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.
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).