Example of Dual Category That Makes No Sense (to me) Example of Dual Category That Seems to Makes No Sense
I have read the several entries on mathstackexchange concerning the definition of dual categories.  For a purely theoretical point of view, the definitions seem straight forward - until I try to apply them to some simple categories and then the results make no sense.
First of all, when dealing with morphs that are functions, "turning the arrows around" to form the dual category does not mean using the inverses of the functions (which may not exist) - it means using the exact same functions but with the domain and codomain switched.  (Right?)
Now take this example: suppose the category C contains the two objects A and B, where A represents the positive real numbers and B represents the negative real numbers. One morph from A to B is the function f which maps each positive real number to its negative square root.  So, 4 is mapped to -2, and 9 is mapped to -3, etc.
Now to form the dual of the category C which contains A, B, and f we just turn the arrows/morphs/functions around.  This means for numbers in B (which are negative) we take their negative square roots, because that is what the function f does.  But the square root of a negative number cannot be a positive real number, i.e, cannot be in A.  So the arrow, i.e. function, from B to A in the dual category cannot exist, or makes no sense.
Perhaps my problem is the language used in the definitions.  But what am I misunderstanding?
 A: No, we don't use "the analogous function" (e.g. "Take the negative square root") - we use literally the same function.
The morphism from $B$ to $A$ in the category $C^{op}$ is the function $$f: A\rightarrow B: a\mapsto-\sqrt{b}.$$ It doesn't matter that this is a function from $A$ to $B$, and not from $B$ to $A$; a morphism from one object to another does not in any way have to be a function from the underlying set of the first to the underlying set of the second (nor, in fact, do objects need to have underlying sets in the first place).

For a "bigger" but less ad-hoc example, that might be easier to think about: in the category Set$^{op}$, a morphism from $A$ to $B$ is a function in the usual sense from $B$ to $A$. 
More natural, but more technical, examples include the homotopy category and the cobordism category.

It might be helpful at this point to consider a purely algebraic description of a category.
First, remember what (say) a group is: it's a set $G$ together with a binary operation $*$ on $G$, with some properties. It doesn't matter what the elements of $G$ are, and in fact we never talk about this - only the structure relating them is relevant.
OK, so along these lines, here's how to think about a category. First, let's ditch the objects for simplicity: we don't actually need them (objects can be conflated with their identity morphisms). A category, for us, is going to be a set $C$ (the set of morphisms) together with a partial (= not always defined) binary operation $*$ (composition), with the following properties:


*

*Associativity: If $(a*b)*c$ is defined, then so is $a*(b*c)$, and they are equal; and if $a*(b*c)$ is defined, then so is $(a*b)*c$, and they are equal.

*"Source" identities: For any $a$, there is a unique $e$ such that 


*

*$a*e$ is defined ($e$ is a morphism to the source of $a$).

*$e*e$ is defined ($e$'s source and target are the same).

*For all $b$, if $b*e$ is defined then $b*e=b$; and if $e*b$ is defined then $e*b=b$ ($e$ acts like an identity morphism).


*"Target" identities: the same as the above, but the first bullet says instead "$e*a$ is defined."
Note that this is purely algebraic, and we don't need to give any meaning to the elements of the set $C$.
Besides building intuition, this is a very fruitful idea: for example, it lets us define a group as a one-object category where every morphism has an inverse! If you unwind the picture above, you'll see that this means exactly that "$*$" is defined everywhere (since there's only one object), so there is only one identity morphism $e$, and associativity and inverses mean that this is just a group! This way of thinking of classical algebraic structures as categories with certain properties is incredibly fruitful, and leads to a number of useful generalizations; for instance, it motivates the definition of a groupoid as a group but with multiple objects, which in turn lets us define fundamental groupoids which are often better to work with than fundamental groups.
A: Let $\mathscr{C}$ be a category. The opposite category of $\mathscr{C}$ is the category $\mathscr{C}^{\mathrm{op}}$ where


*

*$\DeclareMathOperator{Ob}{Ob}\DeclareMathOperator{id}{id}\Ob(\mathscr{C}^{\mathrm{op}})=\Ob(\mathscr{C})$

*$\DeclareMathOperator{Hom}{Hom}\Hom_{\mathscr{C}^{\mathrm{op}}}(X, Y)=\Hom_{\mathscr{C}}(Y,X)$
That is, $\mathscr{C}^{\mathrm{op}}$ has the same objects as $\mathscr{C}$ and the same morphisms as $\mathscr{C}$. The statement "$f$ is a morphism $X\to Y$ in $\mathscr{C}^{\mathrm{op}}$" is the same statement as "$f$ is a morphism $Y\to X$ in $\mathscr{C}$."
For example, consider $\mathscr{C}=\mathsf{Set}$. Here, we have a function $f\in\Hom_{\mathsf{Set}}(\Bbb R^+,\Bbb R)$ given by $f(x)=\log(x)$. The function $f$ is also a morphism in $\mathsf{Set}^{\mathrm{op}}$, except now $f\in\Hom_{\mathsf{Set}^{\mathrm{op}}}(\Bbb R,\Bbb R^+)$ since $\Hom_{\mathsf{Set}^{\mathrm{op}}}(\Bbb R,\Bbb R^+)=\Hom_{\mathsf{Set}}(\Bbb R^+,\Bbb R)$. 
In short, declaring that $f:X\to Y$ in $\mathsf{Set}^{\mathrm{op}}$ is to declare that $f$ is a map of sets with domain $Y$ and target $X$.
In your example, we have a category $\mathscr{C}$ where $\Ob(\mathscr{C})=\{\Bbb R^+,\Bbb R^-\}$ and we are told that one morphism $f\in\Hom_{\mathscr{C}}(\Bbb R^+,\Bbb R^-)$ is the map of sets $f:\Bbb R^+\to\Bbb R^-$ given by the formula $f(x)=-\sqrt{x}$.
The opposite category $\mathscr{C}^{\mathrm{op}}$ also contains this morphism except now $f\in\Hom_{\mathscr{C}^{\mathrm{op}}}(\Bbb R^-,\Bbb R^+)$. 
It is possible that this $f$ is the only nontrivial morphism in $\mathscr{C}$, in which case we have an explicit description of $\mathscr{C}$:
\begin{align*}
\Ob(\mathscr{C}) &= \{\Bbb R^+, \Bbb R^-\} &
\Hom_{\mathscr{C}}(\Bbb R^+,\Bbb R^+) &= \{\id_{\Bbb R^+}\} &
\Hom_{\mathscr{C}}(\Bbb R^-, \Bbb R^-) &= \{\id_{\Bbb R^-}\} \\
&& \Hom_{\mathscr{C}}(\Bbb R^+,\Bbb R^-)&=\{f\} &
\Hom_{\mathscr{C}}(\Bbb R^-,\Bbb R^+) &=\varnothing
\end{align*}
This gives the explicit description of $\mathscr{C}^{\mathrm{op}}$:
\begin{align*}
\Ob(\mathscr{C^{\mathrm{op}}}) &= \{\Bbb R^+, \Bbb R^-\} &
\Hom_{\mathscr{C^{\mathrm{op}}}}(\Bbb R^+,\Bbb R^+) &= \{\id_{\Bbb R^+}\} &
\Hom_{\mathscr{C^{\mathrm{op}}}}(\Bbb R^-, \Bbb R^-) &= \{\id_{\Bbb R^-}\} \\
&& \Hom_{\mathscr{C^{\mathrm{op}}}}(\Bbb R^+,\Bbb R^-)&=\varnothing &
\Hom_{\mathscr{C^{\mathrm{op}}}}(\Bbb R^-,\Bbb R^+) &=\{f\}
\end{align*}
Note that here $\id_{\Bbb R^{\pm}}$ need not be a map of sets!
A: I may be a little late to the party, but I'm having a tough time with categories as well (I've tried digging into the subject a few times over the past few years, I feel that I'm actually making some progress with "Category Theory for Programmers").  The notion that "the dual category simply is" doesn't really cut it for me, so with SET I'm thinking about it in the following way, someone please correct me if I'm wrong.
Given SET is sets and functions between them, I'm thinking that the morphisms in SET$^{OP}$ are equivalence classes indexed by the dual-sources.  While not every function has an inverse, the psuedo-inverse given by $f^{-1}(x) = \{y \,|\, f(y) = x\}$ is a well-defined, and seems to do the trick.  The new initial object is the equivalence where all elements are equivalent to one another, and the terminal object is the equivalence where nothing is equivalent to anything (not even itself).
Am I one the right track?  Is it more like "the dual category of SET is isomorphic to what I've described" or something?  I know you're "not supposed to care" about the morphisms, but when they're something as nice as functions I'd rather just grab 'em by the... unctions and have a more concrete intuition about it.
