Keith wrote:

> Think of it this way, the functor \$$\Delta\$$ is a category copying machine. It takes a category and copies or *duplicates* it.

> \$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C}, \$

Level slip? I wouldn't say this \$$\Delta\$$ is duplicating the category. I'd say it's duplicating each object _in_ the category. Let's call it \$$\Delta_{\mathcal{C}}\$$. It has

$\Delta_{\mathcal{C}}(c) = (c,c)$

for each object \$$c\$$ in \$$\mathcal{C}\$$. Of course, there is also something that's duplicating the category \$$\mathcal{C}\$$. There's a functor

$\Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat} \times \mathbf{Cat}$

and we have

$\Delta_{\mathbf{Cat}}(\mathcal{C}) = (\mathcal{C}, \mathcal{C})$

There's also a functor

$\times_{\mathbf{Cat}} : \mathbf{Cat} \times \mathbf{Cat} \to \mathbf{Cat}$

that takes the product of any two categories:

$\times_{\mathbf{Cat}} (\mathcal{C}, \mathcal{D}) = \mathcal{C} \times \mathcal{D} .$

We can compose these two functors and get a functor

$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$

which does this:

$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} (\mathcal{C}) = \mathcal{C} \times \mathcal{C}$

So, it sends any category \$$\mathcal{C}\$$ to the category \$$\mathcal{C} \times \mathcal{C} \$$.

Finally, there's a natural transformation from the identity functor

$1_\mathbf{Cat} : \mathbf{Cat} \to \mathbf{Cat}$

to this composite functor

$\times_{\mathbf{Cat}} \circ \Delta_{\mathbf{Cat}} : \mathbf{Cat} \to \mathbf{Cat}$

To each category \$$\mathcal{C}\$$, this natural transformation assigns a functor from \$$\mathcal{C}\$$ to \$$\mathcal{C} \times \mathcal{C} \$$. And what is this functor? It's

$\Delta_{\mathcal{C}}: \mathcal{C} \to \mathcal{C} \times \mathcal{C}.$

Hey, we've come full circle!