I'm looking at the skeletal FinSet category ; there is one object for each natural number $n$ (including $n=0$), and a morphism from $m$ to $n$ is an $m$-tuple $(f_0, \ldots, f_{m-1})$ of numbers satisfying $0 \leq f_i \lt n. $ (stolen from http://ncatlab.org/nlab/show/FinSet)
If I understand it correctly it, my objects are $\emptyset, \{0\}, \{0,1\}, \{0,1,2\}, \ldots$ and so on. But I'm having trouble understanding the morphisms. Exactly what happens when I go from a object $m$ to an object $n$, when $\left\vert{n}\right\vert >\left\vert{m}\right\vert$? Do the $f_i$:s in my $m$-tuple actually do anything to the members of $m$ or do I just simply end up in $n$, in a sense making $m$ as big as $n$ in a transformation? Am I Missing Something?
Any help would be highly appreciated!
Remember that in a category $C$, the morphisms do not need to be functions. Also, you can define categories in an object-free (if I'm correct, this is cited in Mac Lane, Categories for the Working Mathematician).
In the category $\mathbf{FinSet}$, what is done is "identifying" each finite set $X$ with its size $|X|$ (or the object corresponding to it). We do not lose information of $X$ (as a set) because two finite sets are isomorphic (in the usual sense) iff they have the same number of elements.
So, given an object $n$ of $\mathbf{FinSet}$, we may think of $n$ as a set with $n$ elements, say $n=\left\{0,1,\ldots,n-1\right\}$ (this can actually be done if we work with finite ordinals). In this sense, the set of functions (morphisms) $f:m\to n$ is in bijection with the set of $m$-tuples as in your question, by the bijection $f\mapsto (f(0),\ldots,f(m-1))$.