Let's try to pin down the idea of the opposite category, which like "mirror image" of a catgory, obtained by systematically reversing "left" and "right" in the construction of a category. Where do left and right appear?

Consider a morphism \\(f: A \rightarrow B\\). We may view this as a relationship between \\(A\\) and \\(B\\), which object \\(A\\) on the left and object \\(B\\) on the right. In a mirror image category, we could have a twin morphism \\(f'\\), with the labelling reversed, so that the left object of \\(f'\\) is \\(B\\) and the right object is \\(A\\).

The standard names for 'left' and 'right' are 'domain' and 'codomain'.

The opposite of a category \\(X\\) has the same objects as \\(X\\), and its system of morphisms consists of all the twins \\(f'\\) for morphisms \\(f\\) in \\(X\\).

The reversal is stated by the following equations:

\\[dom(f') = cod(f)\\]

\\[cod(f') = dom(f)\\]

Now suppose that \\(f: A \rightarrow B\\), \\(g: B \rightarrow C\\) in \\(X\\). This is a composable pair, and we may designate the composition by \\(f \triangleright g\\).

To distinguish composition in the opposite category \\(X'\\), let's use a different symbol for the composition operator.

For morphisms \\(u, v\\) in \\(X'\\), we'll write \\(u \triangleleft v\\) for their composition in \\(X'\\).

Having established notation, let's turn our attention back to our composable pair \\(f \triangleright g\\) in \\(X\\).

The question naturally arises: how can we form a composable pair from \\(f'\\) and \\(g'\\) in the opposite category \\(X'\\)?

First let's try \\(f' \triangleleft g'\\). For that to work, as always, in any category, we would require that \\(cod(f') = dom(g')\\).

But \\(cod(f') = dom(f)\\), which is not equal to \\(dom(g') = cod(f)\\). Bzzt.

The only other choice for the composable pair is \\(g' \triangleleft f'\\), and this works well. Indeed, we have that \\(cod(g') = dom(f')\\), which follows from that \\(cod(f) = dom(g)\\).

In a word:

\\[g' \triangleleft f' = (f \triangleright g)'\\]

These three equations define the transformation of a category \\(X\\) into its opposite category \\(X'\\).

Consider a morphism \\(f: A \rightarrow B\\). We may view this as a relationship between \\(A\\) and \\(B\\), which object \\(A\\) on the left and object \\(B\\) on the right. In a mirror image category, we could have a twin morphism \\(f'\\), with the labelling reversed, so that the left object of \\(f'\\) is \\(B\\) and the right object is \\(A\\).

The standard names for 'left' and 'right' are 'domain' and 'codomain'.

The opposite of a category \\(X\\) has the same objects as \\(X\\), and its system of morphisms consists of all the twins \\(f'\\) for morphisms \\(f\\) in \\(X\\).

The reversal is stated by the following equations:

\\[dom(f') = cod(f)\\]

\\[cod(f') = dom(f)\\]

Now suppose that \\(f: A \rightarrow B\\), \\(g: B \rightarrow C\\) in \\(X\\). This is a composable pair, and we may designate the composition by \\(f \triangleright g\\).

To distinguish composition in the opposite category \\(X'\\), let's use a different symbol for the composition operator.

For morphisms \\(u, v\\) in \\(X'\\), we'll write \\(u \triangleleft v\\) for their composition in \\(X'\\).

Having established notation, let's turn our attention back to our composable pair \\(f \triangleright g\\) in \\(X\\).

The question naturally arises: how can we form a composable pair from \\(f'\\) and \\(g'\\) in the opposite category \\(X'\\)?

First let's try \\(f' \triangleleft g'\\). For that to work, as always, in any category, we would require that \\(cod(f') = dom(g')\\).

But \\(cod(f') = dom(f)\\), which is not equal to \\(dom(g') = cod(f)\\). Bzzt.

The only other choice for the composable pair is \\(g' \triangleleft f'\\), and this works well. Indeed, we have that \\(cod(g') = dom(f')\\), which follows from that \\(cod(f) = dom(g)\\).

In a word:

\\[g' \triangleleft f' = (f \triangleright g)'\\]

These three equations define the transformation of a category \\(X\\) into its opposite category \\(X'\\).