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# [section3] The opposite category; contravariant functors

edited February 2020

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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edited February 2020

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:

$T(f \triangleright g) = T(g) \triangleright T(f)$

Comment Source: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)\$ 
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edited February 2020

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 $$T$$.

Comment Source: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\$$. 
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edited August 2020

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 taking place in an opposite category.

Comment Source: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 taking place in an opposite category.