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I'm new to category theory, and I'm trying to understand diagrams.

What's the connection between the pencil-and-paper diagrams that I draw in my workbook, and the technical definition that a diagram is a functor? Does every pencil-and-paper diagram I can draw give rise to a unique diagram in the technical sense, in a unique way? What is this unique way?

Furthermore, what's the intuition behind the idea that if the source of a diagram is a poset, then the diagram is commutative? And is a poset essentially a "commutative category," in the sense that given any two objects, every two paths between them give rise to the same morphism?

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  • $\begingroup$ I don't know much about category theory, but a commutative diagram is just a way of saying that some compositions of functions are equal. For example, you can write $f \circ g = h \circ k$ in a commutative diagram that looks like a square. $\endgroup$ – Paul Gustafson Apr 10 '13 at 1:58
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    $\begingroup$ Although it's clear from Martin Brandenburg's answer (especially his example), it seems worth emphasizing that the diagrams you draw with pencil and paper usually do not contain all the arrows of the diagram considered as a functor. The latter are given by the paths that can be formed from the arrows you draw. In other words, you usually just draw enough arrows to generate all the others by composition. $\endgroup$ – Andreas Blass Apr 10 '13 at 2:36
  • $\begingroup$ @AndreasBlass, what is the precise sense in which theyre generated? Is it just a transitive reflexive closure? $\endgroup$ – goblin Apr 10 '13 at 10:29
  • $\begingroup$ actually I would slightly generalize and say that a commutative diagram is a functor on a preorder. $\endgroup$ – magma Apr 11 '13 at 17:13
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1) The connection should be clear after reading the Wikipedia article carefully, especially the examples. The index category encodes the shape of the diagram, whereas the functor itsself says which morphisms are attached to which edges of the shape. And yes, every shape is realized by some category. In fact, more generally, every graph can be seen as a category, where the vertices become the objects and the paths become the morphisms.

For example, the graph

$$\begin{array}{ccc} a & \rightarrow & b \\ \downarrow & & \downarrow & \\ c & \rightarrow & d \end{array}$$

gives a index or shape category with four objects $a,b,c,d$ and six non-identity morphisms $a \to b, b \to d, a \to c, c \to d, a \to b \to d$ and $a \to c \to d$. This also describes the diagrams on this shape category. It is commutative iff $a \to b \to d$ and $a \to c \to d$ become evaluated to the same morphism.

2) The assertion "a diagram is commutative iff the source is a poset?" is clearly wrong. But "if" is correct, for trivial reasons. In a poset, all parallel morphisms are equal, so the commutativity condition is quite empty.

3) Yes. A category is a poset iff every diagram in that category commutes. When one starts to learn the basic notions of category theory, it is often helpful to look at posets first, because then many things simplify, but the main ideas remain. For example, colimits become suprema, limits become infima, functors become monotonic maps, adjunctions become Galois connections, etc.

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  • $\begingroup$ Wikipedia says that a diagram is categorical analogue of an indexed family in set theory. How is this so? Is it that we have a bunch of diagrams $D_1,D_2, ... : \mathcal{J} \rightarrow \mathcal{C}$ and $D_1, D_2,...$ are the "labels" for the shape $\mathcal{J}$ in $\mathcal{C}$? $\endgroup$ – Jerry Nov 13 '16 at 3:56
  • $\begingroup$ A family of objects is just a diagram of discrete shape. Discrete categories are just sets. In category theory we have non-discrete categories. I don't think that Wikipedia wanted to allude to families of diagrams; just one. $\endgroup$ – Martin Brandenburg Apr 12 '17 at 12:04

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