Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some confirmation: is the opposite category transformation always a functor?

Also, isn't it always the case that $C^{\text{op}} = C$, since the the way we label an arrow does not matter?

share|cite|improve this question
I don't understand what you mean by "always" here. "Opposite category" is a single functor $\text{Cat} \to \text{Cat}$ from the category of (say, small) categories to itself. – Qiaochu Yuan Nov 10 '12 at 21:58
@QiaochuYuan: You're right, "always" should not be there. Thanks for noticing it. – Andŕe Nov 10 '12 at 22:16
up vote 6 down vote accepted

Here is a silly example. Form a category with objects $a,b,c$ and morphisms $f:a\to b,g:a\to c$ (and identities). This category has an initial object, namely $a.$ On the other hand, its opposite category clearly does not have an initial object ($a$ becomes terminal). Thus the two categories must be distinct.

share|cite|improve this answer
Not so silly :) – Manos Nov 10 '12 at 21:57
@Andrew: Thanks, for giving an example helped me "see it". – Andŕe Nov 10 '12 at 22:19
Dear @Andŕe, you're welcome, I hoped as much! – Andrew Nov 10 '12 at 22:22

Yes, going from $C$ to $C^{op}$ is a contra-variant functor.

The equality $C^{op}=C$ is not true. Don't be misled by the fact that these two categories have exactly the same objects. The morphisms differ, and recall that a category is not simply the collection of objects, rather the objects together with the morphisms.

share|cite|improve this answer
Why contravariant? – Berci Nov 10 '12 at 21:46
If $F$ is the functor $C \rightarrow C^{op}$, and $f:A \rightarrow B$ is a morphism in $C$, then $F(f)$ will be a morphism from $B=F(B)$ to $A=F(A)$. By definition of the opposite category, arrows from $A$ to $B$ in $C$ are viewed as arrows from $B$ to $A$ in $C^{op}$. – Manos Nov 10 '12 at 21:52
Ahh... I thought, it is about the $()^{op}$ functor... – Berci Nov 10 '12 at 21:59
Depending on foundations, $C^\circ$ also has the same morphisms as $C$ -- the difference is in the composition, source, and target operations. For the category corresponding to an abelian group, we would even have $C^\circ = C$ as a true equality. – Hurkyl Nov 10 '12 at 22:18
@Hurkyl: and before posting this question, I was looking at $Mat$, the category of matrices. Just to be sure, in this case it is true that $Mat^{\text{op}} = Mat$, right? – Andŕe Nov 10 '12 at 22:22

The objects of $\mathcal C^{\text{op}}$ are exactly the same as the objects of $\mathcal C$.

The morphisms of $\mathcal C^{\text{op}}$ are backwards versions of the ones in $\mathcal C$.

If we have a morphism $A \color{green}{\longrightarrow} B$ in $C$, we have a morphism $B \color{blue}{\longrightarrow} A$ in $\mathcal C^{\text{op}}$.

These are different categories because there might be a morphism $A \color{green}{\longrightarrow} B$ in $C$ but no morphism $A \color{blue}{\longrightarrow} B$ in $C^{\text{op}}$.

There is a contravariant functor $$ \begin{array}{rrcl} &\text{Dual}&:& \mathcal C \longrightarrow \mathcal C^{\text{op}} \\ &\text{Dual}(A)&=& A \\ &\text{Dual}(A \overset{f}{\color{green}\longrightarrow} B)&=& B \overset{f}{\color{blue}{\longrightarrow}} A \end{array}$$ which is an isomorphism of categories though.

share|cite|improve this answer
  1. Yes, there is a functor $Cat\to Cat$ that maps $C\mapsto C^{op}$ and maps a functor $F:C\to D$ to the corresponding $C^{op}\to D^{op}$.
  2. No. Direction does matter. Well, how to say, it is easy to figure out $C^{op}$ once we know $C$, but these are not equivalent, neither identical.
share|cite|improve this answer
But sperners lemma's answer above gave an isomorphism $Dual : \mathcal{C} \to \mathcal{C}^{op}$, so $\mathcal{C}$ must be equivalent to $\mathcal{C}^{op}$. – Dávid Tóth Apr 20 '15 at 17:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.