What is the meaning of $C^{op}$ if $C$ is a category without inverses?

Question

I have read the definition of the opposite category in the abstract, but I don't understand it's interpretation, and the internet didn't help me.

If $C$ is a category where very morphism has an inverse (a groupoid), then I can imagine an interpretation of the opposite category $C^{op}$: We simply swap each function with its inverse.

But what if we have say, a category $C$ consisting of two objects $0$ and $1$, namely $0$ is the empty set, and $1$ is a set containing one element. Then there is one function from $0$ to $1$, and no functions from $1$ to $0$. Apart from the identity functions there are no other morphisms in this category.

What is the opposite category $C^{op}$ here? It is a category with a morphism from $1$ to $0$, but this morphism cannot be a set-theoretic function, so what does this morphism represent?

• How do we interpret this opposite category?

• Is there necessarily a fixed interpretation?

Answer 1

so what does this morphism represent?

This morphism (like all morphisms) represents nothing. A category is just a collection of objects, a collection of morphisms, and a composition operation, which together satisfy some rules. The first examples you learn are often of set-like objects (including spaces and algebraic objects) together with appropriate functions, but there is no reason for a category to work this way.

Example: A poset $(X, \leq)$ becomes a category if we take $X$ to be the collection of objects, and we let there be a unique morphism $x \to x'$ if $x \leq x'$, and no morphism $x \to x'$ if $x \not \leq x'$. What do the morphisms "represent" here? What happens if you take the opposite of this category? (In fact, your category could be seen as an example of this sort of category.)

How do we interpret this opposite category?

It's more or less the same category as you started with: a category with two objects, and only one non-identity morphism which goes from the one object to the other. You could have described this category (as I just did) without any reference to sets and functions at all.

Is there necessarily a fixed interpretation?

Nope, no more than the group $C_2$ has a fixed interpretation: is it $\{0, 1\}$ under addition modulo 2? Is it $\{\pm 1\}$ under multiplication? Is it $O(1)$, the orthogonal operations on the line under composition?

Answer 2

Yes, in your example $C^{op}$ is the category of two objects with a morphism from $1$ to $0$. It doesn't necessarily represent something other than itself - it is just the result of an operation that you can perform on a category to generate another category.

Although morphisms in simple examples of categories often represent functions, this is not always the case. As you go deeper into category theory, many categories have morphisms that are unrelated (or, at best, distantly related) to functions.

Answer 3

In the situation described in your question $\mathcal C^{\mathsf{op}}$ is again a category with objects $0$ and $1$.

Next to the identities on these objects there is a unique arrow $f:1\to0$.

(So actually in this situation $\mathcal C$ and $\mathcal C^{\mathsf{op}}$ are isomorphic categories.)

You might wonder: what exactly is $f$ here?

Well, it can be described still as a function from $0$ to $1$ or equivalently as an arrow in $\mathcal C$ from $0$ to $1$ but secondly as an arrow in $\mathcal C^{\mathsf{op}}$ from $1$ to $0$.

Domain and codomain are interchanged.

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In general $\mathcal C^{\mathsf{op}}$ can be looked at as the category having exactly the same objects and arrows as $\mathcal C$.

The only essential difference is that there is another composition, and this fact causes the interchange of domain and codomain.

If $\circ$ denotes the composition of $\mathcal C$ then there is a composition $\circ^{\mathsf{op}}$ on $\mathcal C^{\mathsf{op}}$ defined by:$$f\circ^{\mathsf{op}}g:=g\circ f$$

$h\in\mathsf{Hom}_{\mathcal C}(a,b)$ iff and only if the compositions $h\circ\mathsf{id}_a$ and $\mathsf{id}_b\circ h$ are well-defined with: $$h\circ\mathsf{id}_a=h=\mathsf{id}_b\circ h\tag1$$

Now note that $(1)$ is equivalent with: $$\mathsf{id}_a\circ^{\mathsf{op}}h=h=h\circ^{\mathsf{op}}\mathsf{id}_b$$which on its turn is equivalent with $h\in\mathsf{Hom}_{\mathcal C^{\mathsf{op}}}(b,a)$.

So we have: $$\mathsf{Hom}_{\mathcal C}(a,b)=\mathsf{Hom}_{\mathcal C^{\mathsf{op}}}(b,a)$$