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S. Lang gives a statement on page x of his 'Algebra':

Most of our diagrams are composed of triangles and squares as above, and to verify that a diagram consisting of triangles and squares is commutative it suffices to verify that each triangle and square in it is commutative.

If we want to prove this statement the problem arises of defining precisely what 'consisting of squares and triangles' means.

The most obvious definitions of such diagrams (every vertex (arrow) belongs to a triangle\square) turn out to be unsatisfactory (the Lang's statement is wrong then, think of a pentagonal diagram with a commutative triangle on each side (all suitably oriented)).

How could the intuitive notion of a 'diagram consisting of squares and triangles' be strictly formulated such that the Lang's statement is always true?

edited Mar 22 2011 at 22:40

asked Mar 22 2011 at 22:35

Dmitry K
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I'm confused; how does the pentagon fail? – Daniel Litt Mar 22 2011 at 23:02

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The homological algebra chapter of Lang's book contains a lot of diagrams that are composed of squares and triangles. It may be helpful to go through these diagrams, assuming squares and triangles are commutative. Then it's obvious that the outer diagrams also commute. And that's exactly what Lang's statement means. Having some practise with such diagrams you probably won't even feel a need to formalize those statement. – Ralph Mar 22 2011 at 23:25

@Daniel, the triangles are on the outside of the pentagon, checking their commutativity doesn't imply the commutativity over pentagon. – Gjergji Zaimi Mar 22 2011 at 23:45

Ah, I misunderstood where the triangles were. – Daniel Litt Mar 23 2011 at 6:17

Gjergji Zaimi · Accepted Answer · 2011-03-22 23:18:14Z UTC

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Think of a diagram as a planar directed graph with an equivalence relation on directed paths which respects concatenations of paths. A commutative diagram is one where this equivalence relation identifies all paths with common sources and sinks.

The remark above says that to check that a diagram is commutative it is enough to check that each face is a commutative diagram. This is a simple combinatorial fact that you can show by induction on the number of faces.

answered Mar 22 2011 at 23:18

Gjergji Zaimi
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Why must it be planar? – Jeff Strom Mar 22 2011 at 23:34

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It doesn't have to be, it just is for most examples. Without planarity a sentence like "made from squares and triangles" doesn't make much sense. – Gjergji Zaimi Mar 22 2011 at 23:37

@Gjergji: What about a tetrahedron? – Martin Brandenburg Mar 23 2011 at 16:46

A tetrahedron is planar as well. So it makes sense to check commutativity of the faces. Now it's true one can make similar statements for more complicated diagrams. For example to check commutativity of a diagram whose arrows form a $\mathbb Z^n$ lattice, it is enough to check commutativity of each unit hypercube, but it seems like that is beyond what's being asked here. – Gjergji Zaimi Mar 23 2011 at 17:33

I think this answer needs, at least, a more precise definition of "face". What happens if you take a noncommutative grid and stick a terminal object in the middle of each square? Depending on what you mean by "face", the faces might all be commutative triangles, but the diagram as a whole does not commute. – Charles Staats Mar 23 2011 at 23:10

Martin Brandenburg · Answer 2 · 2011-03-22 23:06:41Z UTC

Commutative diagrams visualize calculations of the form

$f_1 * ... * f_n = g_1 * ... * g_m$

of morphisms in the given category. A decomposition of the diagram corresponds to a chain of equations leading to the above equation. A triangle corresponds to an equation of the type $f_1 f_2 = g_1$ and a square to an equation of the type $f_1 f_2 = g_1 g_2$.

Diagrams consisting of triangles and squares

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