# Understanding smash product of discrete space.

Let $X = \{x_0, y_0\}$ and let $Y = \{y_0, y_1\}$.

Then $X \times Y = \{ (x_0, y_0), (x_0, y_1), (x_1, y_0), (x_1, y_1) \}$

and let us identify $x_0 \sim y_0$

Then $X \land Y = X \times Y / X \wedge Y$.

I have no idea how to interpret this space. First of all, should I be thinking about the coordinates as vectors. Then if $x_0 \sim y_0$, would have we $(x_0, y_1) \sim (y_0, y_1)$? How would I picture this?

I appreciate any help with some intuition as to how to imagine these spaces.

First, $$X \wedge Y = (X \times Y) / (X \color{red}{\vee} Y),$$ not what you wrote. I suspect this wasn't a mere typo considering the rest of your post... $X \vee Y$ is called the wedge sum. It is in the wedge sum $X \vee Y$ that one has the identification $x_0 \sim y_0$, not in the smash product $X \wedge Y$. The wedge sum is obtained by taking the disjoint union $X \sqcup Y$, and then identifying $x_0 \sim y_0$. So in your case, $$X \vee Y = \{u, x_1, y_1\},$$ where $u = [x_0] = [y_0]$ is the result of identifying $x_0 \sim y_0$.
It doesn't make sense to say $x_0 \sim y_0$ in the smash product. The smash product's definition is more simply $$X \wedge Y = (X \times Y) / (\{x_0\} \times Y \cup X \times \{y_0\}).$$ So in your case, where $X \times Y = \{ (x_0, y_0), (x_0, y_1), (x_1, y_0), (x_1, y_1) \}$, then $$X \wedge Y = \{v, (x_1, y_1)\}$$ where $v = [(x_0, y_0)] = [(x_0, y_1)] = [(x_1, y_0)]$ is the result of quotienting by $(\{x_0\} \times Y \cup X \times \{y_0\})$.
Then why does one say $X \wedge Y = (X \times Y) / (X \vee Y)$? Because the wedge sum embeds in $X \times Y$: there is a map $i : X \sqcup Y \to X \times Y$ given by $i(x) = (x,y_0)$ if $x \in X$, and $i(y) = (x_0,y)$ if $y \in Y$. This map satisfies $i(x_0) = i(y_0) = (x_0,y_0)$, and so there is an induced map $\bar{i} : X \vee Y \to X \times Y$. This last map $\bar{i}$ is injective, and $X \wedge Y$ is the quotient $X \times Y / \bar{i}(X \vee Y)$, more simply denoted as $X \times Y / X \vee Y$ if there's no confusion.