5
$\begingroup$

The natural numbers can be thought of as being the category of finite sets modulo isomorphisms, since:

  • $|A \times B| = |A| \times |B|$
  • $|A + B| = |A| + |B|$
  • $|A^B| = |A|^{|B|}$
  • $|\text{Aut}(X)| = |X|!$

What category can we form that has real numbers as objects and categorical products, sums, exponentials, etc. are the regular operations on real numbers?

$\endgroup$

1 Answer 1

4
$\begingroup$

Unfortunately, this isn't possible, just from the conditions about the product and coproduct. This is a result of numbers having inverses under $+$ and $\times$, which doesn't happen in $\mathbb{N}$

First note that the description of products ensures that $1$ is an terminal object, since for all $x$ the product of $1$ and $x$ is $x$. This is because for any morphism $f:x \rightarrow 1$, by the naturality of the projection morphisms, we have that

$$x \times 1 \xrightarrow{f \times Id} 1 \times 1 \xrightarrow{\pi_2} 1$$

is equal to

$$x \times 1 \xrightarrow{\pi_2} 1 \xrightarrow{Id} 1$$

In other words, by chasing around the diagram and using the fact that $\pi_2: 1 \times 1 \rightarrow 1$ is invertible, the only possible morphism from $x$ to $1$ is $x \cong x \times 1 \xrightarrow{\pi_2} 1$.

Similarly, $0$ must be initial.

Now, for any $x$, we have that $x + (-x) = 0$. For any other $y$, we have that $Hom(x,y) \times Hom((-x),y) = Hom(0,y)$ which is a singleton and so $x$ (and $(-x)$) is initial as well and therefore isomorphic to $0$.

Even if you leave out the negative reals a similar problem occurs with $x \times \frac{1}{x} = 1$ proving that every object (but $0$) is terminal.

$\endgroup$

You must log in to answer this question.