# Is $C^{\infty}$-related an equivalence relation?

I am working on the "A comprehensive introduction to differential geometry" by Michael Spivak. I have a question about an exercise in chap 2, on differential structures.

Let $M$ be a manifold, and $U$ and $V$ be open subsets of $M$. The functions $x:U\rightarrow x(U)\subset\mathbb{R}^n$ and $y:V\rightarrow y(V)\subset\mathbb{R}^n$ are co-ordinate mappings on $M$. Spivak then says two such maps are $C^{\infty}$-related if:

$y\circ x^{-1}:x(U\cap V)\rightarrow y(U\cap V)$

$x\circ y^{-1}:y(U\cap V)\rightarrow x(U\cap V)$

are both $C^{\infty}$

An exercise asks to show that this is not an equivalence relation. I can't see what is wrong with my reasoning, which seems to show that it is an equivalence relation. Here is my proof:

To check it is an equivalence relation we need to prove the 3 properties: Reflexivity, Symmetry, Transitivity. The relation is: $x\sim y$ if $x$ and $y$ are $C^{\infty}$-related

Reflexivity: $x\circ x^{-1} = Id$ which is a $C^{\infty}$ map.

Symmetry: If $x\sim y$ then $y\sim x$ since the $C^{\infty}$-related condition is symmetric.

Transitivity: If $x\sim y$ and $y\sim z$ then $x\circ y^{-1}$ and $y\circ z^{-1}$ are both $C^{\infty}$, so $(x\circ y^{-1})\circ (y\circ z^{-1})=x\circ z^{-1}$ is $C^{\infty}$. Similarly, $y\circ x^{-1}$ and $z\circ y^{-1}$ are both $C^{\infty}$, so $(z\circ y^{-1})\circ (y\circ x^{-1})=z\circ x^{-1}$ is $C^{\infty}$. So the transitive condition also holds. (am I wrong in assuming that the composition of $C^{\infty}$ maps is a $C^{\infty}$ map)

Thanks, Seb

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The domain of $(x\circ y^{-1})\circ (y\circ z^{-1})$ is different from the domain of $x\circ z^{-1}$. – azarel Feb 18 '12 at 6:51

The problem is in the transitivity portion. The point is that $W$ can intersect $U$ in places where $V$ does not intersect $U$. On $W\cap V\cap U$, your argument is valid, but it's not valid in the other case.

Here's a specific example. Define $U = (0,3)$, $V = (2,5)$ and $W = (-1,1)\cup(4,6)$, all subsets of $\mathbb{R}$. Use the identity charts on $U$ and $V$. Finally, on $W$, define a chart map $z:W\rightarrow \mathbb{R}$ by $z(x) = \sqrt[3]{x-\frac{1}{2}}$.

Then the composition of identity maps is smooth so $Id_U$ and $Id_V$ are $C^\infty$ related. Also, $Id_V$ and $z$ are also $C^\infty$ related since $z$ is smooth away from $\frac{1}{2}$.

However, $Id_U$ and $z$ are not $C^\infty$ related since $z$ is not smooth at $\frac{1}{2}$.

This example can be promoted to one where $W$ is connected by moving to $\mathbb{R}^2$ and having $W$ "go around" $U$ and $V$.

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Cool, that makes sense, thanks. – Moss Feb 18 '12 at 7:09
Dear Jason: I suppose you wanted to write $W$ for the domain of $z$? If that is the case, it seems to me that the charts $Id_U$ and $z$ are perfectly compatible since the corresponding change of coordinates is $(0,1)\to (0,1):u\mapsto 1-u$ – Georges Elencwajg Feb 18 '12 at 13:03
@Georges: You're right, of course. That's what I get for typing up an answer when I'm sleepy. I've corrected it by changing $z$ to $|x-\frac{1}{2}|$, which makes the numbers slightly less pretty, but they were ugly to begin with ;-). – Jason DeVito Feb 18 '12 at 13:37
Dear Jason, it seems that your new chart $z$ is not injective since $z(1/4)=z(3/4)=1/4$ . – Georges Elencwajg Feb 18 '12 at 17:45
@Georges: Of course, you're right again. It's fixed. And thank you for your attention to detail! – Jason DeVito Feb 18 '12 at 18:04