# What does it mean to say a diagram commutes?

$\require{AMScd}$ In the context of smooth manifolds, the map $F:M\rightarrow N$ is smooth if $G$ on the below diagram is smooth.

$\begin{CD} M @> F > > N\\ @V \varphi V V @V V\psi V\\ \varphi(U) @> G >=\psi\circ F\circ\varphi^{-1} > \psi(V) \end{CD}$

Some notes I found say "if and only if this diagram commutes" however this diagram always commutes! if $F$ isn't smooth all it means is $G$ wont be smooth.

What does it mean to say a diagram commutes in this case?

• Sure, it commutes. Have you looked at other books on the subject? – Martin Brandenburg Apr 14 '15 at 14:10
• Of course it commutes, I say that. What says the arrows must be smooth, the context? This is the /definition/ of $F$ being smooth @MartinBrandenburg – Alec Teal Apr 14 '15 at 14:11
• @MartinBrandenburg by that I mean if $F$ isn't smooth, it'll still commute! Just $G$ wont be smooth – Alec Teal Apr 14 '15 at 14:13

"The diagram commutes" means exactly what it always means: that the map produced by following any path through the diagram is the same.

The notes you are reading made a mistake and forgot to require that $G$ be smooth.

• So "diagram commutes" means "under [whatever] constraints - the diagram makes sense" - my confusion was because nothing says an arrow has to be smooth (otherwise $F$ couldn't have one). When I use diagrams I must specify things like this. – Alec Teal Apr 14 '15 at 14:21
• @Alec: Yes, I understood that that was your confusion. Now I am confused as to whether you understand my answer. I am saying that "a diagram commutes" means exactly, and only that certain compositions of maps are equal - absolutely nothing about the properties of those maps. I am agreeing with you that yes, it ought to have been said that $G$ was smooth, because the diagram given always commutes. The notes provide an incorrect (and in this case, vacuous) definition. – Zev Chonoles Apr 14 '15 at 14:25
• The following diagram commuting $$\begin{CD} A @> w > > B\\ @V x V V @V Vy V\\ C @> >z > D \end{CD}$$ means precisely that $y\circ w=z\circ x$. – Zev Chonoles Apr 14 '15 at 14:29
• I see what you mean - I came very close to "forcing" that in the context of smooth manifolds, I defined arrows to be smooth and "commutative" to mean smooth too - which you've saved me from doing Before upvoting and ticking, maths.kisogo.com/… is that "how I ought to say it" (and last question while I've got you here, what do the curved and dashed arrows mean? - I've not found a book that defines this notion - you've given a superb and unambiguous definition of commutative) - Thanks – Alec Teal Apr 14 '15 at 14:32

I think part of the confusion is that the sentence below the definition - "In the above diagram, $G := \Psi \circ F \circ \phi^{-1}: \phi(U \cap F^{-1}(V)) \rightarrow \Psi(V)$" - is not meant to be part of the definition, rather it is a clarification. The definition given is that $F$ is smooth iff there exists $G$ (which, from its position in the diagram, is a morphism between open subsets of $\mathbb{R^s}$, that is, is a smooth map*) such that

$\begin{CD} M @> F > > N\\ @V \varphi V V @V V\psi V\\ \varphi(U) @> G > > \psi(V) \end{CD}$

commutes.

*- but I agree this should have been stated explicitly

Does that make things clearer at all?