Maps in Opposite Categories

Given some category ${\mathcal C}$, then the opposite category will consist of the same objects with the morphisms "turned around." Given $f:A\rightarrow B$, for $A,B$ objects of ${\mathcal C}$, then in general do we have a canonical $f^{op}:B\rightarrow A$ which is induced by $f$?

Slight Motivation: This is something that's just been bothering me recently, and I feel like I am thinking about this concept incorrectly. To make this question slightly less vague, if we had, say, the category of rings and we had the mapping $f:{\mathbb Z}\rightarrow {\mathbb Z}/2{\mathbb Z}$ by $f$ sends even numbers to 0 and odd numbers to 1, then what is the corresponding map in the opposite category, $f^{op}:{\mathbb Z}/2{\mathbb Z}\rightarrow {\mathbb Z}$ induced by $f$?

Also, I always see this $f^{op}$ that I've mentioned just written as $f$; is this shorthand, or is there something deeper there?

The conceptual hurdle you have to pass here is to think of a category as really just being a collection of abstract dots called "objects" and abstract arrows between them called "morphisms" satisfying certain axioms. Just as an abstract group isn't defined with reference to a specific set it acts on, a category isn't defined with reference to a specific "implementation" of it as a collection of sets and maps between them, and in fact unlike the case with groups, there exist categories which cannot be implemented as a collection of sets and maps between them; such categories are said not to be concretizable.

So when we take the opposite category, we are literally doing nothing more than reversing the direction of the arrows. $f^{op}$ is just another name for $f$ with its direction reversed: it doesn't necessarily have any more concrete existence than that.

Strictly speaking, this is not an answer to the question, but more of an elaboration on Thomas Andrews' answer which may be helpful. Paralleling the "When in Rome be a Roman", we should repeat aloud "When doing category theory think categorially". So what is the categorial way of defining new objects? Universal properties. Is there a universal property for the opposite category? But why yes, sir. For a category $A$ there is a contravariant functor $A\to A^{\ast}$ universal among contravariant functors: every contravariant functor $A\to B$ factors uniquely through $A\to A^{\ast}$ (via a covariant functor). So what is $A^{\ast}$? Who cares, it is just the categorial thingy we need to reduce contravariant functors to ordinary, covariant ones.

Later edit: at the risk of going off-topic, I have to protest against the abuse of reductionist expressions like "just a formalism". What could it possibly mean? One way to formulate it, is to ask if the concept (e.g. opposite category) is doing any real work for us. The answer, at least on a first appraisal, seems to be a clear no; it is just a convenient shortcut, little more than a word-game. In response to this, one could invoke the duality principle, but this can still be reasonably seen as "just a formality". On the other hand, the universal property of the opposite category tells us that we have a new object, so we can ask all sorts of questions about it, such as:

(A) for a given category $A$ can we characterize $A^{\ast}$? Pragmatically, this means searching for a "more or less concrete" (deliberately vague expression) category $B$ and an equivalence $A^{\ast}\simeq B$.

As soon as we have the concept of opposite category, question (A) imposes itself very naturally. But is it an interesting question? Yes! There is a body of work devoted to it, not just in the general (some relevant buzzwords: dualities, ambimorphic objects, etc.) but also in particular cases.

As has been noted, the opposite category is purely a formalism.

Consider the axioms of a category, we realize that while morphisms have a direction - they go from some $A$ to some $B$ in the category - the direction doesn't affect any of the axioms. So if we invert the sense of direction on a category, we get another category which encodes all the same ideas but in the opposite direction.

Now, let's say that we take the category of rings, and we define for each ring $R$, the partially ordered set $\mathcal{I}(R)$ to be the set of ideals of $R$ ordered under inclusion. So $\mathcal{I}$ maps objects in the category $\mathbb{Ring}$ to objects in the category $\mathbb{Poset}$.

Then if we have a map $f\in \operatorname{Hom}_{Ring}(R,S)$, we get a map $\mathcal{I}(f)\in \operatorname{Hom}_{Poset}(\mathcal{I}(S),\mathcal{I}(R))$, defined as $\mathcal{I}(f)(J)=f^{-1}(J)$, when $J$ is an ideal of $S$.

We could think of this as a map which sends $\mathbb{Ring}$ to $\mathbb{Poset}$ which reverses the morphism directions, or we could think of this as a map from $\mathbb{Ring}$ to $\mathbb{Poset}^{op}$ which preserves the morphism direction (or, equivalently, a map from $\mathbb{Ring}^{op}$ to $\mathbb{Poset}$.)

In category theory, we get lots of these "functors" between categories that reverse directions - so-called "contravariant functors." But the key is that, since we have the opposite category, contravariant functors can be seen as "covariant functors" (direction-preserving functors,) so we don't have to deal with lots of special cases when we are proving theorems about functors in general.

What is the opposite category exactly depends on a chosen formalism. To define $\mathcal D:=\mathcal C^{op}$, first $ob_\mathcal D:=ob_\mathcal C$. The remainder depends on how morphisms are defined:

• A category has a set $mor$ and functions $dom$ (domain), $cod$ (codomain). Then define $mor_\mathcal D:=mor_\mathcal C$, $dom_\mathcal D:=cod_\mathcal C$, $cod_\mathcal D:=dom_\mathcal C$, $f^{op}:=f$.
• A category has a family $hom$ of sets indexed by objects and objects. Then define $hom_\mathcal D(A,B):=hom_\mathcal C(B,A)$, $f^{op}:=f$.

In both cases $-^{op}$ on morphisms is the identity function.

I thought I'd post this answer in response to the OP's secondary question of when the morphisms of opposite categories are maps of sets. The answer, as it turns out, is that there is a faithful embedding $\mathcal{C}^{\text{op}} \to \textbf{Set}$ if and only if there is a faithful embedding $\mathcal{C} \to \textbf{Set}$, but for stupid reasons.

First, let us consider $\mathbf{Set}^{\text{op}}$. This is a concrete category: consider the contravariant power set functor $\mathcal{P} : \textbf{Set}^{\text{op}} \to \textbf{Set}$. It is representable: $\mathcal{P} \cong \text{Hom}_{\textbf{Set}}(-, 2)$, and is clearly faithful: if $f, g : X \to Y$ are different maps, then for some $x \in X$, $f(x) \ne g(x)$, thus, $x \notin g^{-1}(\{f(x)\})$ but $x \in f^{-1}(\{f(x)\})$. Hence, if there is a faithful embedding $U : \mathcal{C} \to \textbf{Set}$, we can obtain a faithful embedding $$\mathcal{C}^{\text{op}} \xrightarrow{U^{\text{op}}} \textbf{Set}^{\text{op}} \xrightarrow{\mathcal{P}} \textbf{Set}$$ but this embedding is rather ugly and it is sometimes possible to do better, e.g. in the case of integral algebras over an algebraically closed field $k$, the contravariant representable functor $\text{Hom}_{k\textbf{-Alg}}(-, k)$ gives us a faithful embedding of the category of irreducible affine varieties over $k$ into the category of sets.

Moreover, any small category $\mathcal{C}$ is concrete, though again for silly reasons: the Yoneda embedding is a faithful functor $\mathbf{y} : \mathcal{C} \to [\mathcal{C}^{\text{op}}, \textbf{Set}]$, and when $\mathcal{C}$ is small, $[\mathcal{C}^{\text{op}}, \textbf{Set}]$ is concrete. Indeed, if $F : \mathcal{C}^{\text{op}} \to \textbf{Set}$ is a contravariant functor, we just map $F$ to the set $$\coprod_{c \in \text{Ob}(\mathcal{C})} F(c)$$ and this is obviously faithful, because a natural transformation $F \to G$ is just a collection of maps $F(c) \to G(c)$. Unfortunately, most interesting categories are not small.