From Wikipedia https://en.wikipedia.org/wiki/Diagram_(category_theory), it says that "diagram is the categorical analogue of an indexed family in set theory." I'm wondering how it works.

Suppose we have a diagram of type $\mathcal{J}$ in a category $\mathcal{C}$. So this means that $\mathcal{J}$ encodes the "shape" we are interested in $\mathcal{C}$. This is analogous to an indexed family in set theory in the following sense: the functors $D_i : \mathcal{J} \rightarrow \mathcal{C}$ index the shape in $\mathcal{C}$. Is this correct?

As an simple example, we just consider the trivial functors $D_X : \mathcal{J} \rightarrow \mathcal{C}$ where all the objects in $\mathcal{J}$ are mapped to the object $X$ in $\mathcal{C}$, and all arrows in $\mathcal{J}$ are mapped to the identity arrow $\text{id}_X$ in $\mathcal{C}$. Then we can have an "indexing set" $\{D_X\}_{X\in Obj(\mathcal{C})}$.


The idea to keep in mind is that an indexed family of sets can be described as a functor $F:\mathcal{I}\to\mathbf{Set}$ where $\mathcal{I}$ is a discrete category (one where the only morphisms are identities). Since this case is really simple, it's enough to write $\{F_I\}_{I\in Obj(\mathcal{I})}$, or even ignore the category structure on $\mathcal{I}$ and just think of it as a set. (Here I've written "$F_I$" for $F(I)$ to make the notation familiar.)

The way that a diagram generalizes this is in allowing $\mathcal{I}$ to be a category that is not discrete, and the codomain of $F$ to be an arbitrary category $\mathcal{C}$. Then part of this diagram still the family $\{F_I\}_{I\in Obj(\mathcal{I})}$, but now there's an additional family $\{F_f:F_I\to F_J\}_{f:I\to J\in Arr(\mathcal{I})}$ that obeys the additional conditions induced by the structure of $\mathcal{I}$ and the functoriality of $F$. That's what it means to say that a diagram picks out objects and morphisms in the "shape" of $\mathcal{I}$: the diagram gives you an indexed structure of objects and morphisms in $\mathcal{C}$ which, to phrase things loosely, must obey all the identities that hold in $\mathcal{I}$.

TL;DR - A diagram is just a functor, and the "shape" of that diagram is just its domain.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.