Are the categories ${\bf{Sets}}/2$ and ${\bf{Sets}} \times {\bf{Sets}}$ isomorphic? Awodey's exercise 
Let $2=\{a,b\}$ be any set with exactly $2$ elements $a$ and $b$. Define a functor $F:
{\bf{Sets}}/2\rightarrow{\bf{Sets}}\times{\bf{Sets}}$ with $F(f:X\rightarrow 2)=(f^{-1}(a),f^{-1}(b))$. Is this an isomorphism of categories?

This is something that I've done: In ${\bf{Sets}}/2$, let $h:X\rightarrow\{a,b\}$ such that $h^{-1}(a)\not=\emptyset$ and $h^{-1}(b)\not=\emptyset$ and let $f:X\rightarrow\{a,b\}$ such that $f^{-1}(a)=X$ and $f^{-1}(b)=\emptyset$. Then there is no arrow between $f$ and $h$ in ${\bf{Sets}}/2$, but there is one arrow between $F(f)=(X,\emptyset)$ and $F(h)=(h^{-1}(a),h^{-1}(b))$ in ${\bf{Sets}}\times{\bf{Sets}}$, so these two categories are not isomorphic.
Is this reasoning ok?
I am a complete novice in the category theory!
 A: Your argument is not correct. In fact, $F$ is fully faithful.
Define a functor $G : \mathsf{Set} \times \mathsf{Set} \to \mathsf{Set} / \{a,b\}$ by mapping two sets $A,B$ to the function $G(A,B) : A \coprod B \to \{a,b\}$ which maps the elmenets of $A$ to $a$ and the elements of $B$ to $b$. Here, $A \coprod B= A \times \{1\} \cup B \times \{2\}$ is the standard construction of the disjoint union. Then we have $FG \cong \mathrm{id}$ and $GF \cong \mathrm{id}$. Hence, $F$ is an equivalence of categories.
But $F$ is not an isomorphism. In fact, $F$ is not surjective on objects. If $A,B$ are two sets which are not disjoint, then $(A,B)$ doesn't lie in the image of $F$. (Notice, however, that $F$ lies in the essential image of $F$.)
Here is how I think about this: The notion of "isomorphism of categories" belongs to $0$-category theory $\approx$ set theory. It is much too strong to explain various "structural equalities of categories". A far better behaved notion is that of an "equivalence of categories". Only this belongs to $1$-category theory. If $A,B$ are two sets, then the property that $A,B$ are disjoint cannot be formulated in category theoretic terms, since this property is not invariant under self-equivalences of $\mathsf{Set} \times \mathsf{Set}$. In fact, we can make $A,B$ disjoint by replacing them by $A^* = A \times \{1\}$ and $B^* = B \times \{2\}$. Then $(A,B) \cong (A^*,B^*)$ in $\mathsf{Set} \times \mathsf{Set}$. In category theory, in order to state that two objects $A,B$ are disjoint, we need a larger object $S$ which contains both. Then their intersection is given by the pullback $A \times_S B$, and $A,B$ are called disjoint when this pullback is an initial object. In set theory, one often imagines the universe $V$ to be some kind of universal object which contains all other objects of consideration. Every set then has a unique embedding into $V$. In category theory, however, one really has to remember all the embeddings. In particular, there is a completely different point of view on subsets. For more on this, see SE/704593 and SE/295800. In my opinion category theory is the better language to explain structural mathematics.
A: I'm afraid your argument is not correct: there are morphisms from $f$ to $h$. 
Hint: Perhaps $F$ is not an $isomorphism$ of categories, but something slightly weaker?
