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Let's try to pin down the idea of the opposite category, which is a kind "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 mirror-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 reversal is stated as follows:

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

The opposite \(X'\) of a category \(X\) has the same objects as \(X\), and its system of morphisms consists of all the twins of the morphisms in \(X\). (The standard nomenclature is \(X^{op}\)).

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

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 this notation, now let's turn our attention back to the 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 in any category, we would need that \(cod(f') = dom(g')\).

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

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

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## Comments

A related idea is that of a contravariant functor, which is defined like a regular functor, except that it maps every arrow to an "arrow going the other way," and it reverses the order of composition.

Here is an example: the powerset functor \(T\), which maps every set \(S\) to its power set \(2^S\) -- all subsets of \(S\) -- and every function \(f: A \rightarrow B\) to the function \(T(f): 2^B \rightarrow 2^A\) which maps a subset of \(B\) to its preimage under \(f\).

So we have \(T(S) = 2^S\), and \(T(f: A \rightarrow B): T(B) \rightarrow T(A)\). There it is, the image of \(f\) goes in the direction opposite to \(f\).

This in turn implies that \(T\) must "reverse the direction of composition." Suppose \(f: A \rightarrow B\), and \(g: B \rightarrow C\). Then the only way composition can work is this:

\[T(f \triangleright g) = T(g) \triangleright T(f)\]

`A related idea is that of a contravariant functor, which is defined like a regular functor, except that it maps every arrow to an "arrow going the other way," and it reverses the order of composition. Here is an example: the powerset functor \\(T\\), which maps every set \\(S\\) to its power set \\(2^S\\) -- all subsets of \\(S\\) -- and every function \\(f: A \rightarrow B\\) to the function \\(T(f): 2^B \rightarrow 2^A\\) which maps a subset of \\(B\\) to its preimage under \\(f\\). So we have \\(T(S) = 2^S\\), and \\(T(f: A \rightarrow B): T(B) \rightarrow T(A)\\). There it is, the image of \\(f\\) goes in the direction opposite to \\(f\\). This in turn implies that \\(T\\) must "reverse the direction of composition." Suppose \\(f: A \rightarrow B\\), and \\(g: B \rightarrow C\\). Then the only way composition can work is this: \\[T(f \triangleright g) = T(g) \triangleright T(f)\\]`

Here is the connection between contravariant functors and the opposite category.

Suppose \(X\) and \(Y\) are categories, and \(X'\) is the opposite of category \(X\). Suppose \(T': X' \rightarrow Y\) is an ordinary functor from \(X'\) to \(Y\).

Then from \(T'\) we can derive a

contravariantfunctor \(T: X \rightarrow Y\).Suppose \(f: A \rightarrow B\) in category \(X\). Now consider the twin morphism \(f'\) in \(X'\), with \(f': B \rightarrow A\).

Define \(T(f) = T'(f')\).

Since \(T'\) is a ordinary functor, \(T'(f'): B \rightarrow A\).

So \(T(f): B \rightarrow A\).

There we see the arrow-reversal, which is required for contravariant \(T\).

`Here is the connection between contravariant functors and the opposite category. Suppose \\(X\\) and \\(Y\\) are categories, and \\(X'\\) is the opposite of category \\(X\\). Suppose \\(T': X' \rightarrow Y\\) is an ordinary functor from \\(X'\\) to \\(Y\\). Then from \\(T'\\) we can derive a _contravariant_ functor \\(T: X \rightarrow Y\\). Suppose \\(f: A \rightarrow B\\) in category \\(X\\). Now consider the twin morphism \\(f'\\) in \\(X'\\), with \\(f': B \rightarrow A\\). Define \\(T(f) = T'(f')\\). Since \\(T'\\) is a ordinary functor, \\(T'(f'): B \rightarrow A\\). So \\(T(f): B \rightarrow A\\). There we see the arrow-reversal, which is required for contravariant \\(T\\).`

To complete the demonstration that \(T\) is a contravariant functor, it only remains to show that it reverses the order of composition:

\[T(f \triangleright g) = T'((f \triangleright g)') = T'(g' \triangleleft f') = T'(g') \triangleright T'(f') = T(g) \triangleright T(f)\]

Note: we only used \(\triangleleft\) in one place here, only as a stylistic reminder that composition is talking place in an opposite category.

`To complete the demonstration that \\(T\\) is a contravariant functor, it only remains to show that it reverses the order of composition: \\[T(f \triangleright g) = T'((f \triangleright g)') = T'(g' \triangleleft f') = T'(g') \triangleright T'(f') = T(g) \triangleright T(f)\\] Note: we only used \\(\triangleleft\\) in one place here, only as a stylistic reminder that composition is talking place in an opposite category.`

Fin

`Fin`