Eq

First, I’d like to start with a quick sentence a friend said a really long time ago while we were taking Calculus I:

A set is just its cardinality.

At first, my reaction was to dismiss that as nonsense. I mean, look at these sets: $\{1\}$, $\{2\}$. They look very different to me, but their cardinality is the same. You mean to tell me that I can uniquely identify sets by their cardinality? I’ll cut the sarcasm now; yes you can do that identification and it’s actually not that crazy.

First, let’s take a stop by the Yoneda Lemma real quick:

Let $\mathcal{C}$ be a category¹, and let $F : \mathcal{C}^{\text{op}} \to \mathbf{Set}$ be a functor. Then, for every object $a \in \mathcal{C}$, there is a natural isomorphism²

$$\text{Nat}\left( \text{hom}_{\mathcal{C}}(-, a), F \right) \cong F(a).$$

Now, what the f**k does that mean Kobe Bryant? For now, let’s just take for granted that this is true (maybe in the future I’ll make a blog post explaining the intuition behind this and proving it). Take an object $x \in \mathcal{C}$¹ and consider the hom-functor $\hom_\mathcal{C} (-, x)$. Well, that hom-functor is of the form $\hom_\mathcal{C}(-, x) : \mathcal{C}^{\text{op}} \to \mathbf{Set}$. So what would happen if you were to apply the Yoneda Lemma to this? Well you get that for every object $a \in \mathcal{C}^{\text{op}}$ there is a natural isomorphism² of the form

$$\text{Nat}\left(\hom_\mathcal{C}(-, a), \hom_\mathcal{C}(-, x)\right) \cong \hom_\mathcal{C}(a, x).$$

That last expression is the set of morphisms from $a$ to $x$. Which implies that you can uniquely transform