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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeOct 22nd 2010

In Basic Category Theory, it says

A category $C$ is given by a collection $C_0$ of objects and a collection $C_1$ of arrows which have the following structure.

• Each arrow has a domain and a codomain which are objects; one writes $f : X \to Y$ or $X\stackrel{f}{\to} Y$ if $X$ is the domain of the arrow $f$, and $Y$ its codomain. One also writes $X = dom(f)$ and $Y = cod(f)$;

• Given two arrows $f$ and $g$ such that $cod(f) = dom(g)$, the composition of $f$ and $g$, written $g f$, is defined and has domain $dom(f)$ and codomain $cod(g)$:

$(X\stackrel{f}{\to} Y\stackrel{g}{\to} Z) \mapsto (X\stackrel{g f}{\to} Z)$
• Composition is associative, that is: given $f : X \to Y$ , $g : Y \to Z$ and $h : Z \to W$, $h(g f) = (h g)f$;

• For every object X there is an identity arrow $id_X : X \to X$, satisfying $id_X g = g$ for every $g : Y \to X$ and $f id_X$ = f$for every$f : X \to Y\$.

The existence of composites (the second bullet point) means we have the following commutative triangles:

$\begin{matrix} \bullet &{}& \stackrel{f\circ g}{\leftarrow} &{}& \bullet \\ {} & \mathllap{f}{\nwarr} & {} & \mathrlap{\quad g}{\swarr} & {} \\ {} & {} & \bullet & {} & {} \end{matrix} \quad\quad\text{and}\quad\quad \begin{matrix} {} &{}& \bullet &{}& {} \\ {} & \mathllap{g}{\swarr} & {} & \mathrlap{\quad h}{\nwarr} & {} \\ \bullet & {} & \underset{g\circ h}{\leftarrow} & {} & \bullet \end{matrix}$

When you stick these two commutative triangles together, the resulting diagram should be commutative, right?

But the commutativity of the compound diagram implies associativity (the third bullet point).

Note: I like that (if its correct), i.e. commutativity (of diagrams) implies associativity (of morphisms)

Does that sound alright?

A situation I can think of that introduces non-associativity would involve commutators, e.g. for example, we can define a map $f_X:Obj(FinVect)\to Obj(FinVect)$ via

$f_X(A) = [X,A] = X A - A X.$

We can compose two such maps resulting in

$f_Y\circ f_X(A) = f_{[Y,X]}(A) + f_X\circ f_Y(A)$

but generally there is no $Z$ such that

$f_Z = f_Y\circ f_X.$

Is one thing we can take away from this is that these maps are not morphisms in a category?

PS: The following equation is cute so I couldn’t resist including it

$[f_Y,f_X](A) = f_{[Y,X]}(A)$
• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeOct 22nd 2010

Note: I like that (if its correct), i.e. commutativity (of diagrams) implies associativity (of morphisms)

Many people have said this before, but also many people have denounced it.

First of all, a little correction: there is no such principle as commutativity of diagrams. Only some diagrams commute. On the other hand, there is a general principle of associativity of composition of morphisms. So your insight is that the general law of associativity in a category can be given by a specific commutative diagram

$\begin{matrix} \bullet &{}& \stackrel{f\circ g}{\leftarrow} &{}& \bullet \\ {} & \mathllap{f}{\nwarr} & {} & \mathllap{g}{\swarr} & {} & \mathrlap{\quad h}{\nwarr} & {} \\ {} & {} & \bullet & {} & \underset{g\circ h}{\leftarrow} & {} & \bullet \end{matrix}$

However, here is why people denounce this idea as nonsense:

What the heck does a diagram even mean if there is no associativity ???

If in a diagram I write

$\begin{matrix} \bullet & \stackrel{f}{\leftarrow} & \bullet & \stackrel{g}{\leftarrow} & \bullet & \stackrel{h}{\leftarrow} & \bullet \end{matrix}$

and we do not already understand the law of assocaitivity, then what does this mean? You could arbitrarily interpret it as either $(f \circ g) \circ h$ or $f \circ (g \circ h)$, at your choice, but this breaks down in a more complicated diagram in which further arrows come in or out of the objects in the middle.

On the other hand, people do draw diagrams in which associativity is only weak, and this has a meaning. However, you still have to stick in tacit associators in various places to give it a meaning, so you need to understand an appropriate version of associativity before you can draw complicated diagrams.

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeOct 22nd 2010

And regarding specific axioms, there’s no a priori reason (given only the first two bullet points) that sticking together two commutative diagrams (assuming one could give meaning to the notion) would result in another commutative diagram.

• CommentRowNumber4.
• CommentAuthorEric
• CommentTimeOct 23rd 2010

Thank you Toby and Mike.

For a second, imagine we can build up diagrams from commuting triangles like a puzzle. I can come up with six distinct classes of a 2-triangle diagrams:

Class 1

$\begin{matrix} \bullet &{}& \rightarrow &{}& \bullet \\ {} & \searr & {} & \swarr & {} & \nwarr & {} \\ {} & {} & \bullet & {} & \leftarrow & {} & \bullet \end{matrix}$

Class 2

$\begin{matrix} \bullet &{}& \leftarrow &{}& \bullet \\ {} & \nwarr & {} & \swarr & {} & \searr & {} \\ {} & {} & \bullet & {} & \rightarrow & {} & \bullet \end{matrix}$

Class 3

$\begin{matrix} \bullet &{}& \leftarrow &{}& \bullet \\ {} & \searr & {} & \swarr & {} & \searr & {} \\ {} & {} & \bullet & {} & \leftarrow & {} & \bullet \end{matrix}$

Class 4

$\begin{matrix} \bullet &{}& \leftarrow &{}& \bullet \\ {} & \nwarr & {} & \swarr & {} & \searr & {} \\ {} & {} & \bullet & {} & \leftarrow & {} & \bullet \end{matrix}$

Class 5

$\begin{matrix} \bullet &{}& \rightarrow &{}& \bullet \\ {} & \searr & {} & \swarr & {} & \searr & {} \\ {} & {} & \bullet & {} & \leftarrow & {} & \bullet \end{matrix}$

Class 6

$\begin{matrix} \bullet &{}& \leftarrow &{}& \bullet \\ {} & \nwarr & {} & \swarr & {} & \nwarr & {} \\ {} & {} & \bullet & {} & \leftarrow & {} & \bullet \end{matrix}$

If we partake in some pre-denounced nonsense and forget about associativity for a moment, then, unless I’m mistaken, for all but the last diagram (Class 6), if the outer two triangles commute then the entire diagram commutes. The Class 6 diagram is, of course, the diagram corresponding to associativity.

So if we take this together with Toby’s remarks about associativity and diagrams in general, then I’m tempted to say that the axioms of a category are what they are so that diagrams make sense. For if they were anything other what they are, we could not unambiguously draw diagrams. Does that make any sense?

Then, because we do have associativity, then any diagram built up from commuting triangles should commute. Neat.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeOct 25th 2010
• (edited Oct 25th 2010)

Toby wrote:

However, here is why people denounce this idea as nonsense:

What the heck does a diagram even mean if there is no associativity ???

It is true that the way Eric asked the question it did not make sense. But there is something that looks similar and does make sense: take the composition of edges to be free composition (in the free category on the directed graph of edges) and think of the triangles above as being filled with 2-isomorphisms that identify a formal concatenation composite with the actual composite. This appears in the free resolution of a 1-category by a strict 2-category. See here.

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeOct 26th 2010

I’m tempted to say that the axioms of a category are what they are so that diagrams make sense

Now this I like. It may even be historically accurate; people were just starting to draw commutative diagrams in algebraic topology when the algebraic topologists Eilenberg and Mac Lane wrote down the axioms of category theory.