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I have been given two definitions of what an opposite category is. The first is one which is in my textbook and is follows:

If $\mathcal{C}$ is a catagory with objects $ob(\mathcal{C})$ and morphisms $C(U,V)$ for $U,V\in ob(\mathcal{C})$ we define the opposite catagory $\mathcal{C}^{op}$ with $ob(\mathcal{C})= ob(\mathcal{C})^{op}$ and we define the morphisms by simply reversing the arrows of $C(U,V)$.

The second definition I have is that $\mathcal{C}^{op}$ is defined as having the same objects as $\mathcal{C}$ and morphisms $C(V,U)$

To me these do not seem the same. If I take the catagory of sets as an example would the morphisms not be different as in the first definition they need not be functions but in the second they are?

Thanks for any help.

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What do you mean by counterexample? It is the same objects and morphisms $Hom(V,U)$. – Secret Math Oct 15 '13 at 13:48
That first paragraph is a mess. Later, you have to define $C^{op}(U,V)$. What is it? Your question is very vague - it's not clear if you are trying to define an alternate category, or saying something completely different. – Thomas Andrews Oct 15 '13 at 13:51
As in the opposite category of sets is different here? Or am misunderstanding? But with the second definition the morphisms of the opposite category have to be functions and in the first they do not? – hmmmm Oct 15 '13 at 13:51
up vote 2 down vote accepted


What you want is probably : $$C^{op}(U,V)= C(V,U). $$

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And if that's not enough, in $\mathit{Set}^{\mathit{op}}$, morphisms $U \to V$ are indeed functions, but from $V$ to $U$. So, seen as relations $U \to V$, they are generally not functional. – Tom Hirschowitz Oct 16 '13 at 7:31

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