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In the definition of the opposite category $\mathcal C^{op}$ of a category $\mathcal C$, it is said that : objects remain the same and arrows' directions are changed (that is, and arrow $f:A \to B$ in $\mathcal C$ is and arrow $f:B \to A$ in $\mathcal C^{op}$, and vice versa. Now I think this can not be a safe definition in general. Here goes the story:

Consider the definition of a product and coproduct of two objects $A$ and $B$ in a category $\mathcal C$. If we convert the diagram of a product in its "dual form" in the opposite category $\mathcal C^{op}$ (that is, by definition of $\mathcal C^{op}$, letting the objects remain the same and convert the arrows' directions) we gain a diagram similar to the diagram of a coproduct in $\mathcal C$ (defined directly in $C$). I bold "similar" since they can not be equal! In fact, if they were equal, we would have $A+B=A\times B$ and we know by certain that they are not. (E.g., consider a poset $P$ and we know that $x \times y=g.l.b(x,y)$ and $x + y=l.u.b(x,y)$ and those are not the same).

In short, one can not gain a coproduct in $\mathcal C$ simply by re-writing a product in $C$ in its dual form in $\mathcal C^{op}$. Preservance of objects makes troubles.

Now given this, don't you think we must have a more comprehensive definition of $\mathcal C^{op}$, for example one that converts the objects to their duals (defined directly and independently from the traditional $\mathcal C^{op}$ , not themselves (which will work for single objects too: their dual is the same as the original)?

$**$ You can see the attached image from Awodey's text, in order to see the products/coproducts and their related diagrams: Awodey - p56

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    $\begingroup$ There may be a confusion in your terminology. What you're getting at is that a product of $\mathcal{C}$ is a coproduct in $\mathcal{C}^\text{op}$. You made the incorrect leap that this object must then be a coproduct in $\mathcal{C}$. $\endgroup$
    – Andrew
    Feb 4 '16 at 8:06
  • $\begingroup$ Sure you are right! A coproduct in $\mathcal C$ is "a" product in $\mathcal C^{op}$. But Awodey states (p 56) that "A coproduct of two objects is therefore "exactly their" product" in the opposite category.". Then he is wrong at this stage? right? $\endgroup$
    – Ak9
    Feb 4 '16 at 8:37
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    $\begingroup$ No, he is not wrong. The statement "a coproduct of two objects is therefore exactly their product in the opposite category" is correct. It is also exactly what you claimed is correct in the sentence before. $\endgroup$
    – Andrew
    Feb 4 '16 at 17:05
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Nobody claims that $A \times B = A + B$ holds in $\mathcal{C}$. But it is true that $A \times^{\mathcal{C}} B = A +^{\mathcal{C}^{op}} B$, where the supscript denotes the category in which we take the (co)product. More generally, $\lim^{\mathcal{C}}=\mathrm{colim}^{\mathcal{C}^{op}}$. If you want to rename your objects in the dual category - that's ok. In this case, we have $(A \times^{\mathcal{C}} B)^{op} = A^{op} +^{\mathcal{C}^{op}} B^{op}$, which may be even abbreviated to $(A \times B)^{op} = A^{op} + B^{op}$.

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  • $\begingroup$ Well, I got where the mistake is in my first post: I was thinking that $A \times B = A + B$ must hold. But I'm not still sure why $A \times^{\mathcal{C}} B = A +^{\mathcal{C}^{op}} B$ and not just $ A \times^{\mathcal{C}} B \cong A +^{\mathcal{C}^{op}} B$? I mean both the objects $A \times^{\mathcal{C}} B$ and $ A +^{\mathcal{C}^{op}} B$ certainly satisfy the definition of a product (and hence are isomorphic), but I don't get it why $=$ holds between them! $\endgroup$
    – Ak9
    Feb 5 '16 at 4:56
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    $\begingroup$ $A \times B$ is only defined up to isomorphism anyway. So for such objects $X,Y$ we actually mean isomorphism classes, and $X=Y$ is an equality of isomorphism classes. $\endgroup$ Feb 5 '16 at 8:33
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There's no problem. The exact definition of $\mathcal{C}^{op}$ is as a quintuple $(\text{ob}\mathcal{C},\text{mor}(\mathcal{C}),t,s,\circ^{op})$ where $\circ^{op}(f,g)=\circ(g,f)$. That is, $\mathcal{C}^{op}$ has the same objects and morphisms, the source and target functions of $\mathcal{C}$ serve as the target and source functions, respectively, of $\mathcal{C}^{op}$, and the composition function is replaced with its reverse. This definition satisfies associativity and the existence of units, and thus gives another category structure on the objects and morphisms of $\mathcal{C}$. Then a product diagram in $\mathcal{C}$ is exactly the same thing as a coproduct diagram in $\mathcal{C}^{op}$.

If it makes you more comfortable, it's certainly possible to introduce symbols $X^{op},f^{op}$ for each object $X$ and morphism $f$ of $\mathcal{C}$, and put a category structure on these to define $\mathcal{C}^{op}$. Such a structure will be isomorphic.

But there's no need to do this, and Awodey's definition is certainly comprehensive.

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  • $\begingroup$ $\circ^{op}(f,g)=g\circ f$ is a bit weird. I would write $\circ^{op}(f,g)=\circ(g,f)$ or $f \circ^{op} g = g \circ f$. $\endgroup$ Feb 4 '16 at 18:19
  • $\begingroup$ Hah, good point, don't know why I did that. $\endgroup$ Feb 4 '16 at 18:40

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