Commuting diagrams

Question

Reading "P. Smith, Category Theory - A Gentle Introduction". There it is defined

A category diagram commutes if for any two directed paths along edges in the diagram from a node X to the node Y, the composite arrow along the first path is equal to the composite arrow along the second path.

Let's assume we have the following diagram:

A ---f---> B
\          |
 \         |
  k        g
   \       |
    \      |
    \/     \/
        C

Now, let's check if this diagram commutes. I will instantiate the above definition with the nodes A and C:

A category diagram commutes if for any two directed paths along edges in the diagram from the node A to the node C, the composite arrow along the first path is equal to the composite arrow along the second path.

We have two paths:

path_1 := A --f--> B --g--> C
path_2 := A --k--> C

And the corresponding composites:

composite along path_1 := g . f
composite along path_2 := k

So we have to show, that those two composites are equal, i.e., g . f = k.

I'd like to understand intuitively, why can I reason that those two composites are equal?

In the article it is stated, "the diagram commutes by the definition of composition".

OK, I know the fact, that due to cod f = dom g we know that there must exist the composite of f with g: g . f :: A -> C but how do I know that this composite is equal to k (which is the condition for the diagram to commute). In other words, why can't there be a "function" k from A to C not being equal to the "function" g .f?

---

Update as reply to Matthew Towers and Peter Smith (2018-11-25):

Matthew Towers and Peter Smith, please don't get me wrong here. I guess all this is very trivial to all of you. I just wanted to have a pragmatic (algorithmic) way of testing if a diagram commutes.

So, is the following correct:

If one wants to check if a diagram commutes, then for /all/ paths from a node X to a node Y the respective composites have to be checked for equality.

Right?

If yes, let me try to apply this to the first triangle of

the triangle made by the arrows $g \circ f\colon A \to C$, and $f\colon A\to B$ and $g\colon B \to C$, and then also the triangle made by the arrows $h \circ g\colon B \to D$, and $g\colon B \to C$ and $h\colon C \to G$.

Please let me also re-ref your (adapted) definition:

A category diagram commutes if for any two directed paths along edges in the diagram from a node X to a node Y, the composite arrow along the first path is equal to the composite arrow along the second path.

So, with $X = A$ and $Y = C$ we have the following paths:

p1 := A --f--> B --g--> C p2 := A --g∘f--> C

And the corresponding composites along those paths:

c1 along p1 := g∘f c2 along p2 := g∘f

(Matthew Towers and Peter Smith: Is c2, the correct composite along the second path (even the path consists of only one edge)?)

So, in order to check if the first triangle commutes, for /all/ paths from A to C we have to check the respective composites are equal. In this case, there are only two paths, so we have to check $c1 = c2$, which is true.

Answer 1

The set of arrows $\text{hom}(A,C)$ is a set and asking if two elements of this set are equal or not has sense. But of course a priori you don't know if two elements are equal or not, it depends on the particular case.

Answer 2

Indeed, there can be a $k:A\to C$ such that $k\ne g\circ f$, only that in such case the diagram is not called commutative.

Answer 3

One of the purposes of category theory is to abstract the data of "set and maps". Set functions associate, so category morphisms are defined to. And when there are two compatible morphisms you can compose them, just like set maps.

Saying $f = g$ as functions is the same as saying:

$$ \begin{matrix} x & \xrightarrow{f} & y \\ \mid{} & \ & \mid{} \\ x & \xrightarrow{g} & y \end{matrix} $$ where the vertical arrows are identity maps. If you drew just two objects, $x, y$ and two arrows between them, you can also say that $f = g$ if and only if that diagram commutes.

So whenever you have two compatible arrows: $a \xrightarrow{f} b \xrightarrow{g} c$ (think "set maps") you should always have a composition of them defined so you immediately get a commuting triangle.

These commutative diagrams seem very important because they occur on every page of advanced math books. They are a very visual way to do math and should be used as much as possible. Long compositions look unreadable on a line of text, but in CD form they look elegant.

Answer 4

You are simply misreading.

What is clearly said to "commute[s] by definition of composition" is NOT the diagram you give, but rather each of the two triangles in the next diagram in the notes -- the triangle made by the arrows $g \circ f\colon A \to C$, and $f\colon A \to B$ and $g\colon B \to C$, and then also the triangle made by the arrows $h \circ g\colon B \to D$, and $g\colon B \to C$ and $h\colon C \to G$.

As others have pointed out, the diagram you give may or may not commute, depending on what $k$ is. But you didn't really need others to tell you that, as five lines after the words you quote I wrote, just to block this sort of misunderstanding, "And note too that merely drawing a diagram with different routes from e.g. $A$ to $D$ in the relevant category doesn't always mean that we have a commutative diagram -- the identity of the composites along the paths in each case has to be argued for!"