The n-Category Café

The decategorification argument goes like this. We’d already seen various theorems about sets that are strongly redolent of ordinary arithmetic, like

$$X \times (Y + Z) \cong (X \times Y) + (X \times Z)$$

and

$$X^{Y \times Z} \cong (X^Y)^Z.$$

Now, in this chapter we gave recursive definitions of addition, multiplication and exponentiation in $\mathbb{N}$, using the universal property of $\mathbb{N}$. (How else? The universal property is the only tool at our disposal.) Of course, we then wanted to prove that they satisfy equations like

$$x(y + z) = xy + xz$$

and

$$(x^y)^z = x^{yz}.$$

We could have done it by induction. But there’s no need! Instead, we argued as follows.

First define, for each $n \in \mathbb{N}$, the set

$$\mathbf{n} = \{ i \in \mathbb{N} : i \lt n \} = \{0, \ldots, n - 1\}.$$

Then show that the passage $n \mapsto \mathbf{n}$ respects addition, multiplication and exponentiation. For example, if we write $p = m + n$ then $\mathbf{p}$ is the coproduct of $\mathbf{m}$ and $\mathbf{n}$. This makes expressions like $\mathbf{m} + \mathbf{n}$ unambiguous: it doesn’t matter whether you add in $\mathbb{N}$ then turn the result into a set, or turn each of $m$ and $n$ into sets and then take the coproduct. Finally, show that

$$\mathbf{m} \cong \mathbf{n} \iff m = n.$$

This done, we can deduce the standard algebraic laws in $\mathbb{N}$ from general isomorphisms for sets. For instance, the fact that

$$\mathbf{m} \times (\mathbf{n} + \mathbf{p}) \cong (\mathbf{m} \times \mathbf{n}) + (\mathbf{m} \times \mathbf{p})$$

immediately implies that

$$m(n + p) = m n + m p,$$

for all $m, n, p \in \mathbb{N}$.

The same technique also works for inequalities, like $m \leq n \implies m + p \leq n + p$.

The only thing it doesn’t work for is cancellation, because that’s specifically a phenomenon for finite sets: it is true for natural numbers that $m + p = n + p \implies m = n$, but it’s not true for general sets (only finite sets) that $X + Z \cong Y + Z \implies X \cong Y$.

The decategorification method of proving the identities for natural numbers doesn’t save a vast amount of work. The inductions aren’t particularly onerous. But since we’d want to prove anyway that sums, products and powers of natural numbers corresponds to coproducts, products and exponentials of sets, it’s an opportunity to get something for free.