y_0)$ and the unique pointed map $a : \ast \to  (X,x_0) \wedge_{prod} (Y,y_0)$ complete $(j,c)$ to a pushout diagram. This is done by verifying the universal property of the pushout using the universal property of the sum.
