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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?

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Opposite categories are, in a sense, a pure formalism, which we use to express certain concepts in a more elegant way. For example, it is (in my opinion) better to have one definition of a functor than separate concepts of a covariant and contravariant functor. Consequently, the arrows and objects in an opposite category no longer represent any "concrete" maps or structures from your original category (this is sometimes only partially true, but let's not get into that). – Miha Habič Jun 11 '11 at 12:58
@Theo: well, sometimes people use "map" as a synonym for "morphism" in a category. What a morphism need not be is a function. – Pete L. Clark Jun 11 '11 at 13:00
@Pete: Right, I do that too. Using function would have resulted in a much better formulation of that remark. The ambiguity only occured to me after your ping. It pays to have native speakers around :) Thanks! – t.b. Jun 11 '11 at 13:10
To make myself perfectly clear here's my comment again implementing @Pete's comment: There's nothing deeper here than the observation that morphisms are not functions. There are no sets assumed to be underlying the objects, so it doesn't even make sense to speak of functions between the objects. Maps=morphisms are simply elements of the Hom-sets. By definition $\hom_{\mathcal{C}^{op}}{(B^{op},A^{op})} = \hom_{\mathcal{C}}(A,B)$ and that's why the morphism $f^{op}: B^{op} \to A^{op}$ corresponding to $f:A \to B$ is often simply denoted by $f$. – t.b. Jun 11 '11 at 13:13
@james: There are some special instances where both the category and its opposite can be implemented as a subcategory of $\mathbf{Set}$. For example, if $k$ is an algebraically closed field, the opposite of the category of integral $k$-algebras is the category of irreducible affine varieties over $k$. – Zhen Lin Jun 11 '11 at 16:31
up vote 17 down vote accepted

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.

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It is worth noting that MacLane refers to Objects and Arrows. – Baby Dragon Aug 22 '14 at 20:34

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.

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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.

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I appreciate your later edit, since this really addresses my problem with how I usually see the opposite category defined. It's also nice to know that there's a universal property for this, but I don't think I've seen opposite categories done this way. Have I been overlooking it, or is there a nice paper/text which treats them like this? – james Jun 13 '11 at 0:53
Not that I know of; but then again, it is a good exercise to do it yourself. – G. Rodrigues Jun 13 '11 at 22:36

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.

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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.

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The definition $f^{op}:=f$ causes me some discomfort, it feels a bit too strict, against the categorical spirit of preferring 'essentially equal' to 'equal'. E.g. in your second definition I would say something like: for each $A,B$ we take a set $S$ with a bijection $\phi:Hom_{\mathcal C}(B,A)\to S$, and then let $Hom_{\mathcal D}(A,B):=S$ and $f^{op}:=\phi(f)$. Or make no reference to $S,\phi$ and fit it into a universal property, for which this proves existence. Of course it doesn't matter that much. – wildildildlife Jun 12 '11 at 16:41
@wildildildlife: I see “essentially equal” rather as annoyance. It makes proofs bigger. BTW, use “@” when discussing. – beroal Jul 6 '11 at 15:14
I was under the impression that authors are automatically notified about comments on their own postings. This happened to me. And now that I included "@beroal" in this comment, it automatically disappears! – wildildildlife Jul 6 '11 at 15:30

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