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Why are the objects of $\mathbf{Mat}_\mathbb K$ the natural numbers? I came across this in the context of a functor $F:\mathbf{FdVect}_\mathbb K\rightarrow\mathbf{Mat}_\mathbb K$ where we map arrows (linear maps) to their respective matrices in a given basis and each object $\mathbb K ^n$ just to $n$. This all makes sense to me, except that as far as I can see matrices do not have type $m\rightarrow n$ where $n,m\in\mathbb N$. The text (Categories for the Practising Physicist) makes the following comment:

it strongly emphasizes that objects are but labels with no internal structure.

however if we merely need a set with the right cardinality wouldn't $\mathbb N$ do just as well as $\{\mathbb K ^n:n\in\mathbb N\}$ for $\mathbf{FdVect}_\mathbb K$?

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Sure. But using $n$ instead of $K^n$ "strongly emphasizes that objects are but labels with no internal structure." – Zhen Lin Jan 13 '13 at 13:31
up vote 3 down vote accepted

This is only definition.

How do you mean, 'matrices do not have type $m\to n$'? Actually, they have type $m\times n$, but that's it all about: interpret an $m\times n$ matrix as an $m\to n$ arrow in $\bf Mat_{\Bbb K}$.

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I mean type in the sense of domain and codomain. The arrows in $\mathbf {FdVect}_\mathbb K$ are linear maps $A:\mathbb K^n-> \mathbb K^m$ which makes sense to me since those the $\mathbb K^n$ are the objects of $\mathbf {FdVect}_\mathbb K$. On the other hand I don't know what it would even mean for something (let alone a matrix) to have domain $n$ and codomain $m$. – Sean D Jan 18 '13 at 15:48
It is just a formalism. A category is a quiver (directed graph) with composition, and a quiver is a pair of functions between any two sets: $c,d:E\to V$. In this case, an 'edge' (element of $E$) is a matrix $A$, say $n\times m$. Then the 'codomain' and 'domain' functions are defined as $c(A):=n$, $d(A):=m$ (or the other way around, I never know:). They are just abstract stuffs. – Berci Jan 18 '13 at 20:47

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