TITLE: Why do these equivalent categories seem to behave differently?
QUESTION [8 upvotes]: Write $\{1\}$ for the terminal category, and let $\Omega$ denote a category with two distinct (but isomorphic objects), call them $1_\Omega$ and $0_\Omega$, and a total of four arrows; two identities, an arrow $1_\Omega \leftarrow 0_\Omega$, and another arrow $0_\Omega \leftarrow 1_\Omega$.
Write $j : \Omega \leftarrow \{1\}$ for the functor given by $j(1)=1_\Omega.$
Then intuitively, the pair $(\Omega,j)$ should be able to classify the full subcategories of any other category. Explicitly, given a category $\mathbf{C}$ and a subcategory $\mathbf{S},$ there is a functor $P_\mathbf{S} : \Omega \leftarrow \mathbf{C}$ given as follows:

$P_\mathbf{S}(X)=1$ iff $X \in \mathbf{S}$
$P_\mathbf{S}(X)=0$ iff $X \notin \mathbf{S}$

The arrows go the only place they can go.
Furthermore, if we restrict to full subcategories, then this process should be reversible.
But this makes no sense!
In particular, the category $\Omega$ is equivalent to the category $\{1\}$, and hence should have no advantages over $\{1\}.$ Yet we cannot classify full subcategories using the arrow $\{1\} \leftarrow \{1\}$.

What on earth is going on here? Why do these equivalent categories seem to behave differently?

REPLY [21 votes]: Your confusion is the following: you have two categories $1$ and $\Omega$ which are equivalent, but not isomorphic. You note that for any category $C$ there is a unique functor $C \to 1$ but that functors $C \to \Omega$ seem to correspond to full subcategories.
There is a tight analogy between categories and topological spaces (so tight it can be realized by a functor, the geometric realization of the nerve) that goes like this:

Categories are like spaces.
Functors are like maps between spaces.
Isomorphisms of categories are like homeomorphisms of spaces.
Natural transformations are like homotopies.
Functor categories are like spaces of maps between spaces.
Equivalences of categories are like homotopy equivalences of spaces. 

In particular, two spaces (say the point and the interval $[0, 1]$, which are quite analogous to the terminal category and $\Omega$) can be homotopy equivalent, but not homeomorphic, and hence the set of maps from some space $X$ into those two spaces can look very different. However, it's always true that if two spaces are homotopy equivalent, then the spaces of maps from $X$ to those two spaces are also homotopy equivalent: for example, the space of maps from a space $X$ into $[0, 1]$ is always contractible. 
The corresponding statement for categories is that if two categories are equivalent, then the categories of functors (and natural transformations) from any other category $C$ to those categories are equivalent. In particular, we conclude that the category of functors $C \to \Omega$ must in fact be equivalent to the terminal category. 
So, what are the morphisms in this category anyway? Suppose $F, G : C \to \Omega$ are two functors. What is a natural transformation between them? Well, it's a collection of maps $\eta_c : F(c) \to G(c)$ such that... but wait. In $\Omega$ there's a unique morphism between any two objects. Hence all of the $\eta_c$ are in fact uniquely determined, and automatically satisfy the necessary compatibility condition to define a natural transformation. Moreover, every such natural transformation is invertible because its inverse is also uniquely determined! 
At this point let me introduce the following useful terminology: a category is contractible if there is a unique morphism between any two objects. This condition is equivalent to being equivalent to the terminal category. What we've written down is more or less a proof that if $\Omega$ is a contractible category, then so is any functor category $[C, \Omega]$.